I would like to create a 2D array look something like this using a loop in python:
[[1,0,0,0,0,0]
[0,1,0,0,0,0]
[0,0,1,0,0,0]
[0,0,0,1,0,0]
[0,0,0,0,1,0]
[0,0,0,0,0,1]]
1's index increase by 1 next row and rest of elements in array is filled 0.
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1I reopened the question because this is not a duplicate to the linked question, despite having a similar title. This question should be closed instead because it is too broad, without a specific question to an attempt.blhsing– blhsing2019-06-04 23:35:19 +00:00Commented Jun 4, 2019 at 23:35
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2 Answers
P.S: There are several ways to construct such a matrix. This is one way of doing it using for loops as you asked. The matrix you need is called an identity matrix
size = 6
arr = [[0]*size for _ in range(size)] # Initialize 6 lists all with zeros
# Now change the value to 1 on the diagonal enteries
for i in range(size):
arr[i][i] = 1
print (arr)
# [[1 0 0 0 0 0]
# [0 1 0 0 0 0]
# [0 0 1 0 0 0]
# [0 0 0 1 0 0]
# [0 0 0 0 1 0]
# [0 0 0 0 0 1]]
Alternative 1 using NumPy: Similar to the above case, initialize a 6x6 matrix of zeros using NumPy and then just replace 0 by 1 on the diagonal.
import numpy as np
size = 6
arr = np.zeros((size, size))
for i in range(size):
arr[i][i] = 1
print (arr)
Alternative 2
import numpy as np
size = 6
arr = np.eye(size)
Alternative 3
np.identity(6)
6 Comments
blhsing
The nested list in alternative 1 is unnecessary and inefficient.
for i in range(size): arr[i][i] = 1 would be enough.
Sheldore
@blhsing: So shall I delete it? I just tried to provide several options to give a better perspective
blhsing
You can improve it instead.
Sheldore
@blhsing : How about now? Thanks for the suggestion :) Appreciated
Sheldore
@blhsing : Oh sorry, I got confused. I see your point. Let me switch back :)
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you can run following code
arr = []
for i in range(6):
arr.append([])
for j in range(6):
if i == j:
arr[i].append(1)
else:
arr[i].append(0)
print(arr)
