快速上手pandas(下)

2021 年 8 月 3 日
笔记
Python

和上文一样，先导入后面会频繁使用到的模块：

In [1]:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

plt.rc('font', family='Arial Unicode MS')
plt.rc('axes', unicode_minus='False')

pd.__version__

Out[1]:

'1.1.3'

注意：我这里是Mac系统，用matplotlib画图时设置字体为Arial Unicode MS支持中文显示，如果是deepin系统可以设置字体为WenQuanYi Micro Hei，即：

In [2]:

# import numpy as np
# import pandas as pd
# import matplotlib.pyplot as plt

# plt.rc('font', family='WenQuanYi Micro Hei')
# plt.rc('axes', unicode_minus='False')

如果其他系统画图时中文乱码，可以用以下几行代码查看系统字体，然后自行寻找支持中文的字体：

In [3]:

# from matplotlib.font_manager import FontManager
# fonts = set([x.name for x in FontManager().ttflist])
# print(fonts)

话不多说，继续pandas的学习。

部分案例参考：
//www.gairuo.com/p/pandas-tutorial

部分数据集来自：
//archive.ics.uci.edu/ml/machine-learning-databases/

数据合并

在实际的业务处理中，往往需要将多个数据集、文档合并后再进行分析。

concat

Signature:
pd.concat(
    objs: Union[Iterable[~FrameOrSeries], Mapping[Union[Hashable, NoneType], ~FrameOrSeries]],
    axis=0,
    join='outer',
    ignore_index: bool = False,
    keys=None,
    levels=None,
    names=None,
    verify_integrity: bool = False,
    sort: bool = False,
    copy: bool = True,
) -> Union[ForwardRef('DataFrame'), ForwardRef('Series')]

Docstring:
Concatenate pandas objects along a particular axis with optional set logic
along the other axes.

Can also add a layer of hierarchical indexing on the concatenation axis,
which may be useful if the labels are the same (or overlapping) on
the passed axis number.

数据准备：

In [4]:

df1 = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'],
                    'B': ['B0', 'B1', 'B2', 'B3'],
                    'C': ['C0', 'C1', 'C2', 'C3'],
                    'D': ['D0', 'D1', 'D2', 'D3']},
                   index=[0, 1, 2, 3])
df1

Out[4]:

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3

In [5]:

df2 = pd.DataFrame({'A': ['A4', 'A5', 'A6', 'A7'],
                    'B': ['B4', 'B5', 'B6', 'B7'],
                    'C': ['C4', 'C5', 'C6', 'C7'],
                    'D': ['D4', 'D5', 'D6', 'D7']},
                   index=[4, 5, 6, 7])
df2

Out[5]:

	A	B	C	D
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7

In [6]:

df3 = pd.DataFrame({'A': ['A8', 'A9', 'A10', 'A11'],
                    'B': ['B8', 'B9', 'B10', 'B11'],
                    'C': ['C8', 'C9', 'C10', 'C11'],
                    'D': ['D8', 'D9', 'D10', 'D11']},
                   index=[8, 9, 10, 11])
df3

Out[6]:

	A	B	C	D
8	A8	B8	C8	D8
9	A9	B9	C9	D9
10	A10	B10	C10	D10
11	A11	B11	C11	D11

基本连接:

In [7]:

# 将三个有相同列的表合并到一起
pd.concat([df1, df2, df3])

Out[7]:

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7
8	A8	B8	C8	D8
9	A9	B9	C9	D9
10	A10	B10	C10	D10
11	A11	B11	C11	D11

可以再给每个表给一个一级索引，形成多层索引：

In [8]:

pd.concat([df1, df2, df3], keys=['x', 'y', 'z'])

Out[8]:

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

也等同于下面这种方式：

In [9]:

pd.concat({'x': df1, 'y': df2, 'z': df3})

Out[9]:

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

合并时不保留原索引，启用新的自然索引：

In [10]:

df4 = pd.DataFrame({'B': ['B2', 'B3', 'B6', 'B7'],
                    'D': ['D2', 'D3', 'D6', 'D7'],
                    'F': ['F2', 'F3', 'F6', 'F7']},
                   index=[2, 3, 6, 7])
df4

Out[10]:

	B	D	F
2	B2	D2	F2
3	B3	D3	F3
6	B6	D6	F6
7	B7	D7	F7

In [11]:

pd.concat([df1, df4], ignore_index=True, sort=False)

Out[11]:

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
4	NaN	B2	NaN	D2	F2
5	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

有没有类似于数据库中的outer join呢？

In [12]:

pd.concat([df1, df4], axis=1, sort=False)

Out[12]:

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3
6	NaN	NaN	NaN	NaN	B6	D6	F6
7	NaN	NaN	NaN	NaN	B7	D7	F7

很自然联想到inner join：

In [13]:

pd.concat([df1, df4], axis=1, join='inner')

Out[13]:

	A	B	C	D	B	D	F
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3

join : {'inner', 'outer'}, default 'outer'
    How to handle indexes on other axis (or axes).

这里并没有看到left join或者right join，那么如何达到left join的效果呢？

In [14]:

pd.concat([df1, df4.reindex(df1.index)], axis=1)

Out[14]:

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3

与序列合并：

In [15]:

s1 = pd.Series(['X0', 'X1', 'X2', 'X3'], name='X')
pd.concat([df1, s1], axis=1)

Out[15]:

	A	B	C	D	X
0	A0	B0	C0	D0	X0
1	A1	B1	C1	D1	X1
2	A2	B2	C2	D2	X2
3	A3	B3	C3	D3	X3

当然，也是可以使用df.assign()来定义一个新列。

如果序列没名称，会自动给自然索引名称，如下：

In [16]:

s2 = pd.Series(['_A', '_B', '_C', '_D'])
s3 = pd.Series(['_a', '_b', '_c', '_d'])
pd.concat([df1, s2, s3], axis=1)

Out[16]:

	A	B	C	D	0	1
0	A0	B0	C0	D0	_A	_a
1	A1	B1	C1	D1	_B	_b
2	A2	B2	C2	D2	_C	_c
3	A3	B3	C3	D3	_D	_d

ignore_index=True会取消原有列名：

In [17]:

pd.concat([df1, s1], axis=1, ignore_index=True)

Out[17]:

	0	1	2	3	4
0	A0	B0	C0	D0	X0
1	A1	B1	C1	D1	X1
2	A2	B2	C2	D2	X2
3	A3	B3	C3	D3	X3

同理，多个Series也可以合并：

In [18]:

s3 = pd.Series(['李寻欢', '令狐冲', '张无忌', '花无缺'])
s4 = pd.Series(['多情剑客无情剑', '笑傲江湖', '倚天屠龙记', '绝代双骄'])
s5 = pd.Series(['小李飞刀', '独孤九剑', '九阳神功', '移花接玉'])
pd.concat([s3, s4, s5], axis=1)

Out[18]:

	0	1	2
0	李寻欢	多情剑客无情剑	小李飞刀
1	令狐冲	笑傲江湖	独孤九剑
2	张无忌	倚天屠龙记	九阳神功
3	花无缺	绝代双骄	移花接玉

也可以指定keys使用新的列名：

In [19]:

pd.concat([s3, s4, s5], axis=1, keys=['name', 'book', 'skill'])

Out[19]:

	name	book	skill
0	李寻欢	多情剑客无情剑	小李飞刀
1	令狐冲	笑傲江湖	独孤九剑
2	张无忌	倚天屠龙记	九阳神功
3	花无缺	绝代双骄	移花接玉

merge

Signature:
pd.merge(
    left,
    right,
    how: str = 'inner',
    on=None,
    left_on=None,
    right_on=None,
    left_index: bool = False,
    right_index: bool = False,
    sort: bool = False,
    suffixes=('_x', '_y'),
    copy: bool = True,
    indicator: bool = False,
    validate=None,
) -> 'DataFrame'

Docstring:
Merge DataFrame or named Series objects with a database-style join.

The join is done on columns or indexes. If joining columns on
columns, the DataFrame indexes *will be ignored*. Otherwise if joining indexes
on indexes or indexes on a column or columns, the index will be passed on.

Parameters
----------
left : DataFrame
right : DataFrame or named Series
    Object to merge with.
how : {'left', 'right', 'outer', 'inner'}, default 'inner'
    Type of merge to be performed.

    * left: use only keys from left frame, similar to a SQL left outer join;
      preserve key order.
    * right: use only keys from right frame, similar to a SQL right outer join;
      preserve key order.
    * outer: use union of keys from both frames, similar to a SQL full outer
      join; sort keys lexicographically.
    * inner: use intersection of keys from both frames, similar to a SQL inner
      join; preserve the order of the left keys.
on : label or list
    Column or index level names to join on. These must be found in both
    DataFrames. If `on` is None and not merging on indexes then this defaults
    to the intersection of the columns in both DataFrames.
left_on : label or list, or array-like
    Column or index level names to join on in the left DataFrame. Can also
    be an array or list of arrays of the length of the left DataFrame.
    These arrays are treated as if they are columns.
right_on : label or list, or array-like
    Column or index level names to join on in the right DataFrame. Can also
    be an array or list of arrays of the length of the right DataFrame.
    These arrays are treated as if they are columns.
left_index : bool, default False
    Use the index from the left DataFrame as the join key(s). If it is a
    MultiIndex, the number of keys in the other DataFrame (either the index
    or a number of columns) must match the number of levels.
right_index : bool, default False
    Use the index from the right DataFrame as the join key. Same caveats as
    left_index.
sort : bool, default False
    Sort the join keys lexicographically in the result DataFrame. If False,
    the order of the join keys depends on the join type (how keyword).
suffixes : list-like, default is ("_x", "_y")
    A length-2 sequence where each element is optionally a string
    indicating the suffix to add to overlapping column names in
    `left` and `right` respectively. Pass a value of `None` instead
    of a string to indicate that the column name from `left` or
    `right` should be left as-is, with no suffix. At least one of the
    values must not be None.
copy : bool, default True
    If False, avoid copy if possible.
indicator : bool or str, default False
    If True, adds a column to the output DataFrame called "_merge" with
    information on the source of each row. The column can be given a different
    name by providing a string argument. The column will have a Categorical
    type with the value of "left_only" for observations whose merge key only
    appears in the left DataFrame, "right_only" for observations
    whose merge key only appears in the right DataFrame, and "both"
    if the observation's merge key is found in both DataFrames.

validate : str, optional
    If specified, checks if merge is of specified type.

    * "one_to_one" or "1:1": check if merge keys are unique in both
      left and right datasets.
    * "one_to_many" or "1:m": check if merge keys are unique in left
      dataset.
    * "many_to_one" or "m:1": check if merge keys are unique in right
      dataset.
    * "many_to_many" or "m:m": allowed, but does not result in checks.

这里的merge与关系型数据库中的join非常类似。下面根据实例看看如何使用。

on：根据某个字段进行连接，必须存在于两个DateFrame中（若未同时存在，则需要分别使用left_on和right_on来设置）

In [20]:

left = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'],
                     'A': ['A0', 'A1', 'A2', 'A3'],
                     'B': ['B0', 'B1', 'B2', 'B3']})

right = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'],
                      'C': ['C0', 'C1', 'C2', 'C3'],
                      'D': ['D0', 'D1', 'D2', 'D3']})

pd.merge(left, right, on='key')

Out[20]:

	key	A	B	C	D
0	K0	A0	B0	C0	D0
1	K1	A1	B1	C1	D1
2	K2	A2	B2	C2	D2
3	K3	A3	B3	C3	D3

也可以有多个连接键：

In [21]:

left = pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
                     'key2': ['K0', 'K1', 'K0', 'K1'],
                     'A': ['A0', 'A1', 'A2', 'A3'],
                     'B': ['B0', 'B1', 'B2', 'B3']})


right = pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
                      'key2': ['K0', 'K0', 'K0', 'K0'],
                      'C': ['C0', 'C1', 'C2', 'C3'],
                      'D': ['D0', 'D1', 'D2', 'D3']})

pd.merge(left, right, on=['key1', 'key2'])

Out[21]:

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K1	K0	A2	B2	C1	D1
2	K1	K0	A2	B2	C2	D2

how: 可以指定数据用哪种方式进行合并，没有的内容会为NaN，默认值inner

上面没有指定how，默认就是inner，下面分别看看left, right, outer的效果。

左外连接：

In [22]:

pd.merge(left, right, how='left', on=['key1', 'key2'])

Out[22]:

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN

右外连接：

In [23]:

pd.merge(left, right, how='right', on=['key1', 'key2'])

Out[23]:

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K1	K0	A2	B2	C1	D1
2	K1	K0	A2	B2	C2	D2
3	K2	K0	NaN	NaN	C3	D3

全外连接：

In [24]:

pd.merge(left, right, how='outer', on=['key1', 'key2'])

Out[24]:

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN
5	K2	K0	NaN	NaN	C3	D3

如果设置indicator为True, 则会增加名为_merge的一列，显示这列是如何而来，其中left_only表示只在左表中, right_only表示只在右表中, both表示两个表中都有：

In [25]:

pd.merge(left, right, how='outer', on=['key1', 'key2'], indicator=True)

Out[25]:

	key1	key2	A	B	C	D	_merge
0	K0	K0	A0	B0	C0	D0	both
1	K0	K1	A1	B1	NaN	NaN	left_only
2	K1	K0	A2	B2	C1	D1	both
3	K1	K0	A2	B2	C2	D2	both
4	K2	K1	A3	B3	NaN	NaN	left_only
5	K2	K0	NaN	NaN	C3	D3	right_only

如果左、右两边连接的字段名称不同时，可以分别设置left_on、right_on：

In [26]:

left = pd.DataFrame({'key1': ['K0', 'K1', 'K2', 'K3'],
                     'A': ['A0', 'A1', 'A2', 'A3'],
                     'B': ['B0', 'B1', 'B2', 'B3']})

right = pd.DataFrame({'key2': ['K0', 'K1', 'K2', 'K3'],
                      'C': ['C0', 'C1', 'C2', 'C3'],
                      'D': ['D0', 'D1', 'D2', 'D3']})

pd.merge(left, right, left_on='key1', right_on='key2')

Out[26]:

	key1	A	B	key2	C	D
0	K0	A0	B0	K0	C0	D0
1	K1	A1	B1	K1	C1	D1
2	K2	A2	B2	K2	C2	D2
3	K3	A3	B3	K3	C3	D3

非关联字段名称相同时，会怎样？

In [27]:

left = pd.DataFrame({'key1': ['K0', 'K1', 'K2', 'K3'],
                     'A': ['A0', 'A1', 'A2', 'A3'],
                     'B': ['B0', 'B1', 'B2', 'B3']})

right = pd.DataFrame({'key2': ['K0', 'K1', 'K2', 'K3'],
                      'C': ['C0', 'C1', 'C2', 'C3'],
                      'B': [30, 50, 70, 90]})

pd.merge(left, right, left_on='key1', right_on='key2')

Out[27]:

	key1	A	B_x	key2	C	B_y
0	K0	A0	B0	K0	C0	30
1	K1	A1	B1	K1	C1	50
2	K2	A2	B2	K2	C2	70
3	K3	A3	B3	K3	C3	90

默认suffixes=('_x', '_y')，也可以自行修改：

In [28]:

pd.merge(left, right, left_on='key1', right_on='key2',
         suffixes=('_left', '_right'))

Out[28]:

	key1	A	B_left	key2	C	B_right
0	K0	A0	B0	K0	C0	30
1	K1	A1	B1	K1	C1	50
2	K2	A2	B2	K2	C2	70
3	K3	A3	B3	K3	C3	90

append

append可以追加数据，并返回一个新对象，也是一种简单常用的数据合并方式。

Signature: 
df.append(other, ignore_index=False, verify_integrity=False, sort=False) -> 'DataFrame'

Docstring:
Append rows of `other` to the end of caller, returning a new object.

参数解释：

other: 要追加的其他DataFrame或Series
ignore_index: 如果为True则重新进行自然索引
verify_integrity: 如果为True则遇到重复索引内容时报错
sort: 是否进行排序

追加同结构的数据:

In [29]:

df1.append(df2)

Out[29]:

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7

追加不同结构的数据，没有的列会增加，没有对应内容的会为NAN：

In [30]:

df1.append(df4, sort=False)

Out[30]:

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
2	NaN	B2	NaN	D2	F2
3	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

追加多个DataFrame:

In [31]:

df1.append([df2, df3])

Out[31]:

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7
8	A8	B8	C8	D8
9	A9	B9	C9	D9
10	A10	B10	C10	D10
11	A11	B11	C11	D11

忽略原索引：

In [32]:

df1.append(df4, ignore_index=True, sort=False)

Out[32]:

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
4	NaN	B2	NaN	D2	F2
5	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

追加Series：

In [33]:

s2 = pd.Series(['X0', 'X1', 'X2', 'X3'],
               index=['A', 'B', 'C', 'D'])

df1.append(s2, ignore_index=True)

Out[33]:

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	X0	X1	X2	X3

追加字典列表：

In [34]:

d = [{'A': 1, 'B': 2, 'C': 3, 'X': 4},
     {'A': 5, 'B': 6, 'C': 7, 'Y': 8}]

df1.append(d, ignore_index=True, sort=False)

Out[34]:

	A	B	C	D	X	Y
0	A0	B0	C0	D0	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN
2	A2	B2	C2	D2	NaN	NaN
3	A3	B3	C3	D3	NaN	NaN
4	1	2	3	NaN	4.0	NaN
5	5	6	7	NaN	NaN	8.0

来个实战案例。在使用Excel的时候，常常会在数据最后，增加一行汇总数据，比如求和，求平均值等。现在用Pandas如何实现呢？

In [35]:

df = pd.DataFrame(np.random.randint(1, 10, size=(3, 4)),
                  columns=['a', 'b', 'c', 'd'])
df

Out[35]:

	a	b	c	d
0	8	5	6	1
1	6	2	8	5
2	7	2	2	3

In [36]:

df.append(pd.Series(df.sum(), name='total'))

Out[36]:

	a	b	c	d
0	8	5	6	1
1	6	2	8	5
2	7	2	2	3
total	21	9	16	9

数据清洗

数据清洗是指发现并纠正数据集中可识别的错误的一个过程，包括检查数据一致性，处理无效值、缺失值等。数据清洗是为了最大限度地提高数据集的准确性。

缺失值

由于数据来源的复杂性、不确定性，数据中难免会存在字段值不全、缺失等情况，下面先介绍如何找出这些缺失的值。

以这里的电影数据举例，其中数据集来自github，为了方便测试，下载压缩后上传到博客园，原始数据链接：
//github.com/LearnDataSci/articles/blob/master/Python%20Pandas%20Tutorial%20A%20Complete%20Introduction%20for%20Beginners/IMDB-Movie-Data.csv

In [37]:

movies = pd.read_csv('//files.cnblogs.com/files/blogs/478024/IMDB-Movie-Data.csv.zip')
movies.tail(2)

Out[37]:

	Rank	Title	Genre	Description	Director	Actors	Year	Runtime (Minutes)	Rating	Votes	Revenue (Millions)	Metascore
998	999	Search Party	Adventure,Comedy	A pair of friends embark on a mission to reuni…	Scot Armstrong	Adam Pally, T.J. Miller, Thomas Middleditch,Sh…	2014	93	5.6	4881	NaN	22.0
999	1000	Nine Lives	Comedy,Family,Fantasy	A stuffy businessman finds himself trapped ins…	Barry Sonnenfeld	Kevin Spacey, Jennifer Garner, Robbie Amell,Ch…	2016	87	5.3	12435	19.64	11.0

上面可以看到Revenue (Millions)列有个NaN，这里的NaN是一个缺值标识。

NaN (not a number) is the standard missing data marker used in pandas.

判断数据集中是否有缺失值：

In [38]:

np.any(movies.isna())

Out[38]:

True

或者判断数据集中是否不含缺失值：

In [39]:

np.all(movies.notna())

Out[39]:

False

统计每列有多少个缺失值：

In [40]:

movies.isna().sum(axis=0)

Out[40]:

Rank                    0
Title                   0
Genre                   0
Description             0
Director                0
Actors                  0
Year                    0
Runtime (Minutes)       0
Rating                  0
Votes                   0
Revenue (Millions)    128
Metascore              64
dtype: int64

统计每行有多少个缺失值：

In [41]:

movies.isna().sum(axis=1)

Out[41]:

0      0
1      0
2      0
3      0
4      0
      ..
995    1
996    0
997    0
998    1
999    0
Length: 1000, dtype: int64

统计一共有多少个缺失值：

In [42]:

movies.isna().sum().sum()

Out[42]:

筛选出有缺失值的列：

In [43]:

movies.isnull().any(axis=0)

Out[43]:

Rank                  False
Title                 False
Genre                 False
Description           False
Director              False
Actors                False
Year                  False
Runtime (Minutes)     False
Rating                False
Votes                 False
Revenue (Millions)     True
Metascore              True
dtype: bool

统计有缺失值的列的个数：

In [44]:

movies.isnull().any(axis=0).sum()

Out[44]:

筛选出有缺失值的行：

In [45]:

movies.isnull().any(axis=1)

Out[45]:

0      False
1      False
2      False
3      False
4      False
       ...  
995     True
996    False
997    False
998     True
999    False
Length: 1000, dtype: bool

统计有缺失值的行的个数：

In [46]:

movies.isnull().any(axis=1).sum()

Out[46]:

查看Metascore列缺失的数据：

In [47]:

movies[movies['Metascore'].isna()][['Rank', 'Title', 'Votes', 'Metascore']]

Out[47]:

	Rank	Title	Votes	Metascore
25	26	Paris pieds nus	222	NaN
26	27	Bahubali: The Beginning	76193	NaN
27	28	Dead Awake	523	NaN
39	40	5- 25- 77	241	NaN
42	43	Don’t Fuck in the Woods	496	NaN
…	…	…	…	…
967	968	The Walk	92378	NaN
969	970	The Lone Ranger	190855	NaN
971	972	Disturbia	193491	NaN
989	990	Selma	67637	NaN
992	993	Take Me Home Tonight	45419	NaN

64 rows × 4 columns

In [48]:

movies.shape

Out[48]:

(1000, 12)

In [49]:

movies.count()

Out[49]:

Rank                  1000
Title                 1000
Genre                 1000
Description           1000
Director              1000
Actors                1000
Year                  1000
Runtime (Minutes)     1000
Rating                1000
Votes                 1000
Revenue (Millions)     872
Metascore              936
dtype: int64

可以看出，每列应该有1000个数据的，count()时缺失值并没有算进来。

对于缺失值的处理，应根据具体的业务场景以及数据完整性的要求选择较合适的方案。常见的处理方案包括删除存在缺失值的数据(dropna)、替换缺失值(fillna)。

删除

某些场景下，有缺失值会认为该样本数据无效，就需要对整行或者整列数据进行删除(dropna)。

统计无缺失值的行的个数：

In [50]:

movies.notna().all(axis=1).sum()

Out[50]:

删除所有含缺失值的行：

In [51]:

data = movies.dropna()
data.shape

Out[51]:

(838, 12)

In [52]:

movies.shape

Out[52]:

(1000, 12)

data里838行数据都是无缺失值的。但是，值得注意的是，dropna()默认并不会在原来的数据集上删除，除非指定dropna(inplace=True)。下面演示一下：

In [53]:

# 复制一份完整的数据做原地删除演示
movies_copy = movies.copy()
print(movies_copy.shape)

# 原地删除
movies_copy.dropna(inplace=True)

# 查看原地删除是否生效
movies_copy.shape

(1000, 12)

Out[53]:

(838, 12)

统计无缺失值的列的个数：

In [54]:

movies.notna().all(axis=0).sum()

Out[54]:

删除含缺失值的列：

In [55]:

data = movies.dropna(axis=1)
data.shape

Out[55]:

(1000, 10)

how : {‘any’, ‘all’}, default ‘any’. Determine if row or column is removed from DataFrame, when we have at least one NA or all NA.

‘any’ : If any NA values are present, drop that row or column.
‘all’ : If all values are NA, drop that row or column.

删除所有值都缺失的行：

In [56]:

# Drop the rows where all elements are missing
data = movies.dropna(how='all')
data.shape

Out[56]:

(1000, 12)

这里的数据不存在所有值都缺失的行，所以how='all'时dropna()对此处的数据集无任何影响。

subset: Define in which columns to look for missing values.

subset参数可以指定在哪些列中寻找缺失值：

In [57]:

# 指定在Title、Metascore两列中寻找缺失值
data = movies.dropna(subset=['Title', 'Metascore'])
data.shape

Out[57]:

(936, 12)

统计Title、Metascore这两列无缺失值的行数。讲道理，肯定是等于上面删除缺失值后的行数。

In [58]:

movies[['Title', 'Metascore']].notna().all(axis=1).sum()

Out[58]:

thresh : int, optional. Require that many non-NA values.

In [59]:

# Keep only the rows with at least 2 non-NA values.
data = movies[['Title', 'Metascore', 'Revenue (Millions)']].dropna(thresh=2)
data.shape

Out[59]:

(970, 3)

由于Title列没有缺失值，相当于删除掉Metascore,Revenue (Millions)两列都为缺失值的行，如下：

In [60]:

data = movies[['Metascore', 'Revenue (Millions)']].dropna(how='all')
data.shape

Out[60]:

(970, 2)

填充

处理缺失值的另外一种常用方法是填充(fillna)。填充值虽然不绝对准确，但对获得真实结果的影响并不大时，可以尝试一用。

先看个简单的例子，然后再应用到上面的电影数据中去。因为电影数据比较多，演示起来并不直观。

In [61]:

hero = pd.DataFrame(data={'score': [97, np.nan, 96, np.nan, 95],
                          'wins': [np.nan, 9, np.nan, 11, 10],
                          'author': ['古龙', '金庸', np.nan, np.nan, np.nan],
                          'book': ['多情剑客无情剑', '笑傲江湖', '倚天屠龙记', '射雕英雄传', '绝代双骄'],
                          'skill': ['小李飞刀', '独孤九剑', '九阳神功', '降龙十八掌', '移花接玉'],
                          'wife': [np.nan, '任盈盈', np.nan, '黄蓉', np.nan],
                          'child': [np.nan, np.nan, np.nan, '郭襄', np.nan]},
                    index=['李寻欢', '令狐冲', '张无忌', '郭靖', '花无缺'])

hero

Out[61]:

	score	wins	author	book	skill	wife	child
李寻欢	97.0	NaN	古龙	多情剑客无情剑	小李飞刀	NaN	NaN
令狐冲	NaN	9.0	金庸	笑傲江湖	独孤九剑	任盈盈	NaN
张无忌	96.0	NaN	NaN	倚天屠龙记	九阳神功	NaN	NaN
郭靖	NaN	11.0	NaN	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95.0	10.0	NaN	绝代双骄	移花接玉	NaN	NaN

全部填充为unknown：

In [62]:

hero.fillna('unknown')

Out[62]:

	score	wins	author	book	skill	wife	child
李寻欢	97	unknown	古龙	多情剑客无情剑	小李飞刀	unknown	unknown
令狐冲	unknown	9	金庸	笑傲江湖	独孤九剑	任盈盈	unknown
张无忌	96	unknown	unknown	倚天屠龙记	九阳神功	unknown	unknown
郭靖	unknown	11	unknown	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95	10	unknown	绝代双骄	移花接玉	unknown	unknown

只替换第一个：

In [63]:

hero.fillna('unknown', limit=1)

Out[63]:

	score	wins	author	book	skill	wife	child
李寻欢	97	unknown	古龙	多情剑客无情剑	小李飞刀	unknown	unknown
令狐冲	unknown	9	金庸	笑傲江湖	独孤九剑	任盈盈	NaN
张无忌	96	NaN	unknown	倚天屠龙记	九阳神功	NaN	NaN
郭靖	NaN	11	NaN	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95	10	NaN	绝代双骄	移花接玉	NaN	NaN

不同列替换不同的值：

In [64]:

hero.fillna(value={'score': 100, 'author': '匿名', 'wife': '保密'})

Out[64]:

	score	wins	author	book	skill	wife	child
李寻欢	97.0	NaN	古龙	多情剑客无情剑	小李飞刀	保密	NaN
令狐冲	100.0	9.0	金庸	笑傲江湖	独孤九剑	任盈盈	NaN
张无忌	96.0	NaN	匿名	倚天屠龙记	九阳神功	保密	NaN
郭靖	100.0	11.0	匿名	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95.0	10.0	匿名	绝代双骄	移花接玉	保密	NaN

method : {‘backfill’, ‘bfill’, ‘pad’, ‘ffill’, None}, default None

上面是填充固定值，此外还能指定填充方法。

pad / ffill: propagate last valid observation forward to next valid

In [65]:

# 使用前一个有效值填充
hero.fillna(method='ffill')

Out[65]:

	score	wins	author	book	skill	wife	child
李寻欢	97.0	NaN	古龙	多情剑客无情剑	小李飞刀	NaN	NaN
令狐冲	97.0	9.0	金庸	笑傲江湖	独孤九剑	任盈盈	NaN
张无忌	96.0	9.0	金庸	倚天屠龙记	九阳神功	任盈盈	NaN
郭靖	96.0	11.0	金庸	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95.0	10.0	金庸	绝代双骄	移花接玉	黄蓉	郭襄

backfill / bfill: use next valid observation to fill gap.

In [66]:

# 使用后一个有效值填充
hero.fillna(method='bfill')

Out[66]:

	score	wins	author	book	skill	wife	child
李寻欢	97.0	9.0	古龙	多情剑客无情剑	小李飞刀	任盈盈	郭襄
令狐冲	96.0	9.0	金庸	笑傲江湖	独孤九剑	任盈盈	郭襄
张无忌	96.0	11.0	NaN	倚天屠龙记	九阳神功	黄蓉	郭襄
郭靖	95.0	11.0	NaN	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95.0	10.0	NaN	绝代双骄	移花接玉	NaN	NaN

对score、wins两列的缺失值用平均值来填充：

In [67]:

hero.fillna(hero[['score', 'wins']].mean())

Out[67]:

	score	wins	author	book	skill	wife	child
李寻欢	97.0	10.0	古龙	多情剑客无情剑	小李飞刀	NaN	NaN
令狐冲	96.0	9.0	金庸	笑傲江湖	独孤九剑	任盈盈	NaN
张无忌	96.0	10.0	NaN	倚天屠龙记	九阳神功	NaN	NaN
郭靖	96.0	11.0	NaN	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95.0	10.0	NaN	绝代双骄	移花接玉	NaN	NaN

指定列填充为unknown，并原地替换:

In [68]:

hero.child.fillna('unknown', inplace=True)
hero

Out[68]:

	score	wins	author	book	skill	wife	child
李寻欢	97.0	NaN	古龙	多情剑客无情剑	小李飞刀	NaN	unknown
令狐冲	NaN	9.0	金庸	笑傲江湖	独孤九剑	任盈盈	unknown
张无忌	96.0	NaN	NaN	倚天屠龙记	九阳神功	NaN	unknown
郭靖	NaN	11.0	NaN	射雕英雄传	降龙十八掌	黄蓉	郭襄
花无缺	95.0	10.0	NaN	绝代双骄	移花接玉	NaN	unknown

再回到上面的实际的电影数据案例。现在用平均值替换缺失值：

In [69]:

filled_movies = movies.fillna(
    movies[['Revenue (Millions)', 'Metascore']].mean())

np.any(filled_movies.isna())

Out[69]:

False

可见，填充后的电影数据中已经不存在缺失值了。

数据替换

数据替换常用于数据清洗整理、枚举转换、数据修正等场景。

先看下replace()方法的介绍：

Signature:
replace(
    to_replace=None,
    value=None,
    inplace=False,
    limit=None,
    regex=False,
    method='pad',
)

Docstring:
Replace values given in `to_replace` with `value`.

再看几个例子：

In [70]:

s = pd.Series(['a', 'b', 'c', 'd', 'e'])
s

Out[70]:

0    a
1    b
2    c
3    d
4    e
dtype: object

In [71]:

s.replace('a', 'aa')

Out[71]:

0    aa
1     b
2     c
3     d
4     e
dtype: object

In [72]:

s.replace({'d': 'dd', 'e': 'ee'})

Out[72]:

0     a
1     b
2     c
3    dd
4    ee
dtype: object

In [73]:

s.replace(['a', 'b', 'c'], ['aa', 'bb', 'cc'])

Out[73]:

0    aa
1    bb
2    cc
3     d
4     e
dtype: object

In [74]:

# 将c替换为它前一个值
s.replace('c', method='ffill')

Out[74]:

0    a
1    b
2    b
3    d
4    e
dtype: object

In [75]:

# 将c替换为它后一个值
s.replace('c', method='bfill')

Out[75]:

0    a
1    b
2    d
3    d
4    e
dtype: object

In [76]:

df = pd.DataFrame({'A': [0, -11, 2, 3, 35],
                   'B': [0, -20, 2, 5, 16]})
df

Out[76]:

	A	B
0	0	0
1	-11	-20
2	2	2
3	3	5
4	35	16

In [77]:

df.replace(0, 5)

Out[77]:

	A	B
0	5	5
1	-11	-20
2	2	2
3	3	5
4	35	16

In [78]:

df.replace([0, 2, 3, 5], 10)

Out[78]:

	A	B
0	10	10
1	-11	-20
2	10	10
3	10	10
4	35	16

In [79]:

df.replace([0, 2], [100, 200])

Out[79]:

	A	B
0	100	100
1	-11	-20
2	200	200
3	3	5
4	35	16

In [80]:

df.replace({0: 10, 2: 22})

Out[80]:

	A	B
0	10	10
1	-11	-20
2	22	22
3	3	5
4	35	16

In [81]:

df.replace({'A': 0, 'B': 2}, 100)

Out[81]:

	A	B
0	100	0
1	-11	-20
2	2	100
3	3	5
4	35	16

In [82]:

df.replace({'A': {2: 200, 3: 300}})

Out[82]:

	A	B
0	0	0
1	-11	-20
2	200	2
3	300	5
4	35	16

对一些极端值，如过大或者过小，可以使用df.clip(lower, upper)来修剪，当数据大于upper时，使用upper的值，小于lower时用lower 的值，类似numpy.clip的方法。

在修剪之前，再看一眼原始数据：

In [83]:

df

Out[83]:

	A	B
0	0	0
1	-11	-20
2	2	2
3	3	5
4	35	16

In [84]:

# 修剪成最小为2，最大为10
df.clip(2, 10)

Out[84]:

	A	B
0	2	2
1	2	2
2	2	2
3	3	5
4	10	10

In [85]:

# 对每列元素的最小值和最大值进行不同的限制

# 将A列数值修剪成[-3, 3]之间
# 将B列数值修剪成[-5, 5]之间
df.clip([-3, -5], [3, 5], axis=1)

Out[85]:

	A	B
0	0	0
1	-3	-5
2	2	2
3	3	5
4	3	5

In [86]:

# 对每行元素的最小值和最大值进行不同的限制

# 将第1行数值修剪成[5, 10]之间
# 将第2行数值修剪成[-15, -12]之间
# 将第3行数值修剪成[6, 10]之间
# 将第4行数值修剪成[4, 10]之间
# 将第5行数值修剪成[20, 30]之间
df.clip([5, -15, 6, 4, 20],
        [10, -12, 10, 10, 30],
        axis=0)

Out[86]:

	A	B
0	5	5
1	-12	-15
2	6	6
3	4	5
4	30	20

另外，可以将无效值先替换为nan，再做缺失值处理。这样就能应用上前面讲到的缺失值处理相关的知识。

比如这里的df，我们认为小于0的数据都是无效数据，可以：

In [87]:

df.replace([-11, -20], np.nan)

Out[87]:

	A	B
0	0.0	0.0
1	NaN	NaN
2	2.0	2.0
3	3.0	5.0
4	35.0	16.0

当然，也可以像下面这样把无效数据变为nan:

In [88]:

df[df >= 0]

Out[88]:

	A	B
0	0.0	0.0
1	NaN	NaN
2	2.0	2.0
3	3.0	5.0
4	35.0	16.0

此时，上面讲到的缺失值处理就能派上用场了。

文本内容比较复杂时，可以使用正则进行匹配替换。下面看几个例子：

In [89]:

df = pd.DataFrame({'A': ['bat', 'foo', 'bait'],
                   'B': ['abc', 'bar', 'xyz']})
df

Out[89]:

	A	B
0	bat	abc
1	foo	bar
2	bait	xyz

In [90]:

# 利用正则将ba开头且总共3个字符的文本替换为new
df.replace(to_replace=r'^ba.$', value='new', regex=True)

Out[90]:

	A	B
0	new	abc
1	foo	new
2	bait	xyz

In [91]:

# 如果多列正则不同的情况下可以按以下格式对应传入
df.replace({'A': r'^ba.$'}, {'A': 'new'}, regex=True)

Out[91]:

	A	B
0	new	abc
1	foo	bar
2	bait	xyz

In [92]:

df.replace(regex=r'^ba.$', value='new')

Out[92]:

	A	B
0	new	abc
1	foo	new
2	bait	xyz

In [93]:

# 不同正则替换不同的值
df.replace(regex={r'^ba.$': 'new', 'foo': 'xyz'})

Out[93]:

	A	B
0	new	abc
1	xyz	new
2	bait	xyz

In [94]:

# 多个正则替换为同一个值
df.replace(regex=[r'^ba.$', 'foo'], value='new')

Out[94]:

	A	B
0	new	abc
1	new	new
2	bait	xyz

重复值

重复值在数据清洗中可能需要删除。下面介绍Pandas如何识别重复值以及如何删除重复值。

Signature:
df.duplicated(
    subset: Union[Hashable, Sequence[Hashable], NoneType] = None,
    keep: Union[str, bool] = 'first',
) -> 'Series'

Docstring:
Return boolean Series denoting duplicate rows.

Considering certain columns is optional.

Parameters
----------
subset : column label or sequence of labels, optional
    Only consider certain columns for identifying duplicates, by
    default use all of the columns.
keep : {'first', 'last', False}, default 'first'
    Determines which duplicates (if any) to mark.

    - ``first`` : Mark duplicates as ``True`` except for the first occurrence.
    - ``last`` : Mark duplicates as ``True`` except for the last occurrence.
    - False : Mark all duplicates as ``True``.

看官方给的例子：

In [95]:

df = pd.DataFrame({
    'brand': ['Yum Yum', 'Yum Yum', 'Indomie', 'Indomie', 'Indomie'],
    'style': ['cup', 'cup', 'cup', 'pack', 'pack'],
    'rating': [4, 4, 3.5, 15, 5]
})
df

Out[95]:

	brand	style	rating
0	Yum Yum	cup	4.0
1	Yum Yum	cup	4.0
2	Indomie	cup	3.5
3	Indomie	pack	15.0
4	Indomie	pack	5.0

In [96]:

# 默认情况下，对于每行重复的值，第一次出现都设置为False，其他为True
df.duplicated()

Out[96]:

0    False
1     True
2    False
3    False
4    False
dtype: bool

In [97]:

# 将每行重复值的最后一次出现设置为False，其他为True
df.duplicated(keep='last')

Out[97]:

0     True
1    False
2    False
3    False
4    False
dtype: bool

In [98]:

# 所有重复行都为True
df.duplicated(keep=False)

Out[98]:

0     True
1     True
2    False
3    False
4    False
dtype: bool

In [99]:

# 参数subset可以在指定列上查找重复值
df.duplicated(subset=['brand'])

Out[99]:

0    False
1     True
2    False
3     True
4     True
dtype: bool

再看如何删除重复值：

Signature:
df.drop_duplicates(
    subset: Union[Hashable, Sequence[Hashable], NoneType] = None,
    keep: Union[str, bool] = 'first',
    inplace: bool = False,
    ignore_index: bool = False,
) -> Union[ForwardRef('DataFrame'), NoneType]
Docstring:
Return DataFrame with duplicate rows removed.

Considering certain columns is optional. Indexes, including time indexes
are ignored.

Parameters
----------
subset : column label or sequence of labels, optional
    Only consider certain columns for identifying duplicates, by
    default use all of the columns.
keep : {'first', 'last', False}, default 'first'
    Determines which duplicates (if any) to keep.
    - ``first`` : Drop duplicates except for the first occurrence.
    - ``last`` : Drop duplicates except for the last occurrence.
    - False : Drop all duplicates.
inplace : bool, default False
    Whether to drop duplicates in place or to return a copy.
ignore_index : bool, default False
    If True, the resulting axis will be labeled 0, 1, …, n - 1.

同样继续官方给的例子：

In [100]:

# By default, it removes duplicate rows based on all columns
df.drop_duplicates()

Out[100]:

	brand	style	rating
0	Yum Yum	cup	4.0
2	Indomie	cup	3.5
3	Indomie	pack	15.0
4	Indomie	pack	5.0

In [101]:

# To remove duplicates on specific column(s), use `subset`
df.drop_duplicates(subset=['brand'])

Out[101]:

	brand	style	rating
0	Yum Yum	cup	4.0
2	Indomie	cup	3.5

In [102]:

# To remove duplicates and keep last occurences, use `keep`
df.drop_duplicates(subset=['brand', 'style'], keep='last')

Out[102]:

	brand	style	rating
1	Yum Yum	cup	4.0
2	Indomie	cup	3.5
4	Indomie	pack	5.0

分组与聚合

在数据统计与分析中，分组与聚合非常常见。如果是SQL，对应的就是Group By和聚合函数(Aggregation Functions)。下面看看pandas是怎么玩的。

Signature:
df.groupby(
    by=None,
    axis=0,
    level=None,
    as_index: bool = True,
    sort: bool = True,
    group_keys: bool = True,
    squeeze: bool = <object object at 0x7f3df810e750>,
    observed: bool = False,
    dropna: bool = True,
) -> 'DataFrameGroupBy'

Docstring:
Group DataFrame using a mapper or by a Series of columns.

groupby()方法可以按指定字段对DataFrame进行分组，生成一个分组器对象，然后再把这个对象的各个字段按一定的聚合方法输出。

其中by为分组字段，由于是第一个参数可以省略，可以按列表给多个。会返回一个DataFrameGroupBy对象，如果不给聚合方法，不会返回 DataFrame。

准备演示数据：

In [103]:

df = pd.read_csv('//files.cnblogs.com/files/blogs/478024/team.csv.zip')
df

Out[103]:

	name	team	Q1	Q2	Q3	Q4
0	Liver	E	89	21	24	64
1	Arry	C	36	37	37	57
2	Ack	A	57	60	18	84
3	Eorge	C	93	96	71	78
4	Oah	D	65	49	61	86
…	…	…	…	…	…	…
95	Gabriel	C	48	59	87	74
96	Austin7	C	21	31	30	43
97	Lincoln4	C	98	93	1	20
98	Eli	E	11	74	58	91
99	Ben	E	21	43	41	74

100 rows × 6 columns

In [104]:

# 按team分组后对应列求和
df.groupby('team').sum()

Out[104]:

	Q1	Q2	Q3	Q4
team
A	1066	639	875	783
B	975	1218	1202	1136
C	1056	1194	1068	1127
D	860	1191	1241	1199
E	963	1013	881	1033

In [105]:

# 按team分组后对应列求平均值
df.groupby('team').mean()

Out[105]:

	Q1	Q2	Q3	Q4
team
A	62.705882	37.588235	51.470588	46.058824
B	44.318182	55.363636	54.636364	51.636364
C	48.000000	54.272727	48.545455	51.227273
D	45.263158	62.684211	65.315789	63.105263
E	48.150000	50.650000	44.050000	51.650000

In [106]:

# 按team分组后不同列使用不同的聚合方式
df.groupby('team').agg({'Q1': sum,  # 求和
                        'Q2': 'count',  # 计数
                        'Q3': 'mean',  # 求平均值
                        'Q4': max})  # 求最大值

Out[106]:

	Q1	Q2	Q3	Q4
team
A	1066	17	51.470588	97
B	975	22	54.636364	99
C	1056	22	48.545455	98
D	860	19	65.315789	99
E	963	20	44.050000	98

If by is a function, it’s called on each value of the object’s index.

In [107]:

# team在C之前(包括C)分为一组，C之后的分为另外一组
df.set_index('team').groupby(lambda team: 'team1' if team <= 'C' else 'team2')['name'].count()

Out[107]:

team1    61
team2    39
Name: name, dtype: int64

或者下面这种写法也行：

In [108]:

df.groupby(lambda idx: 'team1' if df.loc[idx]['team'] <= 'C' else 'team2')['name'].count()

Out[108]:

team1    61
team2    39
Name: name, dtype: int64

In [109]:

# 按name的长度(length)分组，并取出每组中name的第一个值和最后一个值
df.groupby(df['name'].apply(lambda x: len(x))).agg({'name': ['first', 'last']})

Out[109]:

	name
	first	last
name
3	Ack	Ben
4	Arry	Leon
5	Liver	Aiden
6	Harlie	Jamie0
7	William	Austin7
8	Harrison	Lincoln4
9	Alexander	Theodore3

In [110]:

# 只对部分分组
df.set_index('team').groupby({'A': 'A组', 'B': 'B组'})['name'].count()

Out[110]:

A组    17
B组    22
Name: name, dtype: int64

可以将以上方法混合组成列表进行分组：

In [111]:

# 按team，name长度分组，取分组中最后一行
df.groupby(['team', df['name'].apply(lambda x: len(x))]).last()

Out[111]:

		name	Q1	Q2	Q3	Q4
team	name
A	3	Ack	57	60	18	84
	4	Toby	52	27	17	68
	5	Aaron	96	75	55	8
	6	Nathan	87	77	62	13
	7	Stanley	69	71	39	97
B	3	Kai	66	45	13	48
	4	Liam	2	80	24	25
	5	Lewis	4	34	77	28
	6	Jamie0	39	97	84	55
	7	Albert0	85	38	41	17
	8	Grayson7	59	84	74	33
C	4	Adam	90	32	47	39
	5	Calum	14	91	16	82
	6	Connor	62	38	63	46
	7	Austin7	21	31	30	43
	8	Lincoln4	98	93	1	20
	9	Sebastian	1	14	68	48
D	3	Oah	65	49	61	86
	4	Ezra	16	56	86	61
	5	Aiden	20	31	62	68
	6	Reuben	70	72	76	56
	7	Hunter3	38	80	82	40
	8	Benjamin	15	88	52	25
	9	Theodore3	43	7	68	80
E	3	Ben	21	43	41	74
	4	Leon	38	60	31	7
	5	Roman	73	1	25	44
	6	Dexter	73	94	53	20
	7	Zachary	12	71	85	93
	8	Jackson5	6	10	15	33

We can groupby different levels of a hierarchical index using the level parameter.

In [112]:

arrays = [['Falcon', 'Falcon', 'Parrot', 'Parrot'],
          ['Captive', 'Wild', 'Captive', 'Wild']]

index = pd.MultiIndex.from_arrays(arrays, names=('Animal', 'Type'))

df = pd.DataFrame({'Max Speed': [390., 350., 30., 20.]},
                  index=index)
df

Out[112]:

		Max Speed
Animal	Type
Falcon	Captive	390.0
Falcon	Wild	350.0
Parrot	Captive	30.0
Parrot	Wild	20.0

In [113]:

# df.groupby(level=0).mean()
df.groupby(level="Animal").mean()

Out[113]:

	Max Speed
Animal
Falcon	370.0
Parrot	25.0

In [114]:

# df.groupby(level=1).mean()
df.groupby(level="Type").mean()

Out[114]:

	Max Speed
Type
Captive	210.0
Wild	185.0

We can also choose to include NA in group keys or not by setting dropna parameter, the default setting is True.

In [115]:

l = [[1, 2, 3], [1, None, 4], [2, 1, 3], [1, 2, 2]]
df = pd.DataFrame(l, columns=["a", "b", "c"])
df

Out[115]:

	a	b	c
0	1	2.0	3
1	1	NaN	4
2	2	1.0	3
3	1	2.0	2

In [116]:

df.groupby(by=["b"]).sum()

Out[116]:

	a	c
b
1.0	2	3
2.0	2	5

In [117]:

df.groupby(by=["b"], dropna=False).sum()

Out[117]:

	a	c
b
1.0	2	3
2.0	2	5
NaN	1	4

上面体验了一下pandas分组聚合的基本使用后，接下来看看分组聚合的一些过程细节。

分组

有以下动物最大速度数据：

In [118]:

df = pd.DataFrame([('bird', 'Falconiformes', 389.0),
                   ('bird', 'Psittaciformes', 24.0),
                   ('mammal', 'Carnivora', 80.2),
                   ('mammal', 'Primates', np.nan),
                   ('mammal', 'Carnivora', 58)],
                  index=['falcon', 'parrot', 'lion',
                         'monkey', 'leopard'],
                  columns=('class', 'order', 'max_speed'))

df

Out[118]:

	class	order	max_speed
falcon	bird	Falconiformes	389.0
parrot	bird	Psittaciformes	24.0
lion	mammal	Carnivora	80.2
monkey	mammal	Primates	NaN
leopard	mammal	Carnivora	58.0

In [119]:

# 分组数
df.groupby('class').ngroups

Out[119]:

In [120]:

# 查看分组
df.groupby('class').groups

Out[120]:

{'bird': ['falcon', 'parrot'], 'mammal': ['lion', 'monkey', 'leopard']}

In [121]:

df.groupby('class').size()

Out[121]:

class
bird      2
mammal    3
dtype: int64

In [122]:

# 查看鸟类分组内容
df.groupby('class').get_group('bird')

Out[122]:

	class	order	max_speed
falcon	bird	Falconiformes	389.0
parrot	bird	Psittaciformes	24.0

获取分组中的第几个值：

In [123]:

# 第一个
df.groupby('class').nth(1)

Out[123]:

	order	max_speed
class
bird	Psittaciformes	24.0
mammal	Primates	NaN

In [124]:

# 最后一个
df.groupby('class').nth(-1)

Out[124]:

	order	max_speed
class
bird	Psittaciformes	24.0
mammal	Carnivora	58.0

In [125]:

# 第一个，第二个
df.groupby('class').nth([1, 2])

Out[125]:

	order	max_speed
class
bird	Psittaciformes	24.0
mammal	Primates	NaN
mammal	Carnivora	58.0

In [126]:

# 每组显示前2个
df.groupby('class').head(2)

Out[126]:

	class	order	max_speed
falcon	bird	Falconiformes	389.0
parrot	bird	Psittaciformes	24.0
lion	mammal	Carnivora	80.2
monkey	mammal	Primates	NaN

In [127]:

# 每组最后2个
df.groupby('class').tail(2)

Out[127]:

	class	order	max_speed
falcon	bird	Falconiformes	389.0
parrot	bird	Psittaciformes	24.0
monkey	mammal	Primates	NaN
leopard	mammal	Carnivora	58.0

In [128]:

# 分组序号
df.groupby('class').ngroup()

Out[128]:

falcon     0
parrot     0
lion       1
monkey     1
leopard    1
dtype: int64

In [129]:

# 返回每个元素在所在组的序号的序列
df.groupby('class').cumcount(ascending=False)

Out[129]:

falcon     1
parrot     0
lion       2
monkey     1
leopard    0
dtype: int64

In [130]:

# 按鸟类首字母分组
df.groupby(df['class'].str[0]).groups

Out[130]:

{'b': ['falcon', 'parrot'], 'm': ['lion', 'monkey', 'leopard']}

In [131]:

# 按鸟类第一个字母和第二个字母分组
df.groupby([df['class'].str[0], df['class'].str[1]]).groups

Out[131]:

{('b', 'i'): ['falcon', 'parrot'], ('m', 'a'): ['lion', 'monkey', 'leopard']}

In [132]:

# 在组内的排名
df.groupby('class').rank()

Out[132]:

	max_speed
falcon	2.0
parrot	1.0
lion	2.0
monkey	NaN
leopard	1.0

聚合

对数据进行分组后，接下来就可以收获果实了，给分组给定统计方法，最终得到分组聚合的结果。除了常见的数学统计方法，还可以使用 agg()和transform()等函数进行操作。

In [133]:

# 描述性统计
df.groupby('class').describe()

Out[133]:

	max_speed
	count	mean	std	min	25%	50%	75%	max
class
bird	2.0	206.5	258.093975	24.0	115.25	206.5	297.75	389.0
mammal	2.0	69.1	15.697771	58.0	63.55	69.1	74.65	80.2

In [134]:

# 一列使用多个聚合方法
df.groupby('class').agg({'max_speed': ['min', 'max', 'sum']})

Out[134]:

	max_speed
	min	max	sum
class
bird	24.0	389.0	413.0
mammal	58.0	80.2	138.2

In [135]:

df.groupby('class')['max_speed'].agg(
    Max='max', Min='min', Diff=lambda x: x.max() - x.min())

Out[135]:

	Max	Min	Diff
class
bird	389.0	24.0	365.0
mammal	80.2	58.0	22.2

In [136]:

df.groupby('class').agg(max_speed=('max_speed', 'max'),
                        count_order=('order', 'count'))

Out[136]:

	max_speed	count_order
class
bird	389.0	2
mammal	80.2	3

In [137]:

df.groupby('class').agg(
    max_speed=pd.NamedAgg(column='max_speed', aggfunc='max'),
    count_order=pd.NamedAgg(column='order', aggfunc='count')
)

Out[137]:

	max_speed	count_order
class
bird	389.0	2
mammal	80.2	3

transform类似于agg，但不同的是它返回的是一个DataFrame，每个会将原来的值一一替换成统计后的值，比如按组计算平均值，那么返回的新DataFrame中每个值就是它所在组的平均值。

In [138]:

df.groupby('class').agg(np.mean)

Out[138]:

	max_speed
class
bird	206.5
mammal	69.1

In [139]:

df.groupby('class').transform(np.mean)

Out[139]:

	max_speed
falcon	206.5
parrot	206.5
lion	69.1
monkey	69.1
leopard	69.1

分组后筛选原始数据：

Signature:
DataFrameGroupBy.filter(func, dropna=True, *args, **kwargs) 

Docstring:
    Return a copy of a DataFrame excluding filtered elements.

    Elements from groups are filtered if they do not satisfy the
    boolean criterion specified by func.

In [140]:

# 筛选出 按class分组后，分组内max_speed平均值大于100的元素
df.groupby(['class']).filter(lambda x: x['max_speed'].mean() > 100)

Out[140]:

	class	order	max_speed
falcon	bird	Falconiformes	389.0
parrot	bird	Psittaciformes	24.0

In [141]:

# 取出分组后index
df.groupby('class').apply(lambda x: x.index.to_list())

Out[141]:

class
bird             [falcon, parrot]
mammal    [lion, monkey, leopard]
dtype: object

In [142]:

# 取出分组后每组中max_speed最大的前N个
df.groupby('class').apply(lambda x: x.sort_values(
    by='max_speed', ascending=False).head(1))

Out[142]:

		class	order	max_speed
class
bird	falcon	bird	Falconiformes	389.0
mammal	lion	mammal	Carnivora	80.2

In [143]:

df.groupby('class').apply(lambda x: pd.Series({
    'speed_max': x['max_speed'].max(),
    'speed_min': x['max_speed'].min(),
    'speed_mean': x['max_speed'].mean(),
}))

Out[143]:

	speed_max	speed_min	speed_mean
class
bird	389.0	24.0	206.5
mammal	80.2	58.0	69.1

按分组导出Excel文件：

In [144]:

for group, data in df.groupby('class'):
    data.to_excel(f'data/{group}.xlsx')

In [145]:

# 每组去重值后数量
df.groupby('class').order.nunique()

Out[145]:

class
bird      2
mammal    2
Name: order, dtype: int64

In [146]:

# 每组去重后的值
df.groupby("class")['order'].unique()

Out[146]:

class
bird      [Falconiformes, Psittaciformes]
mammal              [Carnivora, Primates]
Name: order, dtype: object

In [147]:

# 统计每组数据值的数量
df.groupby("class")['order'].value_counts()

Out[147]:

class   order         
bird    Falconiformes     1
        Psittaciformes    1
mammal  Carnivora         2
        Primates          1
Name: order, dtype: int64

In [148]:

# 每组最大的1个
df.groupby("class")['max_speed'].nlargest(1)

Out[148]:

class         
bird    falcon    389.0
mammal  lion       80.2
Name: max_speed, dtype: float64

In [149]:

# 每组最小的2个
df.groupby("class")['max_speed'].nsmallest(2)

Out[149]:

class          
bird    parrot      24.0
        falcon     389.0
mammal  leopard     58.0
        lion        80.2
Name: max_speed, dtype: float64

In [150]:

# 每组值是否单调递增
df.groupby("class")['max_speed'].is_monotonic_increasing

Out[150]:

class
bird      False
mammal    False
Name: max_speed, dtype: bool

In [151]:

# 每组值是否单调递减
df.groupby("class")['max_speed'].is_monotonic_decreasing

Out[151]:

class
bird       True
mammal    False
Name: max_speed, dtype: bool

堆叠与透视

实际生产中，我们拿到的原始数据的表现形状可能并不符合当前需求，比如说不是期望的维度、数据不够直观、表现力不够等等。此时，可以对原始数据进行适当的变形，比如堆叠、透视、行列转置等。

堆叠

看个简单的例子就能明白讲的是什么：

In [152]:

df = pd.DataFrame([[19, 136, 180, 98], [21, 122, 178, 96]], index=['令狐冲', '李寻欢'],
                  columns=['age', 'weight', 'height', 'score'])
df

Out[152]:

	age	weight	height	score
令狐冲	19	136	180	98
李寻欢	21	122	178	96

In [153]:

# 有点像宽表变高表, 我是这样觉得的
df.stack()

Out[153]:

令狐冲  age        19
     weight    136
     height    180
     score      98
李寻欢  age        21
     weight    122
     height    178
     score      96
dtype: int64

In [154]:

# 有点像高表变宽表
df.stack().unstack()

Out[154]:

	age	weight	height	score
令狐冲	19	136	180	98
李寻欢	21	122	178	96

透视表

Signature:
df.pivot(index=None, columns=None, values=None) -> 'DataFrame'

Docstring:
    Return reshaped DataFrame organized by given index / column values.

    Reshape data (produce a "pivot" table) based on column values. Uses
    unique values from specified `index` / `columns` to form axes of the
    resulting DataFrame. This function does not support data
    aggregation, multiple values will result in a MultiIndex in the
    columns.

In [155]:

df = pd.DataFrame({'name': ['江小鱼', '江小鱼', '江小鱼', '花无缺', '花无缺',
                            '花无缺'],
                   'bug_level': ['A', 'B', 'C', 'A', 'B', 'C'],
                   'bug_count': [2, 3, 5, 1, 5, 6]})
df

Out[155]:

	name	bug_level	bug_count
0	江小鱼	A	2
1	江小鱼	B	3
2	江小鱼	C	5
3	花无缺	A	1
4	花无缺	B	5
5	花无缺	C	6

把上面的bug等级与bug数统计表变形如下，还是原来的数据，但是不是更加直观呢？

In [156]:

df.pivot(index='name', columns='bug_level', values='bug_count')

Out[156]:

bug_level	A	B	C
name
江小鱼	2	3	5
花无缺	1	5	6

如果原始数据中有重复的统计呢？就比如说上面的例子中来自不同产品线的bug统计,就可能出现两行这样的数据['江小鱼','B',3]、['江小鱼','B',4]，先试下用pivot会怎样？

In [157]:

df = pd.DataFrame({'name': ['江小鱼', '江小鱼', '江小鱼', '江小鱼', '江小鱼', '花无缺', '花无缺',
                            '花无缺', '花无缺', '花无缺', ],
                   'bug_level': ['A', 'B', 'C', 'B', 'C', 'A', 'B', 'C', 'A', 'B'],
                   'bug_count': [2, 3, 5, 4, 6, 1, 5, 6, 3, 1],
                   'score': [70, 80, 90, 76, 86, 72, 82, 88, 68, 92]})
df

Out[157]:

	name	bug_level	bug_count	score
0	江小鱼	A	2	70
1	江小鱼	B	3	80
2	江小鱼	C	5	90
3	江小鱼	B	4	76
4	江小鱼	C	6	86
5	花无缺	A	1	72
6	花无缺	B	5	82
7	花无缺	C	6	88
8	花无缺	A	3	68
9	花无缺	B	1	92

In [158]:

try:
    df.pivot(index='name', columns='bug_level', values='bug_count')
except ValueError as e:
    print(e)

Index contains duplicate entries, cannot reshape

原来，pivot()只能将数据进行reshape,不支持聚合。遇到上面这种含重复值需进行聚合计算,应使用pivot_table()。它能实现类似Excel那样的高级数据透视功能。

In [159]:

# 统计员工来自不同产品线不同级别的bug总数
df.pivot_table(index=['name'], columns=['bug_level'],
               values='bug_count', aggfunc=np.sum)

Out[159]:

bug_level	A	B	C
name
江小鱼	2	7	11
花无缺	4	6	6

当然，这里的聚合可以非常灵活：

In [161]:

df.pivot_table(index=['name'], columns=['bug_level'], aggfunc={
               'bug_count': np.sum, 'score': [max, np.mean]})

Out[161]:

	bug_count			score
	sum			max			mean
bug_level	A	B	C	A	B	C	A	B	C
name
江小鱼	2	7	11	70	80	90	70	78	88
花无缺	4	6	6	72	92	88	70	87	88

还可以给每列每行加个汇总，如下所示：

In [162]:

df.pivot_table(index=['name'], columns=['bug_level'],
               values='bug_count', aggfunc=np.sum, margins=True, margins_name='汇总')

Out[162]:

bug_level	A	B	C	汇总
name
江小鱼	2	7	11	20
花无缺	4	6	6	16
汇总	6	13	17	36

交叉表

交叉表是用于统计分组频率的特殊透视表。简单来说，就是将两个或者多个列重中不重复的元素组成一个新的 DataFrame，新数据的行和列交叉的部分值为其组合在原数据中的数量。

还是来个例子比较直观。有如下学生选专业数据：

In [163]:

df = pd.DataFrame({'name': ['杨过', '小龙女', '郭靖', '黄蓉', '李寻欢', '孙小红', '张无忌',
                            '赵敏', '令狐冲', '任盈盈'],
                   'gender': ['男', '女', '男', '女', '男', '女', '男', '女', '男', '女'],
                   'major': ['机械工程', '软件工程', '金融工程', '工商管理', '机械工程', '金融工程', '软件工程', '工商管理', '软件工程', '工商管理']})
df

Out[163]:

	name	gender	major
0	杨过	男	机械工程
1	小龙女	女	软件工程
2	郭靖	男	金融工程
3	黄蓉	女	工商管理
4	李寻欢	男	机械工程
5	孙小红	女	金融工程
6	张无忌	男	软件工程
7	赵敏	女	工商管理
8	令狐冲	男	软件工程
9	任盈盈	女	工商管理

若想了解学生选专业是否与性别有关，可以做如下统计：

In [164]:

pd.crosstab(df['gender'], df['major'])

Out[164]:

major	工商管理	机械工程	软件工程	金融工程
gender
女	3	0	1	1
男	0	2	2	1

同时，回忆一下上篇讲到的 //www.cnblogs.com/bytesfly/p/pandas-1.html#画图

In [165]:

# 男、女生填报专业比例饼状图
pd.crosstab(df['gender'], df['major']).T.plot(
    kind='pie', subplots=True, figsize=(12, 8), autopct="%.0f%%")
plt.show()

换个角度看下：

In [166]:

# 各专业男女生填报人数柱状图
pd.crosstab(df['gender'], df['major']).T.plot(
    kind='bar', stacked=True, rot=0, title='各专业男女生填报人数柱状图', xlabel='', figsize=(10, 6))
plt.show()

再回到上面所讲的交叉表相关知识。

In [167]:

# 对交叉结果进行归一化
pd.crosstab(df['gender'], df['major'], normalize=True)

Out[167]:

major	工商管理	机械工程	软件工程	金融工程
gender
女	0.3	0.0	0.1	0.1
男	0.0	0.2	0.2	0.1

In [168]:

# 对交叉结果按行进行归一化
pd.crosstab(df['gender'], df['major'], normalize='index')

Out[168]:

major	工商管理	机械工程	软件工程	金融工程
gender
女	0.6	0.0	0.2	0.2
男	0.0	0.4	0.4	0.2

In [169]:

# 对交叉结果按列进行归一化
pd.crosstab(df['gender'], df['major'], normalize='columns')

Out[169]:

major	工商管理	机械工程	软件工程	金融工程
gender
女	1.0	0.0	0.333333	0.5
男	0.0	1.0	0.666667	0.5

同样，也可以给每列每行加个汇总，如下：

In [170]:

pd.crosstab(df['gender'], df['major'], margins=True, margins_name='汇总')

Out[170]:

major	工商管理	机械工程	软件工程	金融工程	汇总
gender
女	3	0	1	1	5
男	0	2	2	1	5
汇总	3	2	3	2	10

Tags: Python

快速上手pandas(下)

数据合并

concat

merge

append

数据清洗

缺失值

删除

填充

数据替换

重复值

分组与聚合

分组

聚合

堆叠与透视

堆叠

透视表

交叉表

VirMach 便宜 VPS

QNews

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7
8	A8	B8	C8	D8
9	A9	B9	C9	D9
10	A10	B10	C10	D10
11	A11	B11	C11	D11

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
4	NaN	B2	NaN	D2	F2
5	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3
6	NaN	NaN	NaN	NaN	B6	D6	F6
7	NaN	NaN	NaN	NaN	B7	D7	F7

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN
5	K2	K0	NaN	NaN	C3	D3

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
2	NaN	B2	NaN	D2	F2
3	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7
8	A8	B8	C8	D8
9	A9	B9	C9	D9
10	A10	B10	C10	D10
11	A11	B11	C11	D11

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
4	NaN	B2	NaN	D2	F2
5	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3
6	NaN	NaN	NaN	NaN	B6	D6	F6
7	NaN	NaN	NaN	NaN	B7	D7	F7

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN
5	K2	K0	NaN	NaN	C3	D3

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
2	NaN	B2	NaN	D2	F2
3	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

快速上手pandas(下)

数据合并

concat

merge

append

数据清洗

缺失值

删除

填充

数据替换

重复值

分组与聚合

分组

聚合

堆叠与透视

堆叠

透视表

交叉表

分享此文：

Related Posts

Python|快速掌握单双链表和树

Flask 邮件发送

微信小程序 — 英语词典 (小程序插件)

分组密码（一） — 密码学复习（四）

VirMach 便宜 VPS

QNews

热门文章

热门搜寻

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7
8	A8	B8	C8	D8
9	A9	B9	C9	D9
10	A10	B10	C10	D10
11	A11	B11	C11	D11

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

		A	B	C	D
x	0	A0	B0	C0	D0
	1	A1	B1	C1	D1
	2	A2	B2	C2	D2
	3	A3	B3	C3	D3
y	4	A4	B4	C4	D4
	5	A5	B5	C5	D5
	6	A6	B6	C6	D6
	7	A7	B7	C7	D7
z	8	A8	B8	C8	D8
	9	A9	B9	C9	D9
	10	A10	B10	C10	D10
	11	A11	B11	C11	D11

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
4	NaN	B2	NaN	D2	F2
5	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3
6	NaN	NaN	NaN	NaN	B6	D6	F6
7	NaN	NaN	NaN	NaN	B7	D7	F7

	A	B	C	D	B	D	F
0	A0	B0	C0	D0	NaN	NaN	NaN
1	A1	B1	C1	D1	NaN	NaN	NaN
2	A2	B2	C2	D2	B2	D2	F2
3	A3	B3	C3	D3	B3	D3	F3

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN

	key1	key2	A	B	C	D
0	K0	K0	A0	B0	C0	D0
1	K0	K1	A1	B1	NaN	NaN
2	K1	K0	A2	B2	C1	D1
3	K1	K0	A2	B2	C2	D2
4	K2	K1	A3	B3	NaN	NaN
5	K2	K0	NaN	NaN	C3	D3

	A	B	C	D
0	A0	B0	C0	D0
1	A1	B1	C1	D1
2	A2	B2	C2	D2
3	A3	B3	C3	D3
4	A4	B4	C4	D4
5	A5	B5	C5	D5
6	A6	B6	C6	D6
7	A7	B7	C7	D7

	A	B	C	D	F
0	A0	B0	C0	D0	NaN
1	A1	B1	C1	D1	NaN
2	A2	B2	C2	D2	NaN
3	A3	B3	C3	D3	NaN
2	NaN	B2	NaN	D2	F2
3	NaN	B3	NaN	D3	F3
6	NaN	B6	NaN	D6	F6
7	NaN	B7	NaN	D7	F7