Python: How to initialize a nested list with empty values which i can append to

Question

i'm learning python and hit a wall. I am trying to define a 2d list which i can later use to append values. This corresponding to a grid of width *height

I tried using [] to initialize the empty lists but then the wid is ignored. I tried using None as a placeholder but then i can't append

wid = 3
hgt = 3
l1 = [[]*wid ] * hgt
l = [[None]*wid ] * hgt
l[1][1].append("something")

Result

l1: [[], [], []]

l: [[None, None, None], [None, None, None], [None, None, None]]

Error:

append: AttributeError: 'NoneType' object has no attribute 'append'

desired result: [[[], [], []], [[], [], []], [[], [], []]]

[[[] for i in range(wid)] for j in range(hgt)]?

Joshua Nixon
– Joshua Nixon

2019-02-13 15:35:21 +00:00
Commented Feb 13, 2019 at 15:35 — Joshua Nixon
– Joshua Nixon, Commented Feb 13, 2019 at 15:35
Related: stackoverflow.com/q/240178/1639625

tobias_k
– tobias_k

2019-02-13 15:37:48 +00:00
Commented Feb 13, 2019 at 15:37 — tobias_k
– tobias_k, Commented Feb 13, 2019 at 15:37

chris · Accepted Answer · 2019-02-13 16:04:44Z

7

Try using a list comprehension within a list comprehension:

>>> [ [ [] for i in range(wid) ] for i in range(hgt) ]
[[[], [], []], [[], [], []], [[], [], []]]

Note this is preferred to list multiplication because each of these lists is unique. Compare:

>>> x = [ [[] for i in range(wid)] for i in range(hgt) ]
>>> x[1][1].append('a')
>>> x
[[[], [], []], [[], ['a'], []], [[], [], []]]

vs.

>>> y = [ [[]] * wid for i in range(hgt) ]
>>> y[1][1].append('a')
>>> y
[[[], [], []], [['a'], ['a'], ['a']], [[], [], []]]

vs.

>>> z = [ [[]] * wid ] * hgt
>>> z[1][1].append('a')
>>> z
[[['a'], ['a'], ['a']], [['a'], ['a'], ['a']], [['a'], ['a'], ['a']]]

Where, in the second and third cases, 'a' appears in multiple cells! And using None does not avoid this problem:

>>> m = [ [None] * wid ] * hgt
>>> m
[[None, None, None], [None, None, None], [None, None, None]]
>>> if m[1][1] is None:
...     m[1][1] = ['a']
... else:
...     m[1][1].append('a')
...
>>> m
[[None, ['a'], None], [None, ['a'], None], [None, ['a'], None]]

tl;dr - use the double list comprehension. In my opinion, it's the most readable option anyway.

edited Feb 13, 2019 at 16:04

answered Feb 13, 2019 at 15:37

chris

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3 Comments

souza Over a year ago

And refer to this guy! List comprehension is considerably important in Python and I highly recommend you to search a little more about it :)

tobias_k Over a year ago

This answers only part of the question (OP might still try to append to those None entries)

Jop Over a year ago

Thanks for your solution and explenation! exactly what i wanted to know.

blue note · Accepted Answer · 2019-02-13 15:47:18Z

2

There's two ways to do it

use None as placeholder value, and replace it with l[1][1] = 5
use empty lists and append new elements to them, with l[1].append(5)

Currently, you are mixing the two ways. the l[1][1] returns a None value, not a list, and you are trying to call append on it.

Regardless, there is a common python trap here. When you write

mylist = [[None]*2]*3

this does two things

creates a list [None, None]
uses this list 3 times to create another list

The key is that the outer list will consist of 3 copies of the inner list, not of 3 different by identical lists. so, everytime you modify one of them, the rest will be modified too. So, you need to make copies of the inner list.

For a 2D array however, the best way would be to use numpy

x = np.zeros(m, n)
x[1, 2] = 5

edited Feb 13, 2019 at 15:47

answered Feb 13, 2019 at 15:35

blue note

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4 Comments

tobias_k Over a year ago

This answers only part of the question (the sublists are still all the same instance)

blue note Over a year ago

@tobias_k: correct. the question was badly formatted when I read it. editing answer

Jop Over a year ago

Thanks for the help! I still get confused sometimes when python copies or references to the same object. This helps a lot! Would performace wise numpy be faster then lists?

blue note Over a year ago

@Jop: yes, definitely better performance. but, if you care about performance, you should do matrix operations, avoid for loops

Benoît P · Accepted Answer · 2019-02-13 15:37:52Z

0

>>> x = 5
>>> y = 5
>>> [[None for _ in range(x)] for _ in range(y)]
[[None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None]]

Don't use [[None]*x]*y unless you really know that you just created shared references.

answered Feb 13, 2019 at 15:37

Benoît P

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Comments

kederrac · Accepted Answer · 2019-02-13 16:12:30Z

0

if you are dealing with matrices you can use numpy:

import numpy as np

my_matrix = np.zeros((rows, columns) # with 2 dimentions

answered Feb 13, 2019 at 16:12

kederrac

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