How to input matrix (2D list) in Python?

Question

I tried to create this code to input an m by n matrix. I intended to input [[1,2,3],[4,5,6]] but the code yields [[4,5,6],[4,5,6]. Same things happ

Answer 1

row=list(map(int,input().split())) #input no. of row and column
b=[]
for i in range(0,row[0]):
    print('value of i: ',i)
    a=list(map(int,input().split()))
    print(a)
    b.append(a)
print(b)
print(row)

Output:

2 3

value of i:0
1 2 4 5
[1, 2, 4, 5]
value of i:  1
2 4 5 6
[2, 4, 5, 6]
[[1, 2, 4, 5], [2, 4, 5, 6]]
[2, 3]

Note: this code in case of control.it only control no. Of rows but we can enter any number of column we want i.e row[0]=2 so be careful. This is not the code where you can control no of columns.

Answer 2

I used numpy library and it works fine for me. Its just a single line and easy to understand. The input needs to be in a single size separated by space and the reshape converts the list into shape you want. Here (2,2) resizes the list of 4 elements into 2*2 matrix. Be careful in giving equal number of elements in the input corresponding to the dimension of the matrix.

import numpy as np
a=np.array(list(map(int,input().strip().split(' ')))).reshape(2,2)

print(a)

Input

array([[1, 2],
       [3, 4]])

Output

Answer 3

If your matrix is given in row manner like below, where size is s*s here s=5 5 31 100 65 12 18 10 13 47 157 6 100 113 174 11 33 88 124 41 20 140 99 32 111 41 20

then you can use this

s=int(input())
b=list(map(int,input().split()))
arr=[[b[j+s*i] for j in range(s)]for i in range(s)]

your matrix will be 'arr'

Answer 4

you can accept a 2D list in python this way ...

simply

arr2d = [[j for j in input().strip()] for i in range(n)] 
# n is no of rows

for characters

n = int(input().strip())
m = int(input().strip())
a = [[0]*n for _ in range(m)]
for i in range(n):
    a[i] = list(input().strip())
print(a)

or

n = int(input().strip())
n = int(input().strip())
a = []
for i in range(n):
    a[i].append(list(input().strip()))
print(a)

for numbers

n = int(input().strip())
m = int(input().strip())
a = [[0]*n for _ in range(m)]
for i in range(n):
    a[i] = [int(j) for j in input().strip().split(" ")]
print(a)

where n is no of elements in columns while m is no of elements in a row.

In pythonic way, this will create a list of list

Answer 5

Apart from the accepted answer, you can also initialise your rows in the following manner - matrix[i] = [0]*n

Therefore, the following piece of code will work -

m = int(input('number of rows, m = '))
n = int(input('number of columns, n = '))
matrix = []
# initialize the number of rows
for i in range(0,m):
    matrix += [0]
# initialize the matrix
for i in range (0,m):
    matrix[i] = [0]*n
for i in range (0,m):
    for j in range (0,n):
        print ('entry in row: ',i+1,' column: ',j+1)
        matrix[i][j] = int(input())
print (matrix)

Answer 6

If you want to take n lines of input where each line contains m space separated integers like:

1 2 3
4 5 6 
7 8 9 

Then you can use:

a=[] // declaration 
for i in range(0,n):   //where n is the no. of lines you want 
 a.append([int(j) for j in input().split()])  // for taking m space separated integers as input

Then print whatever you want like for the above input:

print(a[1][1]) 

O/P would be 5 for 0 based indexing