Python 替换熊猫数据框中的部分字符串
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replace part of the string in pandas data frame
提问by Mr.Riply
I have pandas data frame in which I need to replace one part of the vale with another value
我有熊猫数据框，我需要用另一个值替换谷的一部分
for Example. I have
例如。我有
HF - Antartica
HF - America
HF - Asia
out of which I'd like to replace ony the HF -part
thus the result would be
我想更换其中的任何HF -部分，因此结果是
Hi Funny Antartica
Hi Funny America
Hi Funny Asia
I have tried pd.replace()but it doesnt work as I need only one part of the string replaced, rather than the entire string
我试过了，pd.replace()但它不起作用，因为我只需要替换字符串的一部分，而不是整个字符串
回答by jezrael
It seems you need Series.replace:
看来你需要Series.replace：
print (df)
val
0 HF - Antartica
1 HF - America
2 HF - Asia
print (df.val.replace({'HF -':'Hi'}, regex=True))
0 Hi Antartica
1 Hi America
2 Hi Asia
Name: val, dtype: object
Similar solution with str.replace:
类似的解决方案str.replace：
print (df.val.str.replace('HF -', 'Hi'))
0 Hi Antartica
1 Hi America
2 Hi Asia
Name: val, dtype: object
回答by kuriouscoder
To add to @jezrael's answer, you need to include regex=Trueotherwise it would match directly. Also, here it replaces the values across all columns in the data frame. If you don't intend this, you could filter to a column and then replace. For replacing across all values in the data frame, try:
要添加到@jezrael 的答案中，您需要包括regex=True否则它将直接匹配。此外，这里它替换了数据框中所有列的值。如果您不打算这样做，您可以过滤到一列然后替换。要替换数据框中的所有值，请尝试：
df.replace('HF', 'Hi Funny', regex=True)
df.replace('HF', 'Hi Funny', regex=True)
You could also provide a list based patterns and replacement values. The complete set of options are provided in the documentation here.
您还可以提供基于列表的模式和替换值。此处的文档中提供了完整的选项集。
So if the data frame is:
所以如果数据框是：
>df = pd.DataFrame({'Column': ['HF - Antartica', 'HF - America', 'HF - Asia']})
>df.replace('HF', 'Hi Funny', regex=True)
should print:
应该打印：
Column
0 Hi Funny - Antartica
1 Hi Funny - America
2 Hi Funny - Asia
回答by Sarmad Mahar
I would like to share one more thing that is very much important, you can replace full stop with space ". " with "." normal full stop
我想再分享一件非常重要的事情，你可以用空格“.”代替句号“.”。正常句号
df['label']=df.label.replace({"\. ": "."},regex=True)
