Python: count TP, FP, FN и TN

Question

I have dataframe with true class and class, that were predicted by some algorithm.

     true  pred
0       1     0
1       1     1
2       1     1
3       0     0
4       1     1

I try to use

def classification(y_actual, y_hat):
    TP = 0
    FP = 0
    TN = 0
    FN = 0

    for i in range(len(y_hat)): 
        if y_actual[i] == y_hat[i] == 1:
           TP += 1
    for i in range(len(y_hat)): 
        if y_actual[i] == 1 and y_actual != y_hat[i]:
           FP += 1
    for i in range(len(y_hat)): 
        if y_actual[i] == y_hat[i] == 0:
           TN += 1
    for i in range(len(y_hat)): 
        if y_actual[i] == 0 and y_actual != y_hat[i]:
           FN += 1

    return(TP, FP, TN, FN)

but it return me

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all(). How can I fix that or maybe there are the better decision?

Answer 1

The error message happens because Python tries to convert an array to a boolean and fails.

That's because you're comparing y_actual with y_hat[i] .

It should be y_actual[i] != y_hat[i] (2 times in the code)

(I realize that it's just a typo, but the message is cryptic enough for the problem to become interesting)

While we're at it, you could make a more efficient routine by merging all your counters in a sole loop and using enumerate to avoid at least one access by index:

def classification(y_actual, y_hat):
    TP = 0
    FP = 0
    TN = 0
    FN = 0

    for i,yh in enumerate(y_hat): 
        if y_actual[i] == yh == 1:
           TP += 1
        if y_actual[i] == 1 and y_actual[i] != yh:
           FP += 1
        if y_actual[i] == yh == 0:
           TN += 1
        if y_actual[i] == 0 and y_actual[i] != yh:
           FN += 1

    return(TP, FP, TN, FN)

you see that this way it can be even be simplified even more , cutting a lot through tests and branches:

def classification(y_actual, y_hat):
    TP = 0
    FP = 0
    TN = 0
    FN = 0

    for i,yh in enumerate(y_hat): 
        if y_actual[i] == yh:
           if yh == 1:
               TP += 1
           elif yh == 0:
               TN += 1
        else: # y_actual[i] != yh
           if y_actual[i] == 1 and :
              FP += 1
           elif y_actual[i] == 0:
              FN += 1

    return(TP, FP, TN, FN)

Answer 2

我使用sklearn.metrics confusion_matrix sklearn.metrics ，它返回我需要的矩阵。