pandas 在"loc"之后如何“替换"工作? [英] Pandas how can 'replace' work after 'loc'?

问题描述

我已经尝试了很多次，但是在使用"loc"后，"replace"似乎无法正常工作. 例如，我想用正则表达式将'conlumn_a'值为'apple'的行替换为'conlumn_b'.

I have tried many times, but seems the 'replace' can NOT work well after use 'loc'. For example I want to replace the 'conlumn_b' with an regex for the row that the 'conlumn_a' value is 'apple'.

这是我的示例代码:

df.loc[df['conlumn_a'] == 'apple', 'conlumn_b'].replace(r'^11*', 'XXX',inplace=True, regex=True)

示例:

conlumn_a       conlumn_b
apple           123
banana          11
apple           11
orange          33

我期望的'df'结果是:

The result that I expected for the 'df' is:

conlumn_a       conlumn_b
apple           123
banana          11
apple           XXX
orange          33

有人遇到过需要在"loc"之后用正则表达式替换"的问题吗?

Anyone has meet this issue that needs 'replace' with regex after 'loc' ?

或者你们还有其他一些好的解决方案?

OR you guys has some other good solutions ?

非常感谢您的帮助！

推荐答案

我认为您双方都需要过滤器:

I think you need filter in both sides:

m = df['conlumn_a'] == 'apple'
df.loc[m,'conlumn_b'] = df.loc[m,'conlumn_b'].astype(str).replace(r'^(11+)','XXX',regex=True)
print (df)
  conlumn_a conlumn_b
0     apple       123
1    banana        11
2     apple       XXX
3    orange        33

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