在 pd.Dataframe 中选择反向索引

Question

How to select the inverse index in pd.DataFrame by using loc or iloc ?

I tried df.loc[!my_index,my_feature] but fail.

And df.loc[[ind for ind in df.index.tolist() if ind not in my_index],my_feature] looks too dull. Any better idea?

Answer 1

Use difference :

df.loc[df.index.difference(my_index),my_feature]

Alternatively numpy.setdiff1d :

df.loc[np.setdiff1d(df.index, my_index),my_feature]

Sample :

my_index = [5,7]
df = pd.DataFrame({'A': ['a','a','a','b'], 'B': list(range(4)) }, index=[5,7,8,9])
print (df)
   A  B
5  a  0
7  a  1
8  a  2
9  b  3

print(df.loc[df.index.difference(my_index),'A'])
8    a
9    b
Name: A, dtype: object

Answer 2

You may take advantage of index.difference .

idx2 = df.index.difference(my_index)

Or, set.difference

idx2 = set(df.index).difference(my_index) # note, order not guaranteed

df.loc[idx2, ...]

Answer 3

假设 my_index 是您要忽略的行索引，您可以将它们删除在数据帧 df 中存在的位置：

df = df.drop(my_index, errors='ignore')