Assign value to an individual cell in a two dimensional python array

Question

Let's say I have the following empty two dimensional array in Python:

q = [[None]*5]*4

I want to assign a value of 5 to the first row in the first column of q. Instinctively, I do the following:

q[0][0] = 5

However, this produces:

 [[5, None, None, None, None], 
  [5, None, None, None, None], 
  [5, None, None, None, None], 
  [5, None, None, None, None]]

The first element of every array is being initialized to 5, where I thought only the first element of the first array would get the update. I have two questions:

1. Why is Python initializing the first value of every array and not just the first one?
2. Is there a better way to accomplish what I'm trying to do?

Answer 1

This doesn't do what you hoped.

q = [[None]*5]*4

It reuses list objects multiple times. As you can see when you made a change to one cell, which was in a reused list object.

A single list with a value of [None] is used five times.

A single list with a value of [[None]*5] is used four times.

q = [ [ None for i in range(5) ] for j in range(4) ]

Might be more what you're looking for.

This explicitly avoids reusing a list object.

80% of the time, a dictionary is what you really wanted.

q = {}
q[0,0]= 5

Will also work. You don't start with a pre-defined grid of None values. But it's rare to need them in the first place.

In Python 2.7 and higher, you can do this.

q = { (i,j):0 for i in range(5) for j in range(4) }

That will build a grid indexed by 2-tuples.

{(0, 1): 0, (1, 2): 0, (3, 2): 0, (0, 0): 0, (3, 3): 0, (3, 0): 0, (3, 1): 0, (2, 1): 0, (0, 2): 0, (2, 0): 0, (1, 3): 0, (2, 3): 0, (4, 3): 0, (2, 2): 0, (1, 0): 0, (4, 2): 0, (0, 3): 0, (4, 1): 0, (1, 1): 0, (4, 0): 0}

Answer 2

The reason why is you have the list, just duplicated four times! Python isn't regenerating that list every time when you do *4. It's using the same list object.

To get around this, you need for force python to regenrate that list for you every time:

[ [None] * 5 for i1 in range(4) ]

In this case, I'm using a list comprehension.

Answer 3

q = [[None]*5]*4 print(q) q[0][1]=4 print(q)

print('right solution') q = [ [ None for i in range(5) ] for j in range(4) ] q[1][1]=4 print(q)

result :

[[None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None]] [[None, 4, None, None, None], [None, 4, None, None, None], [None, 4, None, None, None], [None, 4, None, None, None]] right solution [[None, None, None, None, None], [None, 4, None, None, None], [None, None, None, None, None], [None, None, None, None, None]]

Answer 4

The answer is simple Never use

q = [[None]*5]*4

as when you do assignment

q[0][1]=5 it assigns value multiple time to multiple rows at 1 column try print(q)

rather use

q = { (i,j):0 for i in range(5) for j in range(4) }

then q[0][1]=5 will assign one time only try

print(q)

Answer 5

Why is Python initializing the first value of every array and not just the first one?

Because they are the same array, referred to multiple times.

Is there a better way to accomplish what I'm trying to do?

Create the structure such that the outer array refers to separate inner arrays instead of reusing one. The other answers provide ways to do so.

Answer 6

The answer to question number 2: Using numpy is an option. See following code.

import numpy as np

# creating 2D array with nans
num_of_rows = 5
num_of_cols = 3
a = np.full((num_of_rows, num_of_cols), np.nan) 
#for zero vals: a = np.zeros((num_of_rows, num_of_cols))

# placing number 5 in row 3, col 1
value = [5]
position_row = 3
position_col = 1
# the put command below flattens the 2D array
position = [int(num_of_cols * position_row + position_col)] 
np.put(a, position, value)

result:

[[ nan  nan  nan]
 [ nan  nan  nan]
 [ nan  nan  nan]
 [ nan   5.  nan]
 [ nan  nan  nan]]

Answer 7

In case you want to use a list and not a dictionary as the others propose you can use this:

q[0] = [5,None,None,None,None]