How to create array from input range

Question

If input is 1 then out is [ [1] ]
If input is 2 then out is [ [1,2],[3,4] ] 2x2 array
if input is 3 then out is [ [1,2,3],[4,5,6],[7,8,9] ] 3x3 array

with respect to number the value will be keep on increasing

# A is input
A = 3
matrix = [[0]*A for i in range(A)]
# matrix = []
for i in range(A):
    for j in range(A):
        matrix.append(matrix[i][j])
matrix

I tried with below code

matrix = [[i+1 for i in range(3)] for i in range(3)]
matrix

I am getting [[1, 2, 3], [1, 2, 3], [1, 2, 3]]

Could this help you - stackoverflow.com/questions/6667201/… — Daniel Hao
– Daniel Hao, Commented Oct 10, 2021 at 11:08

Melvin Abraham · Accepted Answer · 2021-10-10 11:16:18Z

The code is almost correct except the line matrix.append(matrix[i][j]). Let's dig a bit...

Consider the line
```
matrix = [[0]*A for i in range(A)]
```
Here a new matrix of the shape A x A is created which is what's expected in the final answer.

Then we have the two for loops which are iterating through each row and each column.

for i in range(A):      # Iterating through each rows
    for j in range(A):  # Iterating through each columns
        ...

Now, let's look at the erroneous line
```
matrix.append(matrix[i][j])
```
This line adds new values at the end of a perfectly fine matrix which distorts the final shape of the matrix.

What you instead need to do is assign values in the corresponding row and column. Effectively, something like this:

matrix[i][j] = ...

Now, on the right hand side of the above expression goes (i * A) + (j + 1). So you can rewrite your program as follows:

A = 3

# Initialize the matrix
matrix = [[0]*A for i in range(A)]

for i in range(A):
    for j in range(A):
        # Replace the value at the corresponding row and column
        matrix[i][j] = (i * A) + (j + 1)

print(matrix)

Output:

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

This is a pure Python solution, thus, it does not require any external library.

itismoej · Accepted Answer · 2021-10-10 11:14:44Z

1

If you don't want to use numpy, you can use list comprehension:

In [1]: a = 3

In [2]: [[j for j in range(((i-1)*a)+1, (a*i)+1)] for i in range(1, a+1)]
Out[2]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

answered Oct 10, 2021 at 11:14

itismoej

1,85616 silver badges20 bronze badges

Comments

yascho · Accepted Answer · 2021-10-10 12:22:49Z

1

Splitting n^2 numbers into n bins:

np.split(np.arange(n**2)+1, n)

In detail:

import numpy as np
for n in range(1,4):
    print(np.split(np.arange(n**2)+1, n))

Output:

[array([1])]
[array([1, 2]), array([3, 4])]
[array([1, 2, 3]), array([4, 5, 6]), array([7, 8, 9])]

edited Oct 10, 2021 at 12:22

answered Oct 10, 2021 at 11:09

yascho

3121 silver badge6 bronze badges

3 Comments

abd Over a year ago

can we do it in python native method

yascho Over a year ago

Sure, please see other answers. This solution is shorter though.

yascho Over a year ago

Additionally, this solution better reads what you actually want: to split n^2 numbers into n bins.

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