问题
I remember I once seen a operator which is able to decompose a list in python.
for example
[[1],[2],[3]]
by applying that operator, you get
[1], [2], [3]
what is that operator, any help will be appreciated.
回答1:
If you want to pass a list of arguments to a function, you can use *, the splat operator. Here's how it works:
list = [1, 2, 3]
function_that_takes_3_arguments(*list)
If you want to assign the contents of a list to a variable, you can list unpacking:
a, b, c = list # a=1, b=2, c=3
回答2:
You can use the tuple function to convert a list to a tuple. A tuple with three elements isn't really any different from three separate elements, but it gives a handy way to work with all three together.
li = [[1], [2], [3]]
a, b, c = tuple(li)
print a # [1]
回答3:
The correct answer to the OP's question: "what is that operator" which transforms the list [[1],[2],[3]] to [1], [2], [3] is tuple() since [1], [2], [3] is a tuple. The builtin function tuple will convert any sequence or iterable to a tuple, although there is seldom a need to do so since, as already pointed out, unpacking a list is as easy as unpacking a tuple:
a, b, c = [[1],[2],[3]]
gives the same result as
a, b, c = tuple([[1],[2],[3]])
This may not be what the OP wanted but it is the correct answer to the question as asked.
回答4:
This can be achieved by running sum(list_name,[]) as mentioned here.
You may also find this question on flattening shallow lists relevant.
来源：https://stackoverflow.com/questions/6319612/python-decompose-a-list