Pandas: for each row, get the matching column

Question

I have a dataframe like this,

df = pd.DataFrame({'A':[1,-2,3,4,5],'B':[1,2,3,4,-5],'C':[1,2,3,-4,5],'value':[123,1567,1456,764,2456]})
print(df)

is,

   A  B  C  value
0  1  1  1    123
1 -2  2  2   1567
2  3 3  3   1456
3  4  4 -4    764
4  5  -5  5   2456

I know that there is only one negative value per row. I want to know in which column such a value is. I know I can do it like this,

desiredOutput = []
for i,row in df.iterrows():
    if any(row < 0):
        print(row[row < 0].index.to_numpy()[0],row[row[row < 0].index.to_numpy()[0]],row.value)
        desiredOutput.append(row[row < 0].index.to_numpy()[0])
    else:
        desiredOutput.append(None)

print(desiredOutput)

gives,

A -2 1567
C -4 764
B -5 2456
[None, 'A', None, 'C', 'B']

But I imagine there must be a pythonnic way to do it (probably using .apply()?)

Answer 1

Use DataFrame.dot for matrix multiplication:

desiredOutput = df.lt(0).dot(df.columns).mask(lambda x: x.eq(''), None)
print (desiredOutput)
0    None
1       A
2    None
3       C
4       B
dtype: object

---

desiredOutput = df.lt(0).dot(df.columns).replace('',np.nan)
print (desiredOutput)
0    NaN
1      A
2    NaN
3      C
4      B
dtype: object

With apply it is possible, but slowier if filter columns by mask per rows:

desiredOutput = df.lt(0).apply(lambda x: next(iter(x.index[x]), None), axis=1)
print (desiredOutput)
0    None
1       A
2    None
3       C
4       B
dtype: object