列表/矩阵未保存正确的值
我收到的作业有一个奇怪的问题。我们应该实现一个矩阵类。嗯，这并不难，但 Python 就是不会按照我的指示去做。但我确信有一个解释。
问题是，在下面的代码中，我尝试将值（在列表中提供）保存到矩阵中。
class simplematrix:
matrix = [[]]
def __init__(self, * args):
lm = args[0]
ln = args[1]
values = args[2]
self.matrix = [[0]*ln]*lm
for m in range(lm):
for n in range(ln):
self.matrix[m][n] = values[(m*ln)+n]
vals = [0,1,2,3,4,5]
a = simplematrix(2,3,vals)
当我尝试打印矩阵时，我期望得到 [[0,1,2],[3,4,5]]，如果我在一张纸上手动运行它，我就会得到它。如果我从 Python 打印矩阵，我会得到 [[3,4,5],[3,4,5]] 。 谁能告诉我为什么Python会这样，或者我是否在某个地方犯了一些愚蠢的错误？ :)
I have a weird problem with an assignment I got. We are supposed to implement a matrix class. Well, it's not that hard, but Python just won't do as I tell it to. But I'm sure there is an explanation.
The problem is that, in the following code, I try to save values (provided in a list) into a matrix.
class simplematrix:
matrix = [[]]
def __init__(self, * args):
lm = args[0]
ln = args[1]
values = args[2]
self.matrix = [[0]*ln]*lm
for m in range(lm):
for n in range(ln):
self.matrix[m][n] = values[(m*ln)+n]
vals = [0,1,2,3,4,5]
a = simplematrix(2,3,vals)
When I try to print the matrix, I expect to get [[0,1,2],[3,4,5]], which I get if I run it by hand, on a piece of paper. If I print the matrix from Python I get [[3,4,5],[3,4,5]] instead.
Can anyone tell me why Python acts like this, or if I made some stupid mistake somewhere? :)
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问题出在
[[0]*ln]*lm中。结果由lm对同一列表的引用组成，因此当您修改一行时，所有行都会发生更改。尝试：
The problem is in
[[0]*ln]*lm. The result consists oflmreferences to the same list, so when you modify one row, all rows appear to change.Try:
Tim 和 aix 的答案纠正了您的错误，但这一步甚至没有必要，您可以使用列表理解在一行中完成整个事情：
您也可以说：
与您已经拥有的相反。这会整理你的代码并使其更加Pythonic。
The answers by Tim and aix correct your mistake, but that step isn't even necessary, you can do the whole thing in one line using a list comprehension:
You can also say:
as opposed to what you already have. This tidies up your code and makes it more Pythonic.
问题是
self.matrix = [[0]*ln]*lm不会给您一个lm单独子列表的列表，而是一个列表lm引用单个相同列表[[0]*ln]。尝试一下
（如果您使用的是 Python 2，请改用
xrange(lm)）。The problem is that
self.matrix = [[0]*ln]*lmdoesn't give you a list oflmseparate sublists, but a list oflmreferences to the single same list[[0]*ln].Try
(If you're on Python 2, use
xrange(lm)instead).