Python 遍历一个数组并计算最终结果

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This topic created in 3497 days ago, the information mentioned may be changed or developed.

我有一段代码，如下：
result = 0
for item in [3, 2, 5]:
result ^= item
print(result)
以上这个 for 循环能不能用一个函数，或一个语句就计算出来？

result

item

for

11 replies • 2016-11-10 14:38:29 +08:00

Pythagodzilla

Nov 10, 2016

print result*reduce(operator.mul, [3,2,5])
这样？

littleshy

Nov 10, 2016

reduce(operator.ixor,[3,2,5])

woostundy

Nov 10, 2016

l = [3,2,5]
reduce(lambda x,y:x^y,l)

woostundy

Nov 10, 2016

好尴尬，刚发出来刷新发现上面有更好的写法

zungmou

Nov 10, 2016

@Pythagodzilla
@littleshy

reduce 在 python 3.5 没有了啊

BBrother

Nov 10, 2016

@zungmou 你需要
from functools import reduce
import operator

Herobs

Nov 10, 2016 via Android

reduce 或者递归

Pythagodzilla

Nov 10, 2016

@zungmou 惊了，没仔细看发现写的是*...2L 是对的...

staticor

Nov 10, 2016

顺便可以了解一些函数式编程的内容

zungmou

Nov 10, 2016

@BBrother 感谢
@littleshy 好厉害！

Zuckonit

Nov 10, 2016

reduce