[FIXED] How is scikit-learn cross_val_predict accuracy score calculated?

Issue

Does the cross_val_predict (see doc, v0.18) with k-fold method as shown in the code below calculate accuracy for each fold and average them finally or not?

cv = KFold(len(labels), n_folds=20)
clf = SVC()
ypred = cross_val_predict(clf, td, labels, cv=cv)
accuracy = accuracy_score(labels, ypred)
print accuracy

Solution

No, it does not!

According to cross validation doc page, cross_val_predict does not return any scores but only the labels based on a certain strategy which is described here:

The function cross_val_predict has a similar interface to cross_val_score, but returns, for each element in the input, the prediction that was obtained for that element when it was in the test set. Only cross-validation strategies that assign all elements to a test set exactly once can be used (otherwise, an exception is raised).

And therefore by calling accuracy_score(labels, ypred) you are just calculating accuracy scores of labels predicted by aforementioned particular strategy compared to the true labels. This again is specified in the same documentation page:

These prediction can then be used to evaluate the classifier:

predicted = cross_val_predict(clf, iris.data, iris.target, cv=10) 
metrics.accuracy_score(iris.target, predicted)

Note that the result of this computation may be slightly different from those obtained using cross_val_score as the elements are grouped in different ways.

If you need accuracy scores of different folds you should try:

>>> scores = cross_val_score(clf, X, y, cv=cv)
>>> scores                                              
array([ 0.96...,  1.  ...,  0.96...,  0.96...,  1.        ])

and then for the mean accuracy of all folds use scores.mean():

>>> print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2))
Accuracy: 0.98 (+/- 0.03)

---

How to calculate Cohen kappa coefficient and confusion matrix for each fold?

For calculating Cohen Kappa coefficient and confusion matrix I assumed you mean kappa coefficient and confusion matrix between true labels and each fold's predicted labels:

from sklearn.model_selection import KFold
from sklearn.svm.classes import SVC
from sklearn.metrics.classification import cohen_kappa_score
from sklearn.metrics import confusion_matrix

cv = KFold(len(labels), n_folds=20)
clf = SVC()
for train_index, test_index in cv.split(X):
    clf.fit(X[train_index], labels[train_index])
    ypred = clf.predict(X[test_index])
    kappa_score = cohen_kappa_score(labels[test_index], ypred)
    confusion_matrix = confusion_matrix(labels[test_index], ypred)

---

What does cross_val_predict return?

It uses KFold to split the data to k parts and then for i=1..k iterations:

• takes i'th part as the test data and all other parts as training data
• trains the model with training data (all parts except i'th)
• then by using this trained model, predicts labels for i'th part (test data)

In each iteration, label of i'th part of data gets predicted. In the end cross_val_predict merges all partially predicted labels and returns them as the final result.

This code shows this process step by step:

X = np.array([[0], [1], [2], [3], [4], [5]])
labels = np.array(['a', 'a', 'a', 'b', 'b', 'b'])

cv = KFold(len(labels), n_folds=3)
clf = SVC()
ypred_all = np.chararray((labels.shape))
i = 1
for train_index, test_index in cv.split(X):
    print("iteration", i, ":")
    print("train indices:", train_index)
    print("train data:", X[train_index])
    print("test indices:", test_index)
    print("test data:", X[test_index])
    clf.fit(X[train_index], labels[train_index])
    ypred = clf.predict(X[test_index])
    print("predicted labels for data of indices", test_index, "are:", ypred)
    ypred_all[test_index] = ypred
    print("merged predicted labels:", ypred_all)
    i = i+1
    print("=====================================")
y_cross_val_predict = cross_val_predict(clf, X, labels, cv=cv)
print("predicted labels by cross_val_predict:", y_cross_val_predict)

The result is:

iteration 1 :
train indices: [2 3 4 5]
train data: [[2] [3] [4] [5]]
test indices: [0 1]
test data: [[0] [1]]
predicted labels for data of indices [0 1] are: ['b' 'b']
merged predicted labels: ['b' 'b' '' '' '' '']
=====================================
iteration 2 :
train indices: [0 1 4 5]
train data: [[0] [1] [4] [5]]
test indices: [2 3]
test data: [[2] [3]]
predicted labels for data of indices [2 3] are: ['a' 'b']
merged predicted labels: ['b' 'b' 'a' 'b' '' '']
=====================================
iteration 3 :
train indices: [0 1 2 3]
train data: [[0] [1] [2] [3]]
test indices: [4 5]
test data: [[4] [5]]
predicted labels for data of indices [4 5] are: ['a' 'a']
merged predicted labels: ['b' 'b' 'a' 'b' 'a' 'a']
=====================================
predicted labels by cross_val_predict: ['b' 'b' 'a' 'b' 'a' 'a']

Answered By - Omid