根据条件创建一列删除不需要的字符串部分

Question

我是 python 新手，我被困在这里。我有一个如下所示的数据框，我正在尝试创建一个仅包含 Genres 列的宏类型的新列。

数据框：

import pandas as pd
d = {'Genres': ['Finance', 'Arcade', 'Business', 'Photography', 'Entertainment;Brain Games', 'Medical', 'Tools', 'Casual;Brain Games', 'Medical', 'Entertainment'], 
     'Last Updated': ['March 10, 2018', 'May 24, 2018', 'April 11, 2018', 'November 6, 2014', 'March 9, 2018', 'May 17, 2018', 'June 3, 2016', 'April 10, 2016', 'July 16, 2018', 'May 17, 2017']}
df = pd.DataFrame(data=d)
df

                       Genres        Last Updated
0                     Finance      March 10, 2018
1                      Arcade        May 24, 2018
2                    Business      April 11, 2018
3                 Photography    November 6, 2014
4   Entertainment;Brain Games       March 9, 2018
5                     Medical        May 17, 2018
6                       Tools        June 3, 2016
7          Casual;Brain Games      April 10, 2016
8                     Medical       July 16, 2018
9               Entertainment        May 17, 2017

所需的输出类似于：

                       Genres          macro_genres        Last Updated
0                     Finance               Finance      March 10, 2018
1                      Arcade                Arcade        May 24, 2018
2                    Business              Business      April 11, 2018
3                 Photography           Photography    November 6, 2014
4   Entertainment;Brain Games         Entertainment       March 9, 2018
5                     Medical               Medical        May 17, 2018
6                       Tools                 Tools        June 3, 2016
7          Casual;Brain Games                Casual      April 10, 2016
8                     Medical               Medical       July 16, 2018
9               Entertainment         Entertainment        May 17, 2017

我尝试过的：

def macro_genre(i):
    for i in df['Genres']:
        if ';' in i:
            j = i.split(';')[0]
            return j
        else:
            return i
                    
df['macro_genres'] = df['Genres'].apply(macro_genre)

但它不起作用。它会创建列，但会重复整个列的第一个值。

当我在函数外尝试for 部分时，它可以工作。

任何提示将不胜感激！谢谢！！！

Answer 1

一种可能是使用map：

df['macro_games'] = df['Genres'].astype(str).map(lambda x : x.split(';')[0])

输出：

>>> df
                       Genres          macro_genres        Last Updated
0                     Finance               Finance      March 10, 2018
1                      Arcade                Arcade        May 24, 2018
2                    Business              Business      April 11, 2018
3                 Photography           Photography    November 6, 2014
4   Entertainment;Brain Games         Entertainment       March 9, 2018
5                     Medical               Medical        May 17, 2018
6                       Tools                 Tools        June 3, 2016
7          Casual;Brain Games                Casual      April 10, 2016
8                     Medical               Medical       July 16, 2018
9               Entertainment         Entertainment        May 17, 2017

1k 数据帧的运行时比较：

#apply method
>>> %timeit -n 1000 df['Genres'].apply(lambda x: x.split(';')[0])
535 µs ± 12.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#str split method (slowest)
>>> %timeit -n 1000 df['Genres'].str.split(';').str[0]
1.36 ms ± 44.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#map method
>>> %timeit -n 1000 df['Genres'].map(lambda x : x.split(';')[0])
527 µs ± 17.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

10k 数据帧的运行时比较：

#apply method
>>> %timeit -n 1000 df['Genres'].apply(lambda x: x.split(';')[0])
3.62 ms ± 105 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#str split method (slowest)
>>> %timeit -n 1000 df['Genres'].str.split(';').str[0]
10 ms ± 259 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#map method
>>> %timeit -n 1000 df['Genres'].map(lambda x : x.split(';')[0])
3.47 ms ± 59.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

50k 数据帧的运行时比较：

#apply method
>>> %timeit -n 1000 df['Genres'].apply(lambda x: x.split(';')[0])
17 ms ± 133 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#map method 
>>> %timeit -n 1000 df['Genres'].map(lambda x : x.split(';')[0])
16.7 ms ± 278 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

100k 数据帧的运行时比较：

#apply method
>>> %timeit -n 1000 df['Genres'].apply(lambda x: x.split(';')[0])
34.1 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#map method 
>>> %timeit -n 1000 df['Genres'].astype(str).map(lambda x : x.split(';')[0])
35.5 ms ± 596 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer 2

您可以使用str.split(';')。如果; 不存在于字符串中，则不会发生任何事情 -> 返回带有原始字符串的列表（因此您始终可以使用[0]）：

df['macro_genres'] = df['Genres'].apply(lambda x: x.split(';')[0])
print(df)

打印：

                      Genres      Last_Updated   macro_genres
0                    Finance    March 10, 2018        Finance
1                     Arcade      May 24, 2018         Arcade
2                   Business    April 11, 2018       Business
3                Photography  November 6, 2014    Photography
4  Entertainment;Brain_Games     March 9, 2018  Entertainment
5                    Medical      May 17, 2018        Medical
6                      Tools      June 3, 2016          Tools
7         Casual;Brain Games    April 10, 2016         Casual
8                    Medical     July 16, 2018        Medical
9              Entertainment      May 17, 2017  Entertainment