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']

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

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']

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]

如果要考虑出现的次数，则可能需要使用集合之类的方法。计数器:

list_1=["a", "b", "c", "d", "e"]
list_2=["a", "f", "c", "m"] 
from collections import Counter
cnt1 = Counter(list_1)
cnt2 = Counter(list_2)
final = [key for key, counts in cnt2.items() if cnt1[key] != counts]

>>> final
['f', 'm']

如前所述，这也可以处理不同数量的事件作为“差异”:

list_1=["a", "b", "c", "d", "e", 'a']
cnt1 = Counter(list_1)
cnt2 = Counter(list_2)
final = [key for key, counts in cnt2.items() if cnt1[key] != counts]

>>> final
['a', 'f', 'm']

从ser1中删除ser2中的条目。

输入

1 = pd。系列([1、2、3、4、5]) ser2 = pd。系列([4、5、6、7、8])

解决方案

ser1[~ser1.isin（ser2）]