Array preallocation Python

Question

I'm writing a simple code in Python to sort a pre existing array A of dimensions [6][7] to do that I wanted to save the sorted array with a different name let's say B. I've done the following code but doesn't work

B=[[0]*7]*6


For i in range(0,6):
  For j in range(0,7):
    If A[i][j] > 7:
       B[i][j] = 1
    Else 
       B[i][j] = 0

The problem is in the preallocation of array B indeed if I print it it doesn't give the result I need, on the other hand if create B from scratch in a way like this it works

B=[
[0,0,0,0,0,0,0],
.
.
.
.
.
.
]

Can someone explain me why? Thank you very much.

This code is not valid python, and thus doesn't give *any* result. — Scott Hunter, Jul 25 '23 at 19:58
`B` is a list, not an array. If you *really* want an array (a la Numpy), try something like `B = (np.array(A) > 7).astype(int)`. — chepner, Jul 25 '23 at 20:05

score 0 · Accepted Answer · answered Jul 25 '23 at 20:22

The problem is this:

When you do: (I'm going to use lowercase variable names since you seem to have some accidental capitalization going on in your question)

b=[[0]*7]*6

what happens is that every row of b is the same object. It's as though you did this:

b = []
row = [0]*7
for _ in range(6):
    b.append(row)

What this means:

>>> import pprint
>>> b=[[0]*7]*6
>>> pprint.pprint(b)
[[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, 0, 0]]
>>> b[3][2] = 1
>>> pprint.pprint(b)
[[0, 0, 1, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0]]
>>> b[4][3] = 2
>>> pprint.pprint(b)
[[0, 0, 1, 2, 0, 0, 0],
 [0, 0, 1, 2, 0, 0, 0],
 [0, 0, 1, 2, 0, 0, 0],
 [0, 0, 1, 2, 0, 0, 0],
 [0, 0, 1, 2, 0, 0, 0],
 [0, 0, 1, 2, 0, 0, 0]]

So what you need is a different object for each row of b. There are a few ways to get this. Any of the following should work:

b = []
for _ in range(6):
  b.append([0]*7)

or

b = [[0]*7 for _ in range(6)]

or

import copy
b = []
row = [0]*7
for _ in range(6):
    b.append(copy.copy(row))

Thank you very much I got the point! Appreciated! – Nick Reve Jul 26 '23 at 19:42 — Nick Reve, Jul 26 '23 at 19:42

score -1 · Answer 2 · answered Jul 25 '23 at 20:01

Here's a better way to create B:

B=[]
for i in range(6):
    B.append( [int(j > 7) for j in A[i]] )

Here, there is no pre-allocation. We create the B array one item at a time. Remember that True is 1 and False is 0 in Python, so we can just do the comparison and convert the result to integer.

Array preallocation Python

2 Answers2