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拆分python列表而不调用itertools';grouper函数,但得到一个错误_Python_List_Itertools - Fatal编程技术网

拆分python列表而不调用itertools';grouper函数,但得到一个错误

拆分python列表而不调用itertools';grouper函数,但得到一个错误,python,list,itertools,Python,List,Itertools,在这段代码中,grouper函数可以很好地工作,但是如果我不调用该函数。它抛出了一个错误 TypeError:izip#u最长参数#1必须支持迭代 from itertools import * def grouper(n, iterable, fillvalue=None): args = [iter(iterable)] * n return izip_longest(fillvalue=fillvalue, *args) x = [1,2,3] args = [iter(x

在这段代码中,grouper函数可以很好地工作,但是如果我不调用该函数。它抛出了一个错误

TypeError:izip#u最长参数#1必须支持迭代

from itertools import *

def grouper(n, iterable, fillvalue=None):
    args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)


x = [1,2,3]

args = [iter(x)] * 2
l = izip_longest(None , *args )
#l = grouper(2,x)
print [x for x in l]

所有位置参数都应该是可编辑的,而不是fillvalue.:

如果
fillvalue
None
;你可以省略它:

it = izip_longest(*iterables)
it = izip_longest(*iterables)