Creating lists of lists in a pythonic way

Question

I'm using a list of lists to store a matrix in python. I tried to initialise a 2x3 Zero matrix as follows.

mat=[[0]*2]*3

However, when I change the value of one of the items in the matrix, it changes the value of that entry in every row, since the id of each row in mat is the same. For example, after assigning

mat[0][0]=1

mat is [[1, 0], [1, 0], [1, 0]].

I know I can create the Zero matrix using a loop as follows,

mat=[[0]*2]
for i in range(1,3):
    mat.append([0]*2)

but can anyone show me a more pythonic way?

There should be one-- and preferably only one --obvious way to do it. ;-) — Ubiquitous
– Ubiquitous, Commented May 28, 2009 at 21:06

Ben Blank · Accepted Answer · 2009-05-28 21:00:21Z

9

Use a list comprehension:

>>> mat = [[0]*2 for x in xrange(3)]
>>> mat[0][0] = 1
>>> mat
[[1, 0], [0, 0], [0, 0]]

Or, as a function:

def matrix(rows, cols):
    return [[0]*cols for x in xrange(rows)]

answered May 28, 2009 at 21:00

Ben Blank

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Comments

Community · Accepted Answer · 2017-05-23 11:46:51Z

8

Try this:

>>> cols = 6
>>> rows = 3
>>> a = [[0]*cols for _ in [0]*rows]
>>> a
[[0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> a[0][3] = 2
>>> a
[[0, 0, 0, 2, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]

This is also discussed in this answer:

>>> lst_2d = [[0] * 3 for i in xrange(3)]
>>> lst_2d
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> lst_2d[0][0] = 5
>>> lst_2d
[[5, 0, 0], [0, 0, 0], [0, 0, 0]]

edited May 23, 2017 at 11:46

CommunityBot

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answered May 28, 2009 at 20:57

Paolo Bergantino

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2 Comments

duffymo Over a year ago

+1 - nice stuff. I'm just learning Python, so I greatly appreciate seeing snippets of "pythonic" code.

jfs Over a year ago

"[0]*rows" part is misleading; you're creating a list that is not used in any way except its length. Use either xrange(n) or (less likely) itertools.repeat(None, n) to do something n times in Python.

John La Rooy · Accepted Answer · 2010-02-25 10:41:49Z

7

This one is faster than the accepted answer!
Using xrange(rows) instead of [0]*rows makes no difference.

>>> from itertools import repeat
>>> rows,cols = 3,6
>>> a=[x[:] for x in repeat([0]*cols,rows)]

A variation that doesn't use itertools and runs around the same speed

>>> a=[x[:] for x in [[0]*cols]*rows]

From ipython:

In [1]: from itertools import repeat

In [2]: rows=cols=10

In [3]: timeit a = [[0]*cols for _ in [0]*rows]
10000 loops, best of 3: 17.8 us per loop

In [4]: timeit a=[x[:] for x in repeat([0]*cols,rows)]
100000 loops, best of 3: 12.7 us per loop

In [5]: rows=cols=100

In [6]: timeit a = [[0]*cols for _ in [0]*rows]
1000 loops, best of 3: 368 us per loop

In [7]: timeit a=[x[:] for x in repeat([0]*cols,rows)]
1000 loops, best of 3: 311 us per loop

edited Feb 25, 2010 at 10:41

answered Feb 25, 2010 at 9:29

John La Rooy

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Comments

othercriteria · Accepted Answer · 2009-05-28 21:03:09Z

4

I use

mat = [[0 for col in range(3)] for row in range(2)]

although depending on what you do with the matrix after you create it, you might take a look at using a NumPy array.

answered May 28, 2009 at 21:03

othercriteria

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1 Comment

Alasdair Over a year ago

I'm going to explore NumPy at some point, but for my current problem, a list of list is enough.

Nadia Alramli · Accepted Answer · 2009-05-28 20:59:40Z

3

This will work

col = 2
row = 3
[[0] * col for row in xrange(row)]

answered May 28, 2009 at 20:59

Nadia Alramli

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Comments

miku · Accepted Answer · 2009-05-28 21:16:28Z

2

What about:

m, n = 2, 3
>>> A = [[0]*m for _ in range(n)]
>>> A
[[0, 0], [0, 0], [0, 0]]
>>> A[0][0] = 1
[[1, 0], [0, 0], [0, 0]]

Aka List comprehension; from the docs:

List comprehensions provide a concise way to create lists 
without resorting to use of     
map(), filter() and/or lambda. 
The resulting list definition tends often to be clearer    
than lists built using those constructs.

edited May 28, 2009 at 21:16

answered May 28, 2009 at 21:00

miku

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RemcoGerlich · Accepted Answer · 2015-01-20 14:26:58Z

2

If the sizes involved are really only 2 and 3,

mat = [[0, 0], [0, 0], [0, 0]]

is easily best and hasn't been mentioned yet.

answered Jan 20, 2015 at 14:26

RemcoGerlich

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Comments

John La Rooy · Accepted Answer · 2010-02-25 09:15:18Z

1

Is there anything itertools can't do? :)

>>> from itertools import repeat,izip
>>> rows=3
>>> cols=6
>>> A=map(list,izip(*[repeat(0,rows*cols)]*cols))
>>> A
[[0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> A[0][3] = 2
>>> A
[[0, 0, 0, 2, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]

answered Feb 25, 2010 at 9:15

John La Rooy

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