Question

我离开了此example以使用scikit-learn创建分类器图像。

虽然每个图像都属于一个类别，但一切都有效，但每个图像可能属于几个类别，例如：白天狗的照片，夜间猫的照片，夜间猫狗照片等...... 我写道：

target=[[0,1],[0,2],[1,2],[0,2,3]]
target = MultiLabelBinarizer().fit_transform(target)

classifier = svm.SVC(gamma=0.001)
classifier.fit(data, target)

但是我收到了这个错误：

Traceback (most recent call last):
  File "test.py", line 49, in <module>
    classifier.fit(data, target)
  File "/home/mezzo/.local/lib/python2.7/site-packages/sklearn/svm/base.py", line 151, in fit
    y = self._validate_targets(y)
  File "/home/mezzo/.local/lib/python2.7/site-packages/sklearn/svm/base.py", line 514, in _validate_targets
    y_ = column_or_1d(y, warn=True)
  File "/home/mezzo/.local/lib/python2.7/site-packages/sklearn/utils/validation.py", line 551, in column_or_1d
    raise ValueError("bad input shape {0}".format(shape))
ValueError: bad input shape (4, 4)

完整代码

import numpy as np
import PIL
from PIL import Image
import matplotlib.image as mpimg

# The digits dataset
digits = datasets.load_digits()

def normalize(old_im):
    base = 400

    if (old_im.size[0] > old_im.size[1]):
        wpercent = (base/float(old_im.size[0]))
        hsize = int((float(old_im.size[1])*float(wpercent)))
        old_im = old_im.resize((base,hsize), PIL.Image.ANTIALIAS)
    else:
        wpercent = (base/float(old_im.size[1]))
        wsize = int((float(old_im.size[0])*float(wpercent)))
        old_im = old_im.resize((wsize, base), PIL.Image.ANTIALIAS)

    old_size = old_im.size

    new_size = (base, base)
    new_im = Image.new("RGB", new_size)
    new_im.paste(old_im, ((new_size[0]-old_size[0])/2,
                          (new_size[1]-old_size[1])/2))

    #new_im.show()
    new_im.save('prov.jpg')
    return mpimg.imread('prov.jpg')

# To apply a classifier on this data, we need to flatten the image, to
# turn the data in a (samples, feature) matrix:
imgs = np.array([normalize(Image.open('/home/mezzo/Immagini/1.jpg')),normalize(Image.open('/home/mezzo/Immagini/2.jpg')),normalize(Image.open('/home/mezzo/Immagini/3.jpg')),normalize(Image.open('/home/mezzo/Immagini/4.jpg'))])
n_samples = len(imgs)
data = imgs.reshape((n_samples, -1))

target=[[0,1],[0,2],[1,2],[0,2,3]]
target = MultiLabelBinarizer().fit_transform(target)

# Create a classifier: a support vector classifier
classifier = svm.SVC(gamma=0.001)

# We learn the digits on the first half of the digits
classifier.fit(data, target)

# Now predict the value of the digit on the second half:
predicted = classifier.predict(data)

print("Classification report for classifier %s:\n%s\n"
      % (classifier, metrics.classification_report(target, predicted)))
print("Confusion matrix:\n%s" % metrics.confusion_matrix(target, predicted))

Answer 1

Scikit-learn的SVM实现本身不支持多标签分类，although it has various other classifiers that do：

支持多标记：Decision Trees，Random Forests，Nearest Neighbors，Ridge Regression。

通过将每个独特的标签组合视为一个单独的类，也可以使用SVM进行多标签分类。您只需使用一个整数标签which can be done efficiently using np.unique替换target矩阵中的每个唯一行：

d = np.dtype((np.void, target.dtype.itemsize * target.shape[1]))
_, ulabels = np.unique(np.ascontiguousarray(target).view(d), return_inverse=True)

然后，您可以像处理单标签分类问题一样训练SVM：

clf = svm.SVC()
clf.fit(data, ulabels)

一个潜在的警告是，如果您没有大量的训练样例，那么对于罕见的标签组合，您的分类器的表现可能会很差。

Answer 2

这是因为你的目标是：

array([[1, 1, 0, 0],
       [1, 0, 1, 0],
       [0, 1, 1, 0],
       [1, 0, 1, 1]])

您的目标必须是形状（m，），其中m是示例数。解决这个问题的一种方法是将二进制数组转换为标签，如：

for item in target:
    print(sum(1<<i for i, b in enumerate(item) if b))

输出结果如下：

现在您可以使用[3,5,6,13]作为目标。

Scikit - 学习多个目标

2 个答案: