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
Answers:
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))
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) ]
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:
NumPy array initialization (fill with identical values) (SO)
How do I create an empty array and then append to it in NumPy? (SO)
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.
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)]
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]]
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)]
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
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
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))
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) ]
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:
NumPy array initialization (fill with identical values) (SO)
How do I create an empty array and then append to it in NumPy? (SO)
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.
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)]
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]]
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)]
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