Initialize empty matrix in Python

Question

I am trying to convert a MATLAB code in Python. I don't know how to initialize empty matrix in Python.

MATLAB Code:

demod4(1) = [];

I tried in Python

demod4[0] = array([])

but it gives error:

only length-1 arrays can be converted to Python scalars

Answer 1

If you are using numpy arrays, you initialize to 0, by specifying the expected matrix size:

import numpy as np
d = np.zeros((2,3))

>>> d
    [[ 0.  0.  0.]
     [ 0.  0.  0.]]

This would be the equivalent of MATLAB 's:

d = zeros(2,3);

You can also initialize an empty array, again using the expected dimensions/size

d = np.empty((2,3))

If you are not using numpy, the closest somewhat equivalent to MATLAB's d = [] (i.e., a zero-size matrix) would be using an empty list and then

append values (for filling a vector)

d = []
d.append(0)
d.append(1)
>>> d                                                                     
[0, 1]

or append lists (for filling a matrix row or column):

d = []                                                                
d.append(range(0,2))                                                    
d.append(range(2,4))                                                  
>>> d                                                                     
[[0, 1], [2, 3]]

See also:

initialize a numpy array (SO)

NumPy array initialization (fill with identical values) (SO)

How do I create an empty array and then append to it in NumPy? (SO)

NumPy for MATLAB users

Answer 2

You could use a nested list comprehension:

# size of matrix n x m
matrix = [ [ 0 for i in range(n) ] for j in range(m) ]

Answer 3

What about initializing a list, populating it, then converting to an array.

demod4 = []  

Or, you could just populate at initialization using a list comprehension

demod4 = [[func(i, j) for j in range(M)] for i in range(N)]

Or, you could initialize an array of all zeros if you know the size of the array ahead of time.

demod4 = [[0 for j in range(M)] for i in range(N)]

or

demod4 = [[0 for i in range(M)]*N]

Or try using numpy.

import numpy as np

N, M = 100, 5000
np.zeros((N, M))

Answer 4

To init matrix with M rows and N columns you can use following pattern:

M = 3
N = 2
matrix = [[0] * N for _ in range(M)]

Answer 5

If you want to initialize the matrix with 0s then use the below code

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

Answer 6

M=[]
n=int(input())
m=int(input())
for j in range(n):
   l=[]
   for k in range(m):
       l.append(0)
   M.append(l)
print(M)

This is the traditional way of doing it matrix[m,n], However, python offers many cool ways of doing so as mentioned in other answers.

Answer 7

i find that this method is the easies method to create a matrix

    rows = 3
    columns = 4
    matrix = [[0] * columns] * rows

my output:

---

    [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

if you want to print the matrix use this:

    for i in range(rows):
         for j in range(columns):
             print(matrix[i][j], end=' ')
         print()

my output:

   0 0 0 0 
   0 0 0 0 
   0 0 0 0 

Answer 8

rows = 3
columns = 2
M = [[0]*columns]*rows

Or you could also use '' instead of 0

print(M)

Output:

M = [[0, 0], [0, 0], [0, 0]]