我用了两种方法，我发现其中一种方法比另一种更有用。以下是我的回答:

我的输入数据:

crkmod_mpp = ['M13','M18','M19','M24']
testmod_mpp = ['M13','M14','M15','M16','M17','M18','M19','M20','M21','M22','M23','M24']

Method1: np。我喜欢这种方法，因为它保留了位置

test= list(np.setdiff1d(testmod_mpp,crkmod_mpp))
print(test)
['M15', 'M16', 'M22', 'M23', 'M20', 'M14', 'M17', 'M21']

Method2:虽然答案和Method1一样，但是打乱了顺序

test = list(set(testmod_mpp).difference(set(crkmod_mpp)))
print(test)
['POA23', 'POA15', 'POA17', 'POA16', 'POA22', 'POA18', 'POA24', 'POA21']

Method1 np。Setdiff1d完全符合我的要求。 这是信息的答案。

我将这些列表压缩在一起，逐个元素进行比较。

main_list = [b for a, b in zip(list1, list2) if a!= b]

当你有本地方法可用时，不确定为什么上面的解释如此复杂:

main_list = list(set(list_2)-set(list_1))

TL;博士: 解决方案(1)

import numpy as np
main_list = np.setdiff1d(list_2,list_1)
# yields the elements in `list_2` that are NOT in `list_1`

解决方案(2)你想要一个排序的列表

def setdiff_sorted(array1,array2,assume_unique=False):
    ans = np.setdiff1d(array1,array2,assume_unique).tolist()
    if assume_unique:
        return sorted(ans)
    return ans
main_list = setdiff_sorted(list_2,list_1)

解释: (1)可以使用NumPy的setdiff1d (array1,array2,assume_unique=False)。

assume_unique询问用户数组是否已经是唯一的。如果为False，则首先确定唯一元素。 如果为True，函数将假设元素已经是唯一的，并且函数将跳过确定唯一元素。

这将产生array1中不在array2中的唯一值。assume_unique默认为False。

如果你关心的是唯一元素(基于Chinny84的响应)，那么只需使用(其中assume_unique=False =>为默认值):

import numpy as np
list_1 = ["a", "b", "c", "d", "e"]
list_2 = ["a", "f", "c", "m"] 
main_list = np.setdiff1d(list_2,list_1)
# yields the elements in `list_2` that are NOT in `list_1`

（2） 对于那些想要对答案进行排序的人，我已经做了一个自定义函数:

import numpy as np
def setdiff_sorted(array1,array2,assume_unique=False):
    ans = np.setdiff1d(array1,array2,assume_unique).tolist()
    if assume_unique:
        return sorted(ans)
    return ans

要得到答案，运行:

main_list = setdiff_sorted(list_2,list_1)

SIDE NOTES: (a) Solution 2 (custom function setdiff_sorted) returns a list (compared to an array in solution 1). (b) If you aren't sure if the elements are unique, just use the default setting of NumPy's setdiff1d in both solutions A and B. What can be an example of a complication? See note (c). (c) Things will be different if either of the two lists is not unique. Say list_2 is not unique: list2 = ["a", "f", "c", "m", "m"]. Keep list1 as is: list_1 = ["a", "b", "c", "d", "e"] Setting the default value of assume_unique yields ["f", "m"] (in both solutions). HOWEVER, if you set assume_unique=True, both solutions give ["f", "m", "m"]. Why? This is because the user ASSUMED that the elements are unique). Hence, IT IS BETTER TO KEEP assume_unique to its default value. Note that both answers are sorted.

pythonnumpy

你可以使用集合:

main_list = list(set(list_2) - set(list_1))

输出:

>>> list_1=["a", "b", "c", "d", "e"]
>>> list_2=["a", "f", "c", "m"]
>>> set(list_2) - set(list_1)
set(['m', 'f'])
>>> list(set(list_2) - set(list_1))
['m', 'f']

根据@JonClements的评论，这里是一个更整洁的版本:

>>> list_1=["a", "b", "c", "d", "e"]
>>> list_2=["a", "f", "c", "m"]
>>> list(set(list_2).difference(list_1))
['m', 'f']

main_list=[]
list_1=["a", "b", "c", "d", "e"]
list_2=["a", "f", "c", "m"]

for i in list_2:
    if i not in list_1:
        main_list.append(i)

print(main_list)

输出:

['f', 'm']