问题描述
我有一个问题，我还没有找到很好的解决方案.我正在寻找一种更好的方法来将函数输出附加到两个或更多列表，而不使用临时变量.示例如下:
I have a question that I haven't quite found a good solution to. I'm looking for a better way to append function output to two or more lists, without using temp variables. Example below:
def f():
return 5,6
a,b = [], []
for i in range(10):
tmp_a, tmp_b = f()
a.append(tmp_a)
b.append(temp_b)
我尝试过使用 zip(*f()) 之类的东西，但还没有找到这样的解决方案.不过，任何删除这些临时变量的方法都会非常有帮助，谢谢！
I've tried playing around with something like zip(*f()), but haven't quite found a solution that way. Any way to remove those temp vars would be super helpful though, thanks!
编辑以获取更多信息:在这种情况下，函数的输出数将始终等于要附加到的列表数.我希望摆脱临时变量的主要原因是可能有 8-10 个函数输出，并且拥有这么多临时变量会变得混乱(尽管我什至不喜欢有两个).
Edit for additional info: In this situation, the number of outputs from the function will always equal the number of lists that are being appended to. The main reason I'm looking to get rid of temps is for the case where there are maybe 8-10 function outputs, and having that many temp variables would get messy (though I don't really even like having two).
推荐答案
def f():
return 5,6
a,b = zip(*[f() for i in range(10)])
# this will create two tuples of elements 5 and 6 you can change
# them to list by type casting it like list(a), list(b)
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