Pandas：基于多个不同的列创建列

Question

Ok im feeling like this one is very easy and there should also be the right answer in a previous thread, but apparently I couldnt manage to find the answer by myself or in a thread. Here is what I got: I've got a dataframe with different samples belonging to groups

pd.DataFrame({'sample1': [1,2,3], 'sample2':[2,4,6], 'sample3':[4,4,4], 'sample4':[6,6,6], 'divisor':[1,2,1]})
groups=[["sample1","sample2"],["sample3","sample4"]]

I want the code te create a new column for each sample dependent on the sum of the group where this sample is in. Result should be 0 if quotient is below 0 or else should be original value. This first part perfectly does the summing:

for i in range(len(groups)):
    df["groupsum"+str(i)]=df[groups[i]].sum(axis=1)

    for sample in groups[i]:
        df[sample+"_corr"]=""
        df[sample+"_corr"]= df[sample].apply(lambda x: 0 if (df["groupsum"+str(i)]/df["divisor"])<4 else df[sample])

I get the error:

 ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

So whats the right way to handle this? Thanks a lot in advance

Answer 1

just use np.wehere instead of looping the dataframe with apply:

df[sample+"_corr"]= np.where((df["groupsum"+str(i)]/df["divisor"])<4 , 0 , df[sample])

Output:

    sample1 sample2 sample3 sample4 divisor groupsum0   sample1_corr    sample2_corr    groupsum1   sample3_corr    sample4_corr
0   1   2   4   6   1   3   0   0   10  4   6
1   2   4   4   6   2   6   0   0   10  4   6
2   3   6   4   6   1   9   3   6   10  4   6

this is also better performance because apply is very slow solution and should be avoided when possible.