Question

             0         1         2         3         4         5
word                                                            
</s>  0.001129 -0.000896  0.000319  0.001534  0.001106 -0.001404
in    0.070312  0.086914  0.087891  0.062500  0.069336 -0.108887
for  -0.011780 -0.047363  0.044678  0.063477 -0.018188 -0.063965
that -0.015747 -0.028320  0.083496  0.050293 -0.110352  0.031738
is    0.007050 -0.073242  0.171875  0.022583 -0.132812  0.198242

我有这个DataFrame，我想获取其中索引（“单词”）包含特定字符串（大小写不变）的行。我尝试过

df[df.index.str.lower().contains("lebron") == True]

它给出KeyError: False。虽然

df[df.index.str.contains("Lebron") == True]

工作正常。

在这种情况下如何使用lower()？

Answer 1

lower在这里不是必需的，为不区分大小写的搜索在str.contains中添加参数case=False，如果可能缺少值则添加na=False，如果没有则添加regex=False正则表达式搜索：

df[df.index.str.contains("lebron", case=False)]

df[df.index.str.contains("lebron", case=False, na=False, regex=False)]

但是，如果只希望将str.lower()与str.contains()链接起来，就不必与True进行比较：

df[df.index.str.lower().str.contains("lebron")]

df[df.index.str.lower().str.contains("lebron", na=False, regex=False)]

Answer 2

如果您要进行简单的子字符串检查，建议不要使用正则表达式比较（即，不对正则表达式使用str.lower）。您可以在此处使用列表理解：

df[['lebron' in x.lower() for x in df.index]]

如果索引中可能存在NaN，则可以修改解决方案以处理它们：

df[[pd.notna(x) and 'lebron' in x.lower() for x in df.index]]

没有正则表达式，可以相应地使用str.contains：

df[df.index.str.lower().str.contains("lebron", regex=False)]

如果您没有NaN，可以省略结尾的== True。否则，

df[df.index.str.lower().str.contains("lebron", regex=False) == True]

可以正常工作。

过滤熊猫DataFrame时如何使用lower（）方法？

2 个答案: