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Initialize Matrix in Python

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  • Difficulty Level : Basic
  • Last Updated : 19 Feb, 2022
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There are many ways to declare a 2 dimensional array with given number of rows and columns. Let us look at some of them and also at the small but tricky catches that accompany it. 
We can do it using list comprehension, concatenation feature of * operator and few other ways.

Method 0: 2 list comprehensions

Python3




rows = 3
cols = 2
 
mat = [[0 for _ in range(cols)] for _ in range(rows)]
print(f'matrix of dimension {rows} x {cols} is {mat}')
 
# editing the individual elements
mat[0][0], mat[0][1] = 1,2
mat[1][0], mat[1][1] = 3,4
mat[2][0], mat[2][1] = 5,6
print(f'modified matrix is {mat}')
 
# checking the memory address of first element of a row
print(f'addr(mat[0][0]) = {id(mat[0][0])}, addr(mat[0][1]) = {id(mat[0][1])}')
print(f'addr(mat[1][0]) = {id(mat[1][0])}, addr(mat[1][1]) = {id(mat[1][1])}')
print(f'addr(mat[2][0]) = {id(mat[2][0])}, addr(mat[2][1]) = {id(mat[2][1])}')

Output

matrix of dimension 3 x 2 is [[0, 0], [0, 0], [0, 0]]
modified matrix is [[1, 2], [3, 4], [5, 6]]
addr(mat[0][0]) = 11094304, addr(mat[0][1]) = 11094336
addr(mat[1][0]) = 11094368, addr(mat[1][1]) = 11094400
addr(mat[2][0]) = 11094432, addr(mat[2][1]) = 11094464

Method 1: 1 list comprehension  inside and  1 concatenation operation outside

Python3




rows = 3
cols = 2
 
mat = [[0 for _ in range(cols)]]*rows
print(f'matrix with dimension {rows} x {cols} is {mat}')
 
# editing the individual elements
mat[0][0], mat[0][1] = 1,2
mat[1][0], mat[1][1] = 3,4
mat[2][0], mat[2][1] = 5,6
print(f'modified matrix is {mat}')
 
# checking the memory address of first element of a row
print(f'addr(mat[0][0]) = {id(mat[0][0])}, addr(mat[0][1]) = {id(mat[0][1])}')
print(f'addr(mat[1][0]) = {id(mat[1][0])}, addr(mat[1][1]) = {id(mat[1][1])}')
print(f'addr(mat[2][0]) = {id(mat[2][0])}, addr(mat[2][1]) = {id(mat[2][1])}')

Output

matrix with dimension 3 x 2 is [[0, 0], [0, 0], [0, 0]]
modified matrix is [[5, 6], [5, 6], [5, 6]]
addr(mat[0][0]) = 11094432, addr(mat[0][1]) = 11094464
addr(mat[1][0]) = 11094432, addr(mat[1][1]) = 11094464
addr(mat[2][0]) = 11094432, addr(mat[2][1]) = 11094464

Method 2: 1 list comprehension  outside and  1 concatenation operation inside

Python3




rows = 3
cols = 2
 
mat = [[0]*cols for _ in range(rows)]
print(f'matrix with dimension {rows} x {cols} is {mat}')
 
# editing the individual elements
mat[0][0], mat[0][1] = 1,2
mat[1][0], mat[1][1] = 3,4
mat[2][0], mat[2][1] = 5,6
print(f'modified matrix is {mat}')
 
# checking the memory address of first element of a row
print(f'addr(mat[0][0]) = {id(mat[0][0])}, addr(mat[0][1]) = {id(mat[0][1])}')
print(f'addr(mat[1][0]) = {id(mat[1][0])}, addr(mat[1][1]) = {id(mat[1][1])}')
print(f'addr(mat[2][0]) = {id(mat[2][0])}, addr(mat[2][1]) = {id(mat[2][1])}')

Output

matrix with dimension 3 x 2 is [[0, 0], [0, 0], [0, 0]]
modified matrix is [[1, 2], [3, 4], [5, 6]]
addr(mat[0][0]) = 11094304, addr(mat[0][1]) = 11094336
addr(mat[1][0]) = 11094368, addr(mat[1][1]) = 11094400
addr(mat[2][0]) = 11094432, addr(mat[2][1]) = 11094464

Method 3: 2 concatenation operations

Python3




rows = 3
cols = 2
 
mat = [[0]*cols]*rows
print(f'matrix with dimension {rows} x {cols} is {mat}')
 
# editing the individual elements
mat[0][0], mat[0][1] = 1,2
mat[1][0], mat[1][1] = 3,4
mat[2][0], mat[2][1] = 5,6
print(f'modified matrix is {mat}')
 
# checking the memory address of first element of a row
print(f'addr(mat[0][0]) = {id(mat[0][0])}, addr(mat[0][1]) = {id(mat[0][1])}')
print(f'addr(mat[1][0]) = {id(mat[1][0])}, addr(mat[1][1]) = {id(mat[1][1])}')
print(f'addr(mat[2][0]) = {id(mat[2][0])}, addr(mat[2][1]) = {id(mat[2][1])}')

Output

matrix with dimension 3 x 2 is [[0, 0], [0, 0], [0, 0]]
modified matrix is [[5, 6], [5, 6], [5, 6]]
addr(mat[0][0]) = 11094432, addr(mat[0][1]) = 11094464
addr(mat[1][0]) = 11094432, addr(mat[1][1]) = 11094464
addr(mat[2][0]) = 11094432, addr(mat[2][1]) = 11094464

Here we can see that output of Method 1 & Method 3 are *unexpected*. 

We expected all rows in mat to be all different after we assigned them with 1,2,3,4,5,6 respectively. But in Method1 & Method3 they are all equal to [5,6]. This shows that essentially mat[0],mat[1] & mat[2] are all referencing the same memory which can further be seen by checking their addresses using the id function in python. 

 Hence be very careful while using (*)  operator

To understand it further we can use 3 dimensional arrays to and there we will have 2^3 possibilities of arranging list comprehension and concatenation operator.  This is an exercise I leave for the reader to perform.

If working with numpy then we can do it using reshape method.

Python3




import numpy as np
 
rows = 3
cols = 2
size = rows*cols
 
mat = np.array([0]*size).reshape(rows,cols)


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