I have a list of lists, each containing a different number of strings. I'd like to (efficiently) convert these all to ints, but am feeling kind of dense, since I can't get it to work out for the life of me. I've been trying: newVals = [int(x) for x in [row for rows in values]]
我有一个列表列表，每个列表包含不同数量的字符串。我想（有效地）将这些全部转换为整数，但我感觉有点密集，因为我无法让它为我的生活而努力。我一直在尝试：newVals = [int（x）for x in [row for values in values]]
Where 'values' is the list of lists. It keeps saying that x is a list and can therefore not be the argument if int(). Obviously I'm doing something stupid here, what is it? Is there an accepted idiom for this sort of thing?
“值”是列表的列表。它一直说x是一个列表，因此如果是int（）则不能成为参数。显然我在这里做了一些蠢事，这是什么？对于这种事情，有没有被接受的习语？
As usual, thanks a ton in advance.
像往常一样，提前感谢。
5 个解决方案
#1
9
This leaves the ints nested
这使得int嵌套
[map(int, x) for x in values]
If you want them flattened, that's not hard either
如果你想让它们变平，那也不难
for Python3 map() returns an iterator. You could use
for Python3 map（）返回一个迭代器。你可以用
[list(map(int, x)) for x in values]
but you may prefer to use the nested LC's in that case
但在这种情况下你可能更喜欢使用嵌套的LC
[[int(y) for y in x] for x in values]
#2
9
How about:
怎么样：
>>> a = [['1','2','3'],['4','5','6'],['7','8','9']]
>>> [[int(j) for j in i] for i in a]
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
#3
1
You simply use incorrect order and parenthesis - should be:
你只是使用不正确的顺序和括号 - 应该是：
inputVals = [['1','2','3'], ['3','3','2','2']]
[int(x) for row in inputVals for x in row]
Or if you need list of list at the output then:
或者如果您需要输出列表，那么：
map(lambda row: map(int, row), inputVals)
#4
0
an ugly way is to use evalf:
一种丑陋的方式是使用evalf：
>>> eval(str(a).replace("'",""))
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
if you don't mind all your numbers in one array you could go:
如果你不介意一个数组中的所有数字，你可以去：
>>> a = [['1','2','3'],['4','5','6'],['7','8','9']]
>>> map(int,sum(a,[]))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
#5
0
Another workaround
另一种解决方法
a = [[1, 2, 3], [7, 8, 6]]
list(map(lambda i: list(map(lambda j: j - 1, i)), a))
[[0, 1, 2], [6, 7, 5]] #output
#1
9
This leaves the ints nested
这使得int嵌套
[map(int, x) for x in values]
If you want them flattened, that's not hard either
如果你想让它们变平，那也不难
for Python3 map() returns an iterator. You could use
for Python3 map（）返回一个迭代器。你可以用
[list(map(int, x)) for x in values]
but you may prefer to use the nested LC's in that case
但在这种情况下你可能更喜欢使用嵌套的LC
[[int(y) for y in x] for x in values]
#2
9
How about:
怎么样：
>>> a = [['1','2','3'],['4','5','6'],['7','8','9']]
>>> [[int(j) for j in i] for i in a]
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
#3
1
You simply use incorrect order and parenthesis - should be:
你只是使用不正确的顺序和括号 - 应该是：
inputVals = [['1','2','3'], ['3','3','2','2']]
[int(x) for row in inputVals for x in row]
Or if you need list of list at the output then:
或者如果您需要输出列表，那么：
map(lambda row: map(int, row), inputVals)
#4
0
an ugly way is to use evalf:
一种丑陋的方式是使用evalf：
>>> eval(str(a).replace("'",""))
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
if you don't mind all your numbers in one array you could go:
如果你不介意一个数组中的所有数字，你可以去：
>>> a = [['1','2','3'],['4','5','6'],['7','8','9']]
>>> map(int,sum(a,[]))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
#5
0
Another workaround
另一种解决方法
a = [[1, 2, 3], [7, 8, 6]]
list(map(lambda i: list(map(lambda j: j - 1, i)), a))
[[0, 1, 2], [6, 7, 5]] #output