I have a long DataFrame with index values like this:
我有一个长的DataFrame,索引值如下:
| burger10 | ...
| pasta25 | ...
| milk | ...
| yoghurt() | ...
I need to get rid of the trailing digits or parentheses. I am trying to use replace() with regex, but without success. Tried this:
我需要摆脱尾随的数字或括号。我正在尝试使用带有正则表达式的replace(),但没有成功。试过这个:
energy.replace(to_replace='[0-9,\.,\(,\)]+', value='', regex=True, inplace=True)
1 个解决方案
#1
2
You don't need to escape () or use , in character class [], just use them as literal, and if you mean trailing, you need the anchor $ to match the end of string:
你不需要在字符类[]中使用escape()或use,只需将它们用作文字,如果你的意思是尾随,你需要使用anchor $来匹配字符串的结尾:
energy[0].str.replace("[0-9()]+$", "")
#0 burger
#1 pasta
#2 milk
#3 yoghurt
#Name: 0, dtype: object
If the strings are in the index, you can use .index to access, modify it and reassign it back to the data frame:
如果字符串在索引中,则可以使用.index访问,修改它并将其重新分配回数据框:
energy.index = energy.index.str.replace("[0-9()]+$", "")
#1
2
You don't need to escape () or use , in character class [], just use them as literal, and if you mean trailing, you need the anchor $ to match the end of string:
你不需要在字符类[]中使用escape()或use,只需将它们用作文字,如果你的意思是尾随,你需要使用anchor $来匹配字符串的结尾:
energy[0].str.replace("[0-9()]+$", "")
#0 burger
#1 pasta
#2 milk
#3 yoghurt
#Name: 0, dtype: object
If the strings are in the index, you can use .index to access, modify it and reassign it back to the data frame:
如果字符串在索引中,则可以使用.index访问,修改它并将其重新分配回数据框:
energy.index = energy.index.str.replace("[0-9()]+$", "")