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

Creating matrix with prepopulated numbers can be done with list comprehension. It may be hard to read but it gets job done:

rows = int(input('Number of rows: '))
cols = int(input('Number of columns: '))
matrix = [[i + cols * j for i in range(1, cols + 1)] for j in range(rows)]

with 2 rows and 3 columns matrix will be [[1, 2, 3], [4, 5, 6]], with 3 rows and 2 columns matrix will be [[1, 2], [3, 4], [5, 6]] etc.

Answer 2

The problem is on the initialization step.

for i in range (0,m):
  matrix[i] = columns

This code actually makes every row of your matrix refer to the same columns object. If any item in any column changes - every other column will change:

>>> for i in range (0,m):
...     matrix[i] = columns
... 
>>> matrix
[[0, 0, 0], [0, 0, 0]]
>>> matrix[1][1] = 2
>>> matrix
[[0, 2, 0], [0, 2, 0]]

You can initialize your matrix in a nested loop, like this:

matrix = []
for i in range(0,m):
    matrix.append([])
    for j in range(0,n):
        matrix[i].append(0)

or, in a one-liner by using list comprehension:

matrix = [[0 for j in range(n)] for i in range(m)]

or:

matrix = [x[:] for x in [[0]*n]*m]

See also:

• How to initialize a two-dimensional array in Python?

Hope that helps.

Answer 3

a,b=[],[]
n=int(input("Provide me size of squre matrix row==column : "))
for i in range(n):
   for j in range(n):
      b.append(int(input()))
    a.append(b)
    print("Here your {} column {}".format(i+1,a))
    b=[]
for m in range(n):
    print(a[m])

works perfectly

Answer 4

You can make any dimension of list

list=[]
n= int(input())
for i in range(0,n) :
    #num = input()
    list.append(input().split())
print(list)

output: