仅从两个列表中获取唯一元素 [英] Get only unique elements from two lists

问题描述

如果我有两个列表(可能具有不同的len):

If I have two lists (may be with different len):

x = [1,2,3,4]
f = [1,11,22,33,44,3,4]

result = [11,22,33,44]

我正在做

for element in f:
    if element in x:
        f.remove(element)

我要

result = [11,22,33,44,4]

推荐答案

更新:

感谢@ Ahito :

In : list(set(x).symmetric_difference(set(f)))

Out: [33, 2, 22, 11, 44]

本文有一个简洁的图表，该图表解释了对称差别.

This article has a neat diagram that explains what the symmetric difference does.

使用以下 Python的文档设置:

>>> # Demonstrate set operations on unique letters from two words
...
>>> a = set('abracadabra')
>>> b = set('alacazam')
>>> a                                  # unique letters in a
{'a', 'r', 'b', 'c', 'd'}
>>> a - b                              # letters in a but not in b
{'r', 'd', 'b'}
>>> a | b                              # letters in a or b or both
{'a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'}
>>> a & b                              # letters in both a and b
{'a', 'c'}
>>> a ^ b                              # letters in a or b but not both
{'r', 'd', 'b', 'm', 'z', 'l'}

我想出了这段代码来从两个列表中获取唯一元素:

I came up with this piece of code to obtain unique elements from two lists:

(set(x) | set(f)) - (set(x) & set(f))

或稍作修改以返回 list :

list((set(x) | set(f)) - (set(x) & set(f))) #if you need a list

这里:

1. | 运算符以 x ， f 或 both
返回元素
2. & 运算符以两者 x 和 f
的形式返回元素
3. -运算符从 | 中减去& 的结果，并为我们提供仅在列表之一中唯一显示的元素

1. | operator returns elements in x, f or both
2. & operator returns elements in both x and f
3. - operator subtracts the results of & from | and provides us with the elements that are uniquely presented only in one of the lists

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