为逻辑回归生成学习曲线

Question

我有一个包含 260 个显微图像的数据集。我想为逻辑回归算法生成学习曲线。但我收到此错误“'模块'对象不可迭代”。也许我不了解一些基本的东西，因为我是刚学Python的初学者

from sklearn.cross_validation import train_test_split
from imutils import paths
from scipy import misc
import numpy as np
import argparse
import imutils
import cv2
import os
from matplotlib import pyplot as plt
from sklearn.model_selection import learning_curve
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import recall_score
from sklearn.metrics import precision_score
from sklearn.model_selection import cross_val_score





def plot_learning_curve(estimator, title, X, y, ylim=None, cv=None,
                        n_jobs=None, train_sizes=np.linspace(50, 80, 110)):
    """
    Generate a simple plot of the test and training learning curve.

    Parameters
    ----------
    estimator : object type that implements the "fit" and "predict" methods
        An object of that type which is cloned for each validation.

    title : string
        Title for the chart.

    X : array-like, shape (n_samples, n_features)
        Training vector, where n_samples is the number of samples and
        n_features is the number of features.

    y : array-like, shape (n_samples) or (n_samples, n_features), optional
        Target relative to X for classification or regression;
        None for unsupervised learning.


    cv : int, cross-validation generator or an iterable, optional
        Determines the cross-validation splitting strategy.
        Possible inputs for cv are:
          - None, to use the default 3-fold cross-validation,
          - integer, to specify the number of folds.
          - :term:`CV splitter`,
          - An iterable yielding (train, test) splits as arrays of indices.

        For integer/None inputs, if ``y`` is binary or multiclass,
        :class:`StratifiedKFold` used. If the estimator is not a classifier
        or if ``y`` is neither binary nor multiclass, :class:`KFold` is used.

        Refer :ref:`User Guide <cross_validation>` for the various
        cross-validators that can be used here.

    n_jobs : int or None, optional (default=None)
        Number of jobs to run in parallel.
        ``None`` means 1 unless in a :obj:`joblib.parallel_backend` context.
        ``-1`` means using all processors. See :term:`Glossary <n_jobs>`
        for more details.

    train_sizes : array-like, shape (n_ticks,), dtype float or int
        Relative or absolute numbers of training examples that will be used to
        generate the learning curve. If the dtype is float, it is regarded as a
        fraction of the maximum size of the training set (that is determined
        by the selected validation method), i.e. it has to be within (0, 1].
        Otherwise it is interpreted as absolute sizes of the training sets.
        Note that for classification the number of samples usually have to
        be big enough to contain at least one sample from each class.
        (default: np.linspace(0.1, 1.0, 5))
    """
    plt.figure()
    plt.title(title)
    if ylim is not None:
        plt.ylim(*ylim)
    plt.xlabel("Training examples")
    plt.ylabel("Score")
    train_sizes, train_scores, test_scores = learning_curve(
        estimator, X, y, cv=cv, n_jobs=n_jobs, train_sizes=train_sizes)
    train_scores_mean = np.mean(train_scores, axis=1)
    train_scores_std = np.std(train_scores, axis=1)
    test_scores_mean = np.mean(test_scores, axis=1)
    test_scores_std = np.std(test_scores, axis=1)
    plt.grid()

    plt.fill_between(train_sizes, train_scores_mean - train_scores_std,
                     train_scores_mean + train_scores_std, alpha=0.1,
                     color="r")
    plt.fill_between(train_sizes, test_scores_mean - test_scores_std,
                     test_scores_mean + test_scores_std, alpha=0.1, color="g")
    plt.plot(train_sizes, train_scores_mean, 'o-', color="r",
             label="Training score")
    plt.plot(train_sizes, test_scores_mean, 'o-', color="g",
             label="Cross-validation score")

    plt.legend(loc="best")
    return plt






#training  with logistic regression
clfLR = LogisticRegression(random_state=0, solver='lbfgs')
clfLR.fit(trainFeat,trainLabels)
acc = clfLR.score(testFeat, testLabels)
print("accuracy of Logistic regression ",acc)

只有当我想绘制曲线时我才会遇到这个问题。其余代码工作正常。

#plotting the curve
estimator =LogisticRegression()
train_sizes, train_scores, valid_scores = plot_learning_curve(
estimator,'logistic learning curve ', trainFeat, trainLabels, cv=5, n_jobs=4,train_sizes=[50, 80, 110])
print(train_sizes)
plt.show()

错误截图

learning curve

Answer 1

尝试在 Jupyter 在线 IDE IDE 上运行代码。如果您将“%matplotlib”行添加到导入部分，它会自动绘图。

如果您想继续使用此 IDE，请分享您的错误消息，以便我可以帮助您。您可能缺少其中一个导入，或者它可能是 Python2/3 问题。