在数据框中选择行与列中的任何项目均相等的行

Question

Let's say I have the following dataframe and I want to select any row that has any of it's values equal to any item in the list: CodesOfInterest=['A','D']

>>> import pandas as pd
>>> d1=pd.DataFrame([['A','B','C','D'],['D','Q','S', np.nan],['R',np.nan,np.nan,np.nan],[np.nan,'A',np.nan,np.nan]],columns=['Code1','Code2','Code3','Code4'])
>>> d1
  Code1 Code2 Code3 Code4
0     A     B     C     D
1     D     Q     S   NaN
2     R   NaN   NaN   NaN
3   NaN     A   NaN   NaN
>>> 

This can be done pretty easily with one line of code:

>>> CodesOfInterest=['A','D']
>>> d1[(d1.isin(CodesOfInterest)==True).any(1)]
  Code1 Code2 Code3 Code4
0     A     B     C     D
1     D     Q     S   NaN
3   NaN     A   NaN   NaN
>>> 

However say I have the following second dataframe indexed the same as the first that adds a condition to this subset.

>>> d2=pd.DataFrame([[1,0,1,0],[0,1,1, np.nan],[1,np.nan,np.nan,np.nan],[np.nan,1,np.nan,np.nan]],columns=['CodeStatus1','CodeStatus2','CodeStatus3','CodeStatus4'])
>>> d2
   CodeStatus1  CodeStatus2  CodeStatus3  CodeStatus4
0            1            0            1            0
1            0            1            1          NaN
2            1          NaN          NaN          NaN
3          NaN            1          NaN          NaN
>>> 

Now I want to only select rows from my d1 that have any of their values equal to any time in my list AND have their corresponding 'CodeStatus' (from d2) equal to 1. And by corresponding CodeStatus I mean pairs of (Code1, CodeStatus1), (Code2, CodeStatus2), etc.

I have a clunky way of doing this that requires looping through each of the 4 Codes and Code Statuses. See below:

>>> bs=[]    
>>> for Num in range(1,5):
...     Code='Code'+str(Num)
...     CodeStatus='CodeStatus'+str(Num)
...     b=(df[Code].isin(CodesOfInterest))&(df[CodeStatus]==1)
...     bs.append(b)
... 
>>> Matches=pd.concat(bs,1)
>>> 
>>> d1[(Matches==True).any(1)]
  Code1 Code2 Code3 Code4
0     A     B     C     D
3   NaN     A   NaN   NaN
>>> 

As you see, record 1 now gets dropped from the dataframe because although it has a column with code 'D', the Code Status for this code is not equal to 1.

Is there a more elegant way to make this query that doesn't require looping through each column?

Answer 1

您可以通过以下方式实现：

d1[pd.DataFrame((d1.isin(CodesOfInterest)==True).values*(d2==1).values).any(1)]