Question

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Simple python code about double loop

I’m stuck with the well known problem of changing 1 item in a list of lists. I need to use a fixed size list of list.

If I use:

In [21]: a=[[2]*2]*3

In [22]: a
Out[22]: [[2, 2], [2, 2], [2, 2]]

In [23]: a[0][0]=1

In [24]: a
Out[24]: [[1, 2], [1, 2], [1, 2]]

But if I define the list of list in the following way, it works:

In [26]: a = [
   ....: [2,2],
   ....: [2,2],
   ....: [2,2],
   ....: ]

In [27]: a   
Out[27]: [[2, 2], [2, 2], [2, 2]]

In [28]: a[0][0]=1

In [29]: a
Out[29]: [[1, 2], [2, 2], [2, 2]]

To me line 22 and 27 looks identical. So which is the difference?

Can someone explain me how to go around this problem and, in particular, how to change a code that change a single item of a list of list? If this is not possible, any suggestion to move to a different data structure allowing that?
thanks

Answer 1

There is a fundamental difference in the 2 ways you defined the matrix in both code samples. In the first (line 21) when you multiply the outer list by 3 you actually multiply the references to the inner list.

Here’s a parameterized example:

a = [2,2]
b = [a] * 3
b => [a, a, a] => [[2,2], [2,2], [2,2]]

As you can see the numerical representation of b is what you would expect, but it actually contains 3 references to a. So if you change either of the inner lists, they all change, because they are actually the same list.

Instead of multiplication you need to clone the lists doing something like this

new_list = old_list[:]

or

new_list = list(old_list)

In the case of nested lists, this can be done using loops.