ValueError：需要多于 0 个值来解包（python 列表）
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ValueError: need more than 0 values to unpack (python lists)
提问by kunaguvarun
I'm learning python from Google code class. I'm trying out the exercises.
我正在从 Google 代码课学习 python。我正在尝试练习。
def front_x(words):
x_list, ord_list = []
for word in words:
if word[0] == 'x':
x_list.append(word)
else:
ord_list.append(word)
return sorted(x_list) + sorted(ord_list)
I believe the error is thrown because of initializing two empty lists on a single line. If if initialize them on separate lines, no more errors occur. Is this the reason?
我相信错误是由于在一行上初始化两个空列表而引发的。如果在单独的行上初始化它们，则不会发生更多错误。这是原因吗？
采纳答案by Martijn Pieters
You are trying to use tuple assignment:
您正在尝试使用元组分配：
x_list, ord_list = []
you probably meant to use multiple assignment:
你可能打算使用多重赋值：
x_list = ord_list = []
which will not do what you expect it to; use the following instead:
这不会做你期望的事情；请改用以下内容：
x_list, ord_list = [], []
or, best still:
或者，最好还是：
x_list = []
ord_list = []
When using a comma-separated list of variable names, Python expects there to be a sequence of expressions on the right-hand side that matches the number variables; the following would be legal too:
当使用逗号分隔的变量名列表时，Python 期望右侧有一个与数字变量匹配的表达式序列；以下也是合法的：
two_lists = ([], [])
x_list, ord_list = two_lists
This is called tuple unpacking. If, on the other hand, you tried to use multiple assignment with oneempty list literal (x_list = ord_list = []) then both x_listand ord_listwould be pointing to the samelist and any changes made through one variable will be visible on the other variable:
这称为元组解包。如果，另一方面，您尝试使用多任务与一个空列表文字（x_list = ord_list = []两者），然后x_list和ord_list将指向相同的列表，并通过一个变量作中将会在其他变量可见的任何变化：
>>> x_list = ord_list = []
>>> x_list.append(1)
>>> x_list
[1]
>>> ord_list
[1]
Better keep things crystal clear and use two separateassignments, giving each variable their own empty list.
最好让事情一目了然，并使用两个单独的赋值，为每个变量提供自己的空列表。
回答by Abhijit
Change the line
换线
x_list, ord_list = []
to
x_list, ord_list = [], []
回答by RICHA AGGARWAL
return type of function does not match with values expected in function...
函数的返回类型与函数中预期的值不匹配...
check the number of variables returned from function and variables you are expecting
检查从函数和您期望的变量返回的变量数
