Sklearn：使用precision_score 和recall_score 计算时召回率和精度是否交换？

Question

I'm using scikit-learn (version 0.22.1) for a machine learning application.

I'm using a Random Forest algorithm and I have some problems in evaluating the performance of the algorithm using precision and recall. I have the labels of my test set (Y_test) and the labels predicted using the Random Forest algorithm (Y_pred). Both data contains two labels (1 and 0)

In detail, I have this matrix:

print(confusion_matrix(y_true=Y_test, y_pred=Y_pred, labels=[1,0]))

[[78 20]
 [36 41]]

Consequently:

True Positive (tp) =  78
False Negative (fn) =  36
False Positive (fp) =  20

So:

PRECISION =  tp/(tp+fn) = 78/(78+36) = 0.7959183673469388
RECALL =  = tp/(tp+fp) = 78/(78+20) 0.6842105263157895

However, using this code:

precision = precision_score(Y_test, Y_pred, pos_label=1)
recall = recall_score(y_true=Y_test, y_pred=Y_pred, pos_label=1)

print("precision: ",precision)
print("recall: ",recall)

I get the following output:

recall:  0.7959183673469388
precision:  0.6842105263157895

It seems that the values are swapped when they are computed using the standard sklearn functions. Did I do something wrong? Please, can you give me some advice?

Thanks,

Daniele

Answer 1

You are currently calculating those values wrong. The correct calculations are;

Precision Calculation:

precision = tp/(tp+fp)

Recall Calculation:

recall = tp/(tp+fn)

Reference: https://developers.google.com/machine-learning/crash-course/classification/precision-and-recall

Answer 2

Formula is incorrect:

Precision: tp / tp+fp
Recall : tp/tp+fn