当我试图绘制决策边界时出现形状错误

Question

在我的wine-dataset中，我试图在两列之间绘制一个决策边界，该边界由代码片段描述：

X0, X1 = X[:, 10], Y

我从scikit svm plot tutorial中获取了以下代码，并对其进行了修改，以替换为我的变量名/索引。但是，当我运行以下代码时，我收到一个错误消息：

^{pr2}$

错误堆栈为：

Traceback (most recent call last):

File "test-wine.py", line 120, in <module>
    cmap=plt.cm.coolwarm, alpha=0.8)
File "test-wine.py", line 96, in plot_contours
    Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])
File "/home/suhail/anaconda3/envs/ml/lib/python3.5/site-packages/sklearn/svm/base.py", line 548, in predict
    y = super(BaseSVC, self).predict(X)
File "/home/suhail/anaconda3/envs/ml/lib/python3.5/site-packages/sklearn/svm/base.py", line 308, in predict
    X = self._validate_for_predict(X)
File "/home/suhail/anaconda3/envs/ml/lib/python3.5/site-packages/sklearn/svm/base.py", line 459, in _validate_for_predict
    (n_features, self.shape_fit_[1]))
ValueError: X.shape[1] = 2 should be equal to 11, the number of features at training time

我无法理解上述错误的原因。这是我修改过的代码。在

import pandas as pd
from sklearn.svm import SVC
import matplotlib.pyplot as plt
import numpy as np

data = pd.read_csv('winequality-red.csv').values

x_data_shape = data.shape[0]
y_data_shape = data.shape[1]

X = data[:, 0:y_data_shape-1]
Y = data[:, y_data_shape-1]


############### PLOT DECISION BOUNDARY SVM #############

def make_meshgrid(x, y, h=.02):
    """Create a mesh of points to plot in

    Parameters
    ----------
    x: data to base x-axis meshgrid on
    y: data to base y-axis meshgrid on
    h: stepsize for meshgrid, optional

    Returns
    -------
    xx, yy : ndarray
    """
    x_min, x_max = x.min() - 1, x.max() + 1
    y_min, y_max = y.min() - 1, y.max() + 1
    xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
                        np.arange(y_min, y_max, h))
    return xx, yy


def plot_contours(ax, clf, xx, yy, **params):
    """Plot the decision boundaries for a classifier.

    Parameters
    ----------
    ax: matplotlib axes object
    clf: a classifier
    xx: meshgrid ndarray
    yy: meshgrid ndarray
    params: dictionary of params to pass to contourf, optional
    """
    Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])
    Z = Z.reshape(xx.shape)
    out = ax.contourf(xx, yy, Z, **params)
    return out


C = 1.0  # SVM regularization parameter
models = (SVC(kernel='linear', C=C),
        SVC(kernel='rbf', gamma=0.7, C=C),
        SVC(kernel='poly', degree=3, C=C))
models = (clf.fit(X, Y) for clf in models)

titles = ('SVC with linear kernel',
        'SVC with RBF kernel',
        'SVC with polynomial (degree 3) kernel')

fig, sub = plt.subplots(2, 2)
plt.subplots_adjust(wspace=0.4, hspace=0.4)

X0, X1 = X[:, 10], Y
xx, yy = make_meshgrid(X0, X1)

for clf, title, ax in zip(models, titles, sub.flatten()):
    plot_contours(ax, clf, xx, yy,
                cmap=plt.cm.coolwarm, alpha=0.8)
    ax.scatter(X0, X1, c=Y, cmap=plt.cm.coolwarm, s=20, edgecolors='k')
    ax.set_xlim(xx.min(), xx.max())
    ax.set_ylim(yy.min(), yy.max())
    ax.set_xlabel('Alcohol Content')
    ax.set_ylabel('Quality')
    ax.set_xticks(())
    ax.set_yticks(())
    ax.set_title(title)

plt.show()

这个错误的原因是什么？在

Answer 1

你训练了所有11个特征的分类器， {cd2>中的{2>只在cd2>中调用的方法。 要评估使用11个特性训练的分类器，您需要提供所有11个特性。这是您的错误消息所指示的。在

因此，为了使代码片段适合您，您应该将自己限制为两个特性（否则，绘制二维决策边界无论如何都没有意义），例如使用

X = data[:, :2]
Y = data[:, y_data_shape-1]

读取数据时。在

请注意，您提到的example也只使用了两个功能：

^{pr2}$