Initialize an nested list with values (e.g., 0's)

Question

array1 = [[0]*4]*4

array2 = [[0,0,0,0]]*4

array3 = [[0,0,0,0],
          [0,0,0,0],
          [0,0,0,0],
          [0,0,0,0]]

array1[0][0] = 9000
array2[0][0] = 9000
array3[0][0] = 9000

In [77]: array1
Out[77]: [[9000, 0, 0, 0], 
          [9000, 0, 0, 0], 
          [9000, 0, 0, 0],
          [9000, 0, 0, 0]]

In [78]: array2
Out[78]: [[9000, 0, 0, 0],
          [9000, 0, 0, 0],
          [9000, 0, 0, 0],
          [9000, 0, 0, 0]]

In [79]: array3
Out[79]: [[9000, 0, 0, 0],
          [0, 0, 0, 0],
          [0, 0, 0, 0], 
          [0, 0, 0, 0]]

Mutating any row value mutates all row values for the shortcut methods of initialization.
The only way I can get non-mutating behavior is by writing out the whole list.
There must be a better way(?), think of the case where you want to initialize the list based on the length of a different list.

https://docs.python.org/3/faq/programming.html#how-do-i-create-a-multidimensional-list — hiro protagonist, Apr 28 '17 at 15:45
https://stackoverflow.com/questions/3880037/how-to-create-a-list-or-tuple-of-empty-lists-in-python (and other duplicates...) — hiro protagonist, Apr 28 '17 at 15:46

Efron Licht · Accepted Answer · 2017-04-28T15:45:03.493

3

Use a list comprehension.

my_list = [ [0]*4 for _ in range(4)]

Note that lists are mutable but integers are not. Thus [0]*4 behaves like you would expect, but [[0]*4]*4 'leaks' changes.

Using a list comprehension creates a new outer list each time, avoiding the problematic behavior of your example.

edited Apr 28 '17 at 15:45

answered Apr 28 '17 at 15:43

Efron Licht

488
3
13

might want to explain your answer and why it works – MooingRawr Apr 28 '17 at 15:44

Initialize an nested list with values (e.g., 0's)

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