classification_metrics.py 文件源码

def f1_metric(solution, prediction, task=BINARY_CLASSIFICATION):
    """
    Compute the normalized f1 measure.

    The binarization differs
    for the multi-label and multi-class case.
    A non-weighted average over classes is taken.
    The score is normalized.
    :param solution:
    :param prediction:
    :param task:
    :return:
    """

    label_num = solution.shape[1]
    score = np.zeros(label_num)
    bin_prediction = binarize_predictions(prediction, task)
    [tn, fp, tp, fn] = acc_stat(solution, bin_prediction)
    # Bounding to avoid division by 0
    eps = 1e-15
    true_pos_num = sp.maximum(eps, tp + fn)
    found_pos_num = sp.maximum(eps, tp + fp)
    tp = sp.maximum(eps, tp)
    tpr = tp / true_pos_num  # true positive rate (recall)
    ppv = tp / found_pos_num  # positive predictive value (precision)
    arithmetic_mean = 0.5 * sp.maximum(eps, tpr + ppv)
    # Harmonic mean:
    f1 = tpr * ppv / arithmetic_mean
    # Average over all classes
    f1 = np.mean(f1)
    # Normalize: 0 for random, 1 for perfect
    if (task != MULTICLASS_CLASSIFICATION) or (label_num == 1):
        # How to choose the "base_f1"?
        # For the binary/multilabel classification case, one may want to predict all 1.
        # In that case tpr = 1 and ppv = frac_pos. f1 = 2 * frac_pos / (1+frac_pos)
        #     frac_pos = mvmean(solution.ravel())
        #     base_f1 = 2 * frac_pos / (1+frac_pos)
        # or predict random values with probability 0.5, in which case
        #     base_f1 = 0.5
        # the first solution is better only if frac_pos > 1/3.
        # The solution in which we predict according to the class prior frac_pos gives
        # f1 = tpr = ppv = frac_pos, which is worse than 0.5 if frac_pos<0.5
        # So, because the f1 score is used if frac_pos is small (typically <0.1)
        # the best is to assume that base_f1=0.5
        base_f1 = 0.5
    # For the multiclass case, this is not possible (though it does not make much sense to
    # use f1 for multiclass problems), so the best would be to assign values at random to get
    # tpr=ppv=frac_pos, where frac_pos=1/label_num
    else:
        base_f1 = 1. / label_num
    score = (f1 - base_f1) / sp.maximum(eps, (1 - base_f1))
    return score