【Python】详解pandas的isin索引和~反向索引

       有的时候会经常遇到条件过滤的场景，这个时候可能经常使用isin或者是~来进行一步操作，而不是写条件语句的方式，这样来提高效率和简洁度。

1、直接根据条件进行索引，isin()接受一个列表，判断该列中元素是否在列表中

import numpy as np
import pandas as pd
df=pd.DataFrame(np.random.randn(4,4),columns=['A','B','C','D'])
df
Out[189]: 
          A         B         C         D
0  0.289595  0.202207 -0.850390  0.197016
1  0.403254 -1.287074  0.916361  0.055136
2 -0.359261 -1.266615 -0.733625 -0.790208
3  0.164862 -0.649637  0.716620  1.447703
df['E'] = ['aa', 'bb', 'cc', 'cc']
df
Out[191]: 
          A         B         C         D   E
0  0.289595  0.202207 -0.850390  0.197016  aa
1  0.403254 -1.287074  0.916361  0.055136  bb
2 -0.359261 -1.266615 -0.733625 -0.790208  cc
3  0.164862 -0.649637  0.716620  1.447703  cc
df.E.isin(['aa','cc'])
Out[192]: 
0     True
1    False
2     True
3     True
Name: E, dtype: bool
df[df.E.isin(['aa','cc'])]
Out[193]: 
          A         B         C         D   E
0  0.289595  0.202207 -0.850390  0.197016  aa
2 -0.359261 -1.266615 -0.733625 -0.790208  cc
3  0.164862 -0.649637  0.716620  1.447703  cc

2、根据多条件进行索引，此时用&（交集）或者|（并集）进行连接

df[df.E.isin(['aa'])|df.E.isin(['cc'])]
Out[194]: 
          A         B         C         D   E
0  0.289595  0.202207 -0.850390  0.197016  aa
2 -0.359261 -1.266615 -0.733625 -0.790208  cc
3  0.164862 -0.649637  0.716620  1.447703  cc
df[df.E.isin(['aa'])]
Out[195]: 
          A         B        C         D   E
0  0.289595  0.202207 -0.85039  0.197016  aa

3、通过字典的形式传递多个条件{‘某列’:[条件],‘某列’:[条件],}

df['D'] = [1,2,3,4]
df[df.isin({'D':[0,3],'E':['aa','cc']})]
Out[200]: 
    A   B   C    D    E
0 NaN NaN NaN  NaN   aa
1 NaN NaN NaN  NaN  NaN
2 NaN NaN NaN  3.0   cc
3 NaN NaN NaN  NaN   cc

4、~相当于isnotin

df[~(df.E=='cc')]
Out[202]: 
          A         B         C  D   E
0  0.289595  0.202207 -0.850390  1  aa
1  0.403254 -1.287074  0.916361  2  bb