Python non-numpy matrixes issue

Question

I have the following issue:

I have a list of lists with the following declaration:

As = [[0]*3]*3

I then try to change the values of this "matrix" with this:

for i in range(3):
    for j in range(3):
        As[i][j] = calculate(A, i, j)*((-1)**(i+j))

As you may have guessed, this is used in calculating the inverse of a 3x3 matrix.

The function calculate returns the following values:

However, As has the following value:

[[4.0, -10.0, -21.0], [4.0, -10.0, -21.0], [4.0, -10.0, -21.0]], which is unexpected.

What am I missing?

Netwave · Accepted Answer · 2016-10-04 07:36:36Z

2

When you build a list like this [[0]*3]*3 you are creating 3 references to the same list, use a list comprehension instead:

[[0 for _ in xrange(3)] for _ in xrange(3)]

See how in the comprehension here just the [0][0] is modified:

>>> l1 = [[0]*3]*3
>>> l1[0][0] = 10
>>> l1
[[10, 0, 0], [10, 0, 0], [10, 0, 0]]
>>> l2 = [[0 for _ in xrange(3)] for _ in xrange(3)]
>>> l2[0][0] = 10
>>> l2
[[10, 0, 0], [0, 0, 0], [0, 0, 0]]

answered Oct 4, 2016 at 7:36

Netwave

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2 Comments

Will Vousden Over a year ago

There's no need to construct the inner list using a comprehension. In fact, the list will refer to the same int instances anyway, since Python caches integers in [-5, 256]. The result is therefore identical to [0] * 3.

Netwave Over a year ago

@WillVousden, it was just for clarifiying the concept, It usually clears my mind to use for ranges instead of * operator for this purpouse

Will Vousden · Accepted Answer · 2016-10-04 07:51:20Z

1

The problem is in your first line:

As = [[0]*3]*3

This is equivalent to the following:

a = [0] * 3
As = [a] * 3

As therefore contains three references to the same list:

>>> for a in As:
...     print(id(a))
...
4293437996
4293437996
4293437996

So when you change this list, it's reflected in all of the rows of your matrix.

To get around this, you can use a list comprehension to construct the outer list:

As = [[0] * 3 for _ in xrange(3)]

Note that the inner list [0] * 3 is fine, since the list contains only an integer, which is immutable. It therefore doesn't matter that the list refers to the same instance of that integer, since they can't be changed.

edited Oct 4, 2016 at 7:51

answered Oct 4, 2016 at 7:43

Will Vousden

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