相同功能的不同输出

问题描述 投票:0回答:1

我在python中实现了KNN算法。

import math

            #height,width,deepth,thickness,Label
data_set = [(2,9,8,4, "Good"),
            (3,7,7,9, "Bad"),
            (10,3,10,3, "Good"),
            (2,9,6,10, "Good"),
            (3,3,2,5, "Bad"),
            (2,8,5,6, "Bad"),
            (7,2,3,10, "Good"),
            (1,10,8,10, "Bad"),
            (2,8,1,10, "Good")
            ]


A = (3,2,1,5)
B = (8,3,1,2)
C = (6,10,8,3)
D = (9,6,4,1)


distances = []
labels = []

def calc_distance(datas,test):
    for data in datas:
        distances.append(
            ( round(math.sqrt(((data[0] - test[0])**2 + (data[1] - test[1])**2 + (data[2] - test[2])**2 + (data[3] - test[3])**2)), 3), data[4] )) 
    return distances

def most_frequent(list1): 
    return max(set(list1), key = list1.count) 

def get_neibours(k):
    distances.sort()
    print(distances[:k])
    for distance in distances[:k]:
        labels.append(distance[1])
    print("It can be classified as: ", end="")
    print(most_frequent(labels))



calc_distance(data_set,D)
get_neibours(7)

calc_distance(data_set,D)
get_neibours(7)

我在大多数情况下都工作良好,并且得到了正确的标签。例如对于D,我确实得到了标签“ Good”。但是我发现了一个错误,例如,当我两次调用它时:

 calc_distance(data_set,D)
get_neibours(7)

calc_distance(data_set,D)
get_neibours(7)

并且我运行了几次,当我运行几次程序时,得到了不同的输出-“好”和“差”。[enter image description here

我无法找到某个地方一定有一个错误。

python machine-learning data-science knn
1个回答
1
投票

问题是您使用相同的距离和标签,对k个第一个元素进行排序和获取。在函数内部创建列表并返回。检查下面的修改。

import math

data_set = [
    (2,9,8,4, "Good"),
    (3,7,7,9, "Bad"),
    (10,3,10,3, "Good"),
    (2,9,6,10, "Good"),
    (3,3,2,5, "Bad"),
    (2,8,5,6, "Bad"),
    (7,2,3,10, "Good"),
    (1,10,8,10, "Bad"),
    (2,8,1,10, "Good"),
]

A = (3,2,1,5)
B = (8,3,1,2)
C = (6,10,8,3)
D = (9,6,4,1)

def calc_distance(datas, test):
    distances = []
    for data in datas:
        distances.append(
            ( round(math.sqrt(((data[0] - test[0])**2 + (data[1] - test[1])**2 + (data[2] - test[2])**2 + (data[3] - test[3])**2)), 3), data[4] ))
    return distances

def most_frequent(list1):
    return max(set(list1), key = list1.count)

def get_neibours(distances, k):
    labels = []
    distances.sort()
    print(distances[:k])
    for distance in distances[:k]:
        labels.append(distance[1])
    print("It can be classified as: ", end="")
    print(most_frequent(labels))

distances = calc_distance(data_set,D)
get_neibours(distances, 7)

distances = calc_distance(data_set,D)
get_neibours(distances, 7) 

[((7.071,'良好'),(8.062,'不良'),(8.888,'不良'),(9.11,'良好'),(10.1,'良好'),(10.488,'不良') ,(11.958,'Good')]可以分类为:良好

[((7.071,'良好'),(8.062,'不良'),(8.888,'不良'),(9.11,'良好'),(10.1,'良好'),(10.488,'不良') ,(11.958,'Good')]可以分类为:良好

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