use*_*587 5 python list unique
如果我有两个列表(可能有不同的len):
x = [1,2,3,4]
f = [1,11,22,33,44,3,4]
result = [11,22,33,44]
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我正在做:
for element in f:
if element in x:
f.remove(element)
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我越来越
result = [11,22,33,44,4]
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小智 20
如果你想要两个列表中的唯一元素,这应该有效:
x = [1,2,3,4]
f = [1,11,22,33,44,3,4]
res = list(set(x+f))
print(res)
# res = [1, 2, 3, 4, 33, 11, 44, 22]
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Mik*_*tty 13
基于在一个新的(封闭的)问题中澄清这个问题:
如果您希望第二个列表中的所有项目都没有出现在第一个列表中,您可以编写:
x = [1,2,3,4]
f = [1,11,22,33,44,3,4]
result = set(f) - set(x) # correct elements, but not yet in sorted order
print(sorted(result)) # sort and print
# Output: [11, 22, 33, 44]
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Iop*_*eam 13
在集合上使用此Python文档:
>>> # 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'}
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我想出了这段代码来从两个列表中获取唯一元素:
(set(x) | set(f)) - (set(x) & set(f))
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或稍作修改即可返回list:
list((set(x) | set(f)) - (set(x) & set(f))) #if you need a list
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这里:
|在操作者返回元件x,f或两者&运算符在和中都 返回元素xf -运算符&从中减去结果|并为我们提供仅在列表之一中唯一显示的元素输入 :
x = [1,2,3,4]
f = [1,11,22,33,44,3,4]
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代码:
l = list(set(x).symmetric_difference(set(f)))
print(l)
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输出 :
[2, 22, 33, 11, 44]
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小智 6
x = [1, 2, 3, 4]
f = [1, 11, 22, 33, 44, 3, 4]
list(set(x) ^ set(f))
[33, 2, 22, 11, 44]
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如果您只想从两个列表中获取唯一元素,则可以通过获取。
a=[1,2,3,4,5]
b= [2,4,1]
list(set(a) - set(b))
OP:- [3, 5]
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您的方法不会获得唯一元素“2”。关于什么:
list(set(x).intersection(f))
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