Python 创建一列，根据条件删除字符串中不需要的部分
我是python新手，我被困在这里。我有一个如下的数据框，我正在尝试创建一个新的列，其中只包含“类型”列的宏类型 数据帧：Python 创建一列，根据条件删除字符串中不需要的部分,python,pandas,loops,split,Python,Pandas,Loops,Split,我是python新手，我被困在这里。我有一个如下的数据框，我正在尝试创建一个新的列，其中只包含“类型”列的宏类型 数据帧： import pandas as pd d = {'Genres': ['Finance', 'Arcade', 'Business', 'Photography', 'Entertainment;Brain Games', 'Medical', 'Tools', 'Casual;Brain Games', 'Medical', 'Entertainment'],
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
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
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)
部分时，它工作正常
任何提示都将不胜感激！谢谢 您只需使用str.split（“；”）
。如果不存在于字符串中，不会发生任何事情->返回带有原始字符串的列表（因此您可以始终使用[0]
）：
印刷品：
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
您可以只使用str.split（“；”）
。如果不存在于字符串中，不会发生任何事情->返回带有原始字符串的列表（因此您可以始终使用[0]
）：
印刷品：
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
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数据帧上的运行时比较：
#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)
#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)
#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)
#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)
10k数据帧上的运行时比较：
#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)
#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)
#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)
#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)
50k数据帧上的运行时比较：
#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)
#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)
#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)
#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)
100k数据帧上的运行时比较：
#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)
#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)
#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)
#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)
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数据帧上的运行时比较：
#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)
#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)
#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)
#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)
10k数据帧上的运行时比较：
#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)
#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)
#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)
#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)
50k数据帧上的运行时比较：
#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)
#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)
#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)
#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)
100k数据帧上的运行时比较：
#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)
#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)
#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)
#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)
我想说的是df['macro_-genres']=df['genres'].str.split（'；'）.str[0]
可能是一个更好的答案，但是如果你用%timeit
快速回答，性能会更好。它是258µs
而不是474µs
@DavidErickson为什么不使用map
。我认为这比申请要快。查看我的答案在10k数据帧上的运行时比较。@grayrigel我认为您不会看到应用程序和映射程序之间的性能差异，如您的答案所示。@DavidErickson是的。你是对的。我做了不同长度的测试。我收回我的陈述，即map
更快。但是，在小型dfs上稍好一些，在大型dfs上似乎变得更慢。我想说的是df['macro_genres']=df['genres'].str.split（“；”）.str[0]
可能是一个更好的答案，但是使用快速%timeit
的答案，性能会更好。它是258µs
而不是474µs
@DavidErickson为什么不使用map
。我认为这比申请要快。查看我的答案在10k数据帧上的运行时比较。@grayrigel我认为您不会看到应用程序和映射程序之间的性能差异，如您的答案所示。@DavidErickson是的。你是对的。我做了不同长度的测试。我收回我的陈述，即map
更快。在小型dfs上稍微好一些，但是在大型dfs上似乎变慢了。看到不同长度的数据帧进行比较会很好。@AndrejKesely感谢您的投票。用1K、10K、50K、100K数据帧更新了我的答案。我收回我的陈述，即map
更快。在小dfs上稍微好一些，但是在大dfs上似乎变慢了。非常感谢@Grayrigel！我确信有一个干净简单的解决方案，但没有达到目的。它完全奏效了！速度重要吗？使用更干净/更惯用的方法不是更好吗？@AMC我认为速度很重要，尤其是在处理大型数据帧时。我不确定有什么比半行代码更好。你有什么建议？你有其他的方法吗？投票支持基准测试。看到不同长度的数据帧进行比较会很好。@AndrejKesely感谢您的投票。用1K、10K、50K、100K数据帧更新了我的答案。我收回我的陈述，即map
更快。在小dfs上稍微好一些，但是在大dfs上似乎变慢了。非常感谢@Grayrigel！我确信有一个干净简单的解决方案，但没有达到目的。它完全奏效了！速度重要吗？使用更干净/更惯用的方法不是更好吗？@AMC我认为速度很重要，尤其是在处理大型数据帧时。我不确定有什么比半行代码更好。你有什么建议？你有其他的方法吗？请提供一个。很抱歉。我是新来的。你说哪一部分应该是最小可复制的，数据帧本身？我输入它是因为它只是一个更大数据框的一小部分。你说哪一部分应该是最小的可复制的，数据框本身？应该可以复制/粘贴您的代码和数据，并且能够立即运行代码。感谢AMC的提示。虽然已经给出了解决方案，但我已经包含了生成数据帧的代码。请提供一个。对此表示抱歉。我是新来的。你说哪一部分应该是最小可复制的，数据帧本身？我输入它是因为它只是一个更大数据框的一小部分。你说哪一部分应该是最小的可复制的，数据框本身？应该可以复制/粘贴您的代码和数据，并且能够立即运行代码。感谢AMC的提示。虽然已经给出了解决方案，但我已经包含了生成数据帧的代码。