我正在使用混淆矩阵来检查我的分类器的性能。
我正在使用 Scikit-Learn，我有点困惑。我如何解释结果
from sklearn.metrics import confusion_matrix
>>> y_true = [2, 0, 2, 2, 0, 1]
>>> y_pred = [0, 0, 2, 2, 0, 2]
>>> confusion_matrix(y_true, y_pred)
array([[2, 0, 0],
[0, 0, 1],
[1, 0, 2]])
我怎样才能决定这个预测值是好还是不好。
混淆矩阵是一种将错误分类数量制成表格的方法，即根据真实类别最终进入错误分类箱的预测类别数量。
虽然 sklearn.metrics.confusion_matrix 提供了一个数字矩阵，但我发现使用以下内容生成“报告”更有用：
import pandas as pd
y_true = pd.Series([2, 0, 2, 2, 0, 1, 1, 2, 2, 0, 1, 2])
y_pred = pd.Series([0, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 2])
pd.crosstab(y_true, y_pred, rownames=['True'], colnames=['Predicted'], margins=True)
这导致：
Predicted 0 1 2 All
True
0 3 0 0 3
1 0 1 2 3
2 2 1 3 6
All 5 2 5 12
这让我们看到：
y_true和中每个类别的分类总数y_pred，来自“全部”小计此方法也适用于文本标签，并且对于数据集中的大量样本可以扩展以提供百分比报告。
import numpy as np
import pandas as pd
# create some data
lookup = {0: 'biscuit', 1:'candy', 2:'chocolate', 3:'praline', 4:'cake', 5:'shortbread'}
y_true = pd.Series([lookup[_] for _ in np.random.random_integers(0, 5, size=100)])
y_pred = pd.Series([lookup[_] for _ in np.random.random_integers(0, 5, size=100)])
pd.crosstab(y_true, y_pred, rownames=['True'], colnames=['Predicted']).apply(lambda r: 100.0 * r/r.sum())
那么输出是：
Predicted biscuit cake candy chocolate praline shortbread
True
biscuit 23.529412 10 23.076923 13.333333 15.384615 9.090909
cake 17.647059 20 0.000000 26.666667 15.384615 18.181818
candy 11.764706 20 23.076923 13.333333 23.076923 31.818182
chocolate 11.764706 5 15.384615 6.666667 15.384615 13.636364
praline 17.647059 10 30.769231 20.000000 0.000000 13.636364
shortbread 17.647059 35 7.692308 20.000000 30.769231 13.636364
现在的数字代表分类结果的百分比（而不是病例数）。
尽管请注意，sklearn.metrics.confusion_matrix可以使用以下命令直接可视化输出：
import matplotlib.pyplot as plt
conf = sklearn.metrics.confusion_matrix(y_true, y_pred)
plt.imshow(conf, cmap='binary', interpolation='None')
plt.show()
在 y 轴上混淆矩阵具有实际值，在 x 轴上具有预测器给出的值。因此，对角线上的计数是正确预测的数量。对角线的元素是不正确的预测。
在你的情况下：
>>> confusion_matrix(y_true, y_pred)
array([[2, 0, 0], # two zeros were predicted as zeros
[0, 0, 1], # one 1 was predicted as 2
[1, 0, 2]]) # two 2s were predicted as 2, and one 2 was 0
我想以图形方式指定理解这一点的必要性。这是一个简单的矩阵，在得出结论之前需要很好地理解。因此，这是上述答案的简化可解释版本。
0 1 2 <- Predicted 0 [2, 0, 0] TRUE 1 [0, 0, 1] 2 [1, 0, 2] # At 0,0: True value was 0, Predicted value was 0, - 2 times predicted # At 1,1: True value was 1, Predicted value was 1, - 0 times predicted # At 2,2: True value was 2, Predicted value was 2, - 2 times predicted # At 1,2: True value was 1, Predicted value was 2, - 1 time predicted # At 2,0: True value was 2, Predicted value was 0, - 1 time predicted... ...Like that
而且，正如我的朋友@fu DL 所问，代码如下：
from sklearn.metrics import confusion_matrix
Y_true = [0,0,0,1,1,1,2,2,0,1,2]
Y_pred = [0,0,1,1,1,2,2,2,0,0,0]
confusion = confusion_matrix(Y_true, Y_pred)
# PUT YOUR DESIRED LABELS HERE...
row_label = "True"
col_label = "Predicted"
# For printing column label right in the middle
col_space = len(row_label)
index_middle = int(int(len(set(Y_true)))/2)
# Prints first row
print(" "*(col_space + 4), " ".join([str(i) for i in set(Y_true)]), " <- {}".format(col_label))
# Prints rest of the table
for index in range(len(set(Y_true))):
if index == index_middle:
print(row_label, " ", index, confusion[index])
else:
print(" "*(col_space+2), index, confusion[index])