Python 字符串中的第三个字符索引

我有一个python字符串。根据这个字符串，我想编写一个函数，返回整个字符串，直到（不含）第三个逗号

将熊猫作为pd导入
将numpy作为np导入
mystr=pd.系列（['文化冲突，未来，太空战争，太空殖民地，社会'，
‘海洋、毒品滥用、异国小岛、东印度、爱情、叛徒】）
def转换：
索引=0
计数=0
当计数小于3时：
index=s.str.find（'，'索引）
计数=计数+1
指数+=1
返回s.str[0:index-1]
输出=变换（mystr）
出来

这是南。我想：

• “文化冲突、未来、太空战”
• “海洋、毒品滥用、异国情调的岛屿”

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有人能帮我吗？

使用

str.split

Ex:

import pandas as pd

mystr = pd.Series(['culture clash, future, space war, space colony, society', 'ocean, drug abuse, exotic island, east india, love, traitor'])
print(mystr.str.split(",").str[:3].apply(",".join))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

>>> mystr.apply(lambda x : ",".join(x.split(',')[:3]))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

输出：

import pandas as pd

mystr = pd.Series(['culture clash, future, space war, space colony, society', 'ocean, drug abuse, exotic island, east india, love, traitor'])
print(mystr.str.split(",").str[:3].apply(",".join))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

>>> mystr.apply(lambda x : ",".join(x.split(',')[:3]))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

试试这个

>>> mystr = pd.Series(['culture clash, future, space war, space colony, society','ocean, drug abuse, exotic island, east india, love, traitor'])

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输出：

import pandas as pd

mystr = pd.Series(['culture clash, future, space war, space colony, society', 'ocean, drug abuse, exotic island, east india, love, traitor'])
print(mystr.str.split(",").str[:3].apply(",".join))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

>>> mystr.apply(lambda x : ",".join(x.split(',')[:3]))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

说明：

import pandas as pd

mystr = pd.Series(['culture clash, future, space war, space colony, society', 'ocean, drug abuse, exotic island, east india, love, traitor'])
print(mystr.str.split(",").str[:3].apply(",".join))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

>>> mystr.apply(lambda x : ",".join(x.split(',')[:3]))

0    culture clash, future, space war
1    ocean, drug abuse, exotic island
dtype: object

• 按

  ，

  进行拆分，并通过像

  [：3]

  这样的切片方式获取前三个单词，然后用

  ，

  再次将它们连接起来

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如果要考虑性能，列表理解会更快，因为

str

方法在以下方面速度较慢：

pd.Series([','.join(i.split(',')[:3]) for i in mystr])
#pd.Series(','.join(i.split(',')[:3]) for i in mystr)

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