熊猫系列列表到一个系列

Question

我有一个Pandas系列的字符串列表:

0                           [slim, waist, man]
1                                [slim, waistline]
2                                     [santa]

如您所见,列表因长度而异.我想要一种有效的方法将其折叠成一个系列

0 slim
1 waist
2 man
3 slim
4 waistline
5 santa

我知道我可以使用分解列表

series_name.split(' ')

但我很难将这些字符串放回一个列表中.

谢谢!

Answer 1

这是一个只使用pandas函数的简单方法:

import pandas as pd

s = pd.Series([
    ['slim', 'waist', 'man'],
    ['slim', 'waistline'],
    ['santa']])

然后

s.apply(pd.Series).stack().reset_index(drop=True)

给出所需的输出.在某些情况下,您可能希望保存原始索引并添加第二级以索引嵌套元素,例如

0  0         slim
   1        waist
   2          man
1  0         slim
   1    waistline
2  0        santa

如果这是你想要的,只需.reset_index(drop=True)从链中省略即可.

Answer 2

In pandas version 0.25.0 appeared a new method 'explode' for series and dataframes. Older versions do not have such method.

It helps to build the result you need.

For example you have such series:

import pandas as pd

s = pd.Series([
    ['slim', 'waist', 'man'],
    ['slim', 'waistline'],
    ['santa']])

Then you can use

s.explode()

To get such result:

0         slim
0        waist
0          man
1         slim
1    waistline
2        santa

In case of dataframe:

df = pd.DataFrame({
  's': pd.Series([
    ['slim', 'waist', 'man'],
    ['slim', 'waistline'],
    ['santa']
   ]),
   'a': 1
})

You will have such DataFrame:

                    s  a
0  [slim, waist, man]  1
1   [slim, waistline]  1
2             [santa]  1

Applying explode on s column:

df.explode('s')

Will give you such result:

           s  a
0       slim  1
0      waist  1
0        man  1
1       slim  1
1  waistline  1
2      santa  1

If your series, contain empty lists

import pandas as pd

s = pd.Series([
    ['slim', 'waist', 'man'],
    ['slim', 'waistline'],
    ['santa'],
    []
])

Then running explode will introduce NaN values for empty lists, like this:

0         slim
0        waist
0          man
1         slim
1    waistline
2        santa
3          NaN

If this is not desired, you can dropna method call:

s.explode().dropna()

To get this result:

0         slim
0        waist
0          man
1         slim
1    waistline
2        santa

Dataframes also have dropna method:

df = pd.DataFrame({
  's': pd.Series([
    ['slim', 'waist', 'man'],
    ['slim', 'waistline'],
    ['santa'],
    []
   ]),
   'a': 1
})

Running explode without dropna:

df.explode('s')

Will result into:

           s  a
0       slim  1
0      waist  1
0        man  1
1       slim  1
1  waistline  1
2      santa  1
3        NaN  1

with dropna:

df.explode('s').dropna(subset=['s'])

Result:

           s  a
0       slim  1
0      waist  1
0        man  1
1       slim  1
1  waistline  1
2      santa  1

Answer 3

您基本上只是想在这里展平嵌套列表.

你应该只能迭代系列的元素:

slist =[]
for x in series:
    slist.extend(x)

或者更清晰(但更难理解)列表理解:

slist = [st for row in s for st in row]

Answer 4

series_name.sum()

确实满足您的需求。请确保它是一系列列表，否则您的值将被串联（如果是字符串）或添加（如果是int）

Answer 5

您可以尝试使用itertools.chain来简化列表：

In [70]: from itertools import chain
In [71]: import pandas as pnd
In [72]: s = pnd.Series([['slim', 'waist', 'man'], ['slim', 'waistline'], ['santa']])
In [73]: s
Out[73]: 
0    [slim, waist, man]
1     [slim, waistline]
2               [santa]
dtype: object
In [74]: new_s = pnd.Series(list(chain(*s.values)))
In [75]: new_s
Out[75]: 
0         slim
1        waist
2          man
3         slim
4    waistline
5        santa
dtype: object