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在没有sklearn的情况下从数据构建混淆矩阵

[英]Constructing a confusion matrix from data without sklearn

I am trying to construct a confusion matrix without using the sklearn library.我试图在不使用 sklearn 库的情况下构建混淆矩阵。 I am having trouble correctly forming the confusion matrix.我无法正确形成混淆矩阵。 Here's my code:这是我的代码:

def comp_confmat():
    currentDataClass = [1,3,3,2,5,5,3,2,1,4,3,2,1,1,2]    
    predictedClass = [1,2,3,4,2,3,3,2,1,2,3,1,5,1,1]
    cm = []
    classes = int(max(currentDataClass) - min(currentDataClass)) + 1 #find number of classes

    for c1 in range(1,classes+1):#for every true class
        counts = []
        for c2 in range(1,classes+1):#for every predicted class
            count = 0
            for p in range(len(currentDataClass)):
                if currentDataClass[p] == predictedClass[p]:
                    count += 1
            counts.append(count)
        cm.append(counts)
    print(np.reshape(cm,(classes,classes)))

However this returns:然而,这返回:

[[7 7 7 7 7]
[7 7 7 7 7]
[7 7 7 7 7]
[7 7 7 7 7]
[7 7 7 7 7]]

But I don't understand why each iteration results in 7 when I am reseting the count each time and it's looping through different values?但是我不明白为什么每次重置计数时每次迭代都会产生 7 并且循环遍历不同的值?

This is what I should be getting (using the sklearn's confusion_matrix function):这是我应该得到的(使用 sklearn 的 confusion_matrix 函数):

[[3 0 0 0 1]
[2 1 0 1 0]
[0 1 3 0 0]
[0 1 0 0 0]
[0 1 1 0 0]]

You can derive the confusion matrix by counting the number of instances in each combination of actual and predicted classes as follows:您可以通过计算实际类和预测类的每个组合中的实例数来导出混淆矩阵,如下所示:

import numpy as np

def comp_confmat(actual, predicted):

    # extract the different classes
    classes = np.unique(actual)

    # initialize the confusion matrix
    confmat = np.zeros((len(classes), len(classes)))

    # loop across the different combinations of actual / predicted classes
    for i in range(len(classes)):
        for j in range(len(classes)):

           # count the number of instances in each combination of actual / predicted classes
           confmat[i, j] = np.sum((actual == classes[i]) & (predicted == classes[j]))

    return confmat

# sample data
actual = [1, 3, 3, 2, 5, 5, 3, 2, 1, 4, 3, 2, 1, 1, 2]
predicted = [1, 2, 3, 4, 2, 3, 3, 2, 1, 2, 3, 1, 5, 1, 1]

# confusion matrix
print(comp_confmat(actual, predicted))
# [[3. 0. 0. 0. 1.]
#  [2. 1. 0. 1. 0.]
#  [0. 1. 3. 0. 0.]
#  [0. 1. 0. 0. 0.]
#  [0. 1. 1. 0. 0.]]

In your innermost loop, there should be a case distinction: Currently this loop counts agreement, but you only want that if actually c1 == c2 .在您的最内层循环中,应该区分大小写:目前此循环计算协议,但您只需要c1 == c2

Here's another way, using nested list comprehensions:这是另一种方式,使用嵌套列表理解:

currentDataClass = [1,3,3,2,5,5,3,2,1,4,3,2,1,1,2]    
predictedClass = [1,2,3,4,2,3,3,2,1,2,3,1,5,1,1]

classes = int(max(currentDataClass) - min(currentDataClass)) + 1 #find number of classes

counts = [[sum([(currentDataClass[i] == true_class) and (predictedClass[i] == pred_class) 
                for i in range(len(currentDataClass))])
           for pred_class in range(1, classes + 1)] 
           for true_class in range(1, classes + 1)]
counts    
[[3, 0, 0, 0, 1],
 [2, 1, 0, 1, 0],
 [0, 1, 3, 0, 0],
 [0, 1, 0, 0, 0],
 [0, 1, 1, 0, 0]]

Here is my solution using numpy and pandas:这是我使用 numpy 和 pandas 的解决方案:

import numpy as np
import pandas as pd

true_classes = [1, 3, 3, 2, 5, 5, 3, 2, 1, 4, 3, 2, 1, 1, 2]
predicted_classes = [1, 2, 3, 4, 2, 3, 3, 2, 1, 2, 3, 1, 5, 1, 1]

classes = set(true_classes)
number_of_classes = len(classes)

conf_matrix = pd.DataFrame(
    np.zeros((number_of_classes, number_of_classes),dtype=int),
    index=classes,
    columns=classes)

for true_label, prediction in zip(true_classes ,predicted_classes):
    # Each pair of (true_label, prediction) is a position in the confusion matrix (row, column)
    # Basically here we are counting how many times we have each pair.
    # The counting will be placed at the matrix index (true_label/row, prediction/column)
 
    conf_matrix.loc[true_label, prediction] += 1

print(conf_matrix.values)
[[3 0 0 0 1]
 [2 1 0 1 0]
 [0 1 3 0 0]
 [0 1 0 0 0]
 [0 1 1 0 0]]

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