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Editorial Team
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Asked: 2026-05-28T02:11:04+00:00

>>> rows = [[”]*5]*5 >>> rows [[”, ”, ”, ”, ”], [”, ”, ”,

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>>> rows = [['']*5]*5
>>> rows
[['', '', '', '', ''], ['', '', '', '', ''], ['', '', '', '', ''], ['', '', '', '', ''], ['', '', '', '', '']]
>>> rows[0][0] = 'x'

Naturally, I expect rows to become:

[['x', '', '', '', ''], ['', '', '', '', ''], ['', '', '', '', ''], ['', '', '', '', ''], ['', '', '', '', '']]

Instead, I get:

[['x', '', '', '', ''], ['x', '', '', '', ''], ['x', '', '', '', ''], ['x', '', '', '', ''], ['x', '', '', '', '']]

It seems that elements of rows list are pointers to the same old [”]*5 list. Why does it work this way and is this a Python feature?

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1 Answer

  1. Editorial Team

    The behaviour is not specific to the repetition operator (*). For example, if you concatenate two lists using +, the behaviour is the same:

    In [1]: a = [[1]]

In [2]: b = a + a

In [3]: b
Out[3]: [[1], [1]]

In [4]: b[0][0] = 10

In [5]: b
Out[5]: [[10], [10]]

    This has to do with the fact that lists are objects, and objects are stored by reference. When you use * et al, it is the reference that gets repeated, hence the behaviour that you’re seeing.

    The following demonstrates that all elements of rows have the same identity (i.e. memory address in CPython):

    In [6]: rows = [['']*5]*5

In [7]: for row in rows:
   ...:     print id(row)
   ...:     
   ...:     
15975992
15975992
15975992
15975992
15975992

    The following is equivalent to your example except it creates five distinct lists for the rows:

    rows = [['']*5 for i in range(5)]
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