Python

Question

我正在寻找一种方法，可以让我在比较 2 个列表时检查缺少哪些元素。很像这个线程，但我想用 NumPy Python.

来写

Complement of Two Lists?

import numpy as np

numbers = np.array([1,2,3,4,5,6,7,8,9])
A = np.array([2,5,6,9])

def listComplementElements(list1, list2):

    storeResults = []

    for i in list1:
        for j in list2:

            if i != j:
                #write to storeResults if 2 numbers are different
                storeResults.append(i)
            else:
                #if numebrs are equal break out of the loop
                break

    return storeResults            

result = listComplementElements(numbers, A)
print(result) #expected result [1,3,4,7,8]

目前输出如下所示：[1, 1, 1, 1, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 6, 7, 7, 7 , 7, 8, 8, 8, 8, 9, 9, 9]

我做错了什么？

Answer 1

这将为您提供两个数组的补码：

list(set(numbers) - set(A))

[1, 3, 4, 7, 8]

如果您有想要保留的重复值，可以使用以下方法：

from collections import Counter

c1 = Counter(numbers)
c2 = Counter(A)

diff = c1 - c2
print(list(diff.elements()))

Answer 2

这就是您想要的：

import numpy as np

numbers = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
A = np.array([2, 5, 6, 9])

def listComplementElements(list1, list2):
    storeResults = []

    for num in list1:
        if num not in list2: # this will essentially iterate your list behind the scenes
            storeResults.append(num)

    return storeResults

result = listComplementElements(numbers, A)
print(result) #expected result [1,3,4,7,8]

打印：

[1, 3, 4, 7, 8]

你应该只有一个循环，然后说if not in来评估与第二个列表的关系。否则你会不必要地循环，就像你看到的那样，它会多次输出结果！此外，使用像 is in 或 not in 这样的运算符可以使您的代码比试图理解许多嵌套循环更具可读性:)

Python - 查找 2 个列表的补集

Python - finding complement of 2 lists

compare

numpy

list

complement