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2019-11-29

2019-11-29

作者: 元气小地瓜 | 来源:发表于2019-11-29 22:44 被阅读0次

from sklearn.metrics import accuracy_score

help(accuracy_score)

Help on function accuracy_score in module sklearn.metrics.classification:

accuracy_score(y_true, y_pred, normalize=True, sample_weight=None)

    Accuracy classification score.

    In multilabel classification, this function computes subset accuracy:

    the set of labels predicted for a sample must *exactly* match the

    corresponding set of labels in y_true.

    Read more in the :ref:`User Guide <accuracy_score>`.

    Parameters

    ----------

    y_true : 1d array-like, or label indicator array / sparse matrix

        Ground truth (correct) labels.

    y_pred : 1d array-like, or label indicator array / sparse matrix

        Predicted labels, as returned by a classifier.

    normalize : bool, optional (default=True)

        If ``False``, return the number of correctly classified samples.

        Otherwise, return the fraction of correctly classified samples.

ormalize:bool,可选(默认值=True),如果是`False‘,则返回正确分类的样本数。否则,返回正确分类样本的分数。

    sample_weight : array-like of shape = [n_samples], optional

        Sample weights.

    Returns

    -------

    score : float

        If ``normalize == True``, return the fraction of correctly

        classified samples (float), else returns the number of correctly

        classified samples (int).

        The best performance is 1 with ``normalize == True`` and the number

        of samples with ``normalize == False``.

    See also

    --------

    jaccard_score, hamming_loss, zero_one_loss

    Notes

    -----

    In binary and multiclass classification, this function is equal

    to the ``jaccard_score`` function.

    Examples

    --------

from sklearn.metrics import accuracy_score

y_pred = [0, 2, 1, 3]

y_true = [0, 1, 2, 3]

accuracy_score(y_true, y_pred)

    0.5

accuracy_score(y_true, y_pred, normalize=False)

    2

    In the multilabel case with binary label indicators:

import numpy as np

accuracy_score(np.array([[0, 1], [1, 1]]), np.ones((2, 2)))

    0.5

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