Is F1 micro the same as Accuracy? - machine-learning

Question

I have tried many examples with F1 micro and Accuracy in scikit-learn and in all of them, I see that F1 micro is the same as Accuracy. Is this always true?
Script
from sklearn import svm
from sklearn import metrics
from sklearn.cross_validation import train_test_split
from sklearn.datasets import load_iris
from sklearn.metrics import f1_score, accuracy_score
# prepare dataset
iris = load_iris()
X = iris.data[:, :2]
y = iris.target
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
# svm classification
clf = svm.SVC(kernel='rbf', gamma=0.7, C = 1.0).fit(X_train, y_train)
y_predicted = clf.predict(X_test)
# performance
print "Classification report for %s" % clf
print metrics.classification_report(y_test, y_predicted)
print("F1 micro: %1.4f\n" % f1_score(y_test, y_predicted, average='micro'))
print("F1 macro: %1.4f\n" % f1_score(y_test, y_predicted, average='macro'))
print("F1 weighted: %1.4f\n" % f1_score(y_test, y_predicted, average='weighted'))
print("Accuracy: %1.4f" % (accuracy_score(y_test, y_predicted)))
Output
Classification report for SVC(C=1.0, cache_size=200, class_weight=None, coef0=0.0,
decision_function_shape=None, degree=3, gamma=0.7, kernel='rbf',
max_iter=-1, probability=False, random_state=None, shrinking=True,
tol=0.001, verbose=False)
precision recall f1-score support
0 1.00 0.90 0.95 10
1 0.50 0.88 0.64 8
2 0.86 0.50 0.63 12
avg / total 0.81 0.73 0.74 30
F1 micro: 0.7333
F1 macro: 0.7384
F1 weighted: 0.7381
Accuracy: 0.7333
F1 micro = Accuracy

Answer 1

In classification tasks for which every test case is guaranteed to be assigned to exactly one class, micro-F is equivalent to accuracy. It won't be the case in multi-label classification.

Answer 2

This is because we are dealing with a multi class classification , where every test data should belong to only 1 class and not multi label , in such case where there is no TN , we can call True Negatives as True Positives.
Formula wise ,
correction : F1 score is 2* precision* recall / (precision + recall)

Answer 3

Micoaverage precision, recall, f1 and accuracy are all equal for cases in which every instance must be classified into one (and only one) class. A simple way to see this is by looking at the formulas precision=TP/(TP+FP) and recall=TP/(TP+FN). The numerators are the same, and every FN for one class is another classes's FP, which makes the denominators the same as well. If precision = recall, then f1 will also be equal.
For any inputs should should be able to show that:
from sklearn.metrics import accuracy_score as acc
from sklearn.metrics import f1_score as f1
f1(y_true,y_pred,average='micro')=acc(y_true,y_pred)

Answer 4

I had the same issue so I investigated and came up with this:
Just thinking about the theory, it is impossible that accuracy and the f1-score are the very same for every single dataset. The reason for this is that the f1-score is independent from the true-negatives while accuracy is not.
By taking a dataset where f1 = acc and adding true negatives to it, you get f1 != acc.
>>> from sklearn.metrics import accuracy_score as acc
>>> from sklearn.metrics import f1_score as f1
>>> y_pred = [0, 1, 1, 0, 1, 0]
>>> y_true = [0, 1, 1, 0, 0, 1]
>>> acc(y_true, y_pred)
0.6666666666666666
>>> f1(y_true,y_pred)
0.6666666666666666
>>> y_true = [0, 1, 1, 0, 1, 0, 0, 0, 0]
>>> y_pred = [0, 1, 1, 0, 0, 1, 0, 0, 0]
>>> acc(y_true, y_pred)
0.7777777777777778
>>> f1(y_true,y_pred)
0.6666666666666666