Python list assignment -- puzzling behavior

Asked Jan 27 '14 at 18:49

Active Jan 27 '14 at 18:51

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I am having strange behavior with a list of lists in Python. There are two levels, but depending on how I set up the list, editing one of the levels either edits just the correct element, or all elements that share the 2nd reference.

This can be summarized as follows -- b and c are ititially the same (Python shows they are equivalent). Then I edit them the same way by changing the [1][1] entry, however for b, this impacts any entry with 2nd index=1 (ie adding to b[1][1] changes b[2][1]), where for c it only changes c[1][1]. This behavior is very puzzling, clearly something in the assignment of b relates the indexed entities to each other. Anyone know what is going on?

b = 10*[4*[0]]
c = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]];

print(b==c);

b[1][1] += 2;
c[1][1] += 2;

print(b==c);

edited Jan 27 '14 at 18:51

Stefano Sanfilippo

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asked Jan 27 '14 at 18:49

Javed Ahmed

Be aware that the trailing semicolon `;` after `print` is redundant. – Stefano Sanfilippo Jan 27 '14 at 18:51
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@wnnmaw That's not the correct dup. – Ashwini Chaudhary Jan 27 '14 at 18:52
@AshwiniChaudhary, whoops, careless reading on my part – wnnmaw Jan 27 '14 at 18:54
List `b` contains ten references to the same list `[0,0,0,0]`, so modifications of any of `b`'s items will affect all of them. Use `b = [4*[0] for i in range(10)]` to resemble the assignment you made to `c`. – J. Katzwinkel Jan 27 '14 at 18:58
@Stefano Like all of them. – Konrad Rudolph Jan 27 '14 at 19:01
@KonradRudolph Yes, I did not see the other `;` at a first glance. Bottom line: no semicolon in Python, please. – Stefano Sanfilippo Jan 27 '14 at 19:03

Python list assignment -- puzzling behavior

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