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 happen when I input other m by n matrix, the code yields an m by n matrix whose rows are identical.

Perhaps you can help me to find what is wrong with my code.

m = int(input('number of rows, m = '))
n = int(input('number of columns, n = '))
matrix = []; columns = []
# initialize the number of rows
for i in range(0,m):
  matrix += [0]
# initialize the number of columns
for j in range (0,n):
  columns += [0]
# initialize the matrix
for i in range (0,m):
  matrix[i] = columns
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)

Related: [Creating a 2D array from a single list of input integers separated by space](https://stackoverflow.com/questions/47552154/creating-a-2d-array-from-a-single-list-of-input-integers-separated-by-space) — Georgy, Jul 12 '19 at 10:10
Related: [How to create a 2D list from user input with separate prompts for each column](https://stackoverflow.com/questions/57002617/how-to-create-a-2d-list-from-user-input-with-separate-prompts-for-each-column) — Georgy, Jul 12 '19 at 10:19

score 20 · Accepted Answer · edited May 23 '17 at 12:00

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.

Ani · Answer 2 · 2017-03-19T15:21:20.703

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

score 5 · Answer 3 · edited Jul 28 '19 at 14:20

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

Julkar9 · Answer 4 · 2022-03-19T17:29:57.463

5

If the input is formatted like this,

1 2 3
4 5 6
7 8 9

a one liner can be used

mat = [list(map(int,input().split())) for i in range(row)]

explanation with example:

input() takes a string as input. "1 2 3"
split() splits the string by whitespaces and returns a
list of strings. ["1", "2", "3"]
list(map(int, ...)) transforms/maps the list of strings into a list of ints. [1, 2, 3]
All these steps are done row times and these lists are stored in another list.[[1, 2, 3], [4, 5, 6], [7, 8, 9]], row = 3

edited Mar 19 '22 at 17:29

answered Oct 28 '19 at 08:15

Julkar9

1,508
1
12
24

would you please give me any reference link or explain how this statement work?? – Taufiq Ahommed Neloy Mar 19 '22 at 16:40
@TaufiqAhommedNeloy I have added an explanation. – Julkar9 Mar 19 '22 at 17:30

score 2 · Answer 5 · edited Dec 30 '18 at 12:24

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)

score 1 · Answer 6 · answered Jun 15 '18 at 14:25

This code takes number of row and column from user then takes elements and displays as a matrix.

m = int(input('number of rows, m : '))
n = int(input('number of columns, n : '))
a=[]
for i in range(1,m+1):
  b = []
  print("{0} Row".format(i))
  for j in range(1,n+1):
    b.append(int(input("{0} Column: " .format(j))))
  a.append(b)
print(a)

score 1 · Answer 7 · answered Jul 12 '18 at 06:20

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'

score 1 · Answer 8 · answered Feb 01 '19 at 08:55

1

m,n=map(int,input().split()) # m - number of rows; n - number of columns;

matrix = [[int(j) for j in input().split()[:n]] for i in range(m)]

for i in matrix:print(i)

answered Feb 01 '19 at 08:55

Max Amiri

11
1

2

Can you explain why this should answer OP question ? – executable Feb 01 '19 at 09:02

score 1 · Answer 9 · edited Oct 25 '19 at 07:45

1

no_of_rows = 3  # For n by n, and even works for n by m but just give no of rows
matrix = [[int(j) for j in input().split()] for i in range(n)]
print(matrix)

edited Oct 25 '19 at 07:45

Georgy

12,464
7
65
73

answered Oct 25 '19 at 07:07

Mahesh Kumar

11
2

score 1 · Answer 10 · edited Jun 29 '20 at 16:34

1

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:

code in shown with output

edited Jun 29 '20 at 16:34

Roberto Caboni

7,252
10
25
39

answered Jun 29 '20 at 16:17

skabeed hossain

11
1

score 0 · Answer 11 · edited Apr 02 '19 at 17:00

0

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

input

2 3
1 2 3
4 5 6

output [[1, 2, 3], [4, 5, 6]]

edited Apr 02 '19 at 17:00

Sumedh Junghare

381
2
4
21

answered Jul 07 '18 at 06:31

Sitabja Pal

1
1

score 0 · Answer 12 · edited Sep 03 '18 at 14:29

0

a = []
b = []

m=input("enter no of rows: ")
n=input("enter no of coloumns: ")

for i in range(n):
     a = []
     for j in range(m):
         a.append(input())
     b.append(a)

Input : 1 2 3 4 5 6 7 8 9

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

edited Sep 03 '18 at 14:29

Rishikesh Pathak

3
3

answered Jul 16 '18 at 06:45

Deepak Kumar

387
4
6

score 0 · Answer 13 · answered Jul 17 '18 at 04:47

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.

score 0 · Answer 14 · edited Jan 06 '19 at 15:53

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.

score 0 · Answer 15 · answered Mar 04 '19 at 16:01

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

score 0 · Answer 16 · answered May 07 '19 at 07:52

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

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

16 Answers16

Linked