For a 3 by 3 matrix, how can I create a matrix that has values which alternate between 1s and 0s?
table = [ [ 0 for i in range(3) ] for j in range(3) ]
for row in table:
for d1 in range(3):
for d2 in range(3):
table[d1][d2]
print row
Above is edited text of the code I used to create a 3 by 3 matrix with zeros, however, I want something like this
1 0 1
0 1 0
1 0 1
A 3 by 3 matrix that alternates between 1 and zero question beforehand. Is there any way of doing this?
You could slice and reshape.
import numpy as np
n = 3
a = np.zeros(n*n, dtype=int)
a[::2]=1
a = a.reshape(n, n)
If you prefer not using numpy, Peter Wood's suggestion is nice and compact.
values
is an iterator that yields 0 and 1 forever:
>>> from itertools import cycle
>>> values = cycle([0, 1])
>>> [[next(values) for i in range(3)] for j in range(3)]
[[0, 1, 0],
[1, 0, 1],
[0, 1, 0]]
It works well when the number of columns is odd, not when it's even:
>>> values = cycle([0, 1])
>>> [[next(values) for i in range(5)] for j in range(4)]
[[0, 1, 0, 1, 0],
[1, 0, 1, 0, 1],
[0, 1, 0, 1, 0],
[1, 0, 1, 0, 1]]
>>> values = cycle([0, 1])
>>> [[next(values) for i in range(4)] for j in range(5)]
[[0, 1, 0, 1],
[0, 1, 0, 1],
[0, 1, 0, 1],
[0, 1, 0, 1],
[0, 1, 0, 1]]
If you want to avoid using numpy
:
flat_list = [1-i%2 for i in xrange(9)]
print [flat_list[i:i+3] for i in xrange(0,9,3)]
Here is a function to create a nxm 1 0 matrix. If you want to start with 0 1 sequence, consider to modify the code block a litte.
import numpy as np
def my_one_zero_matrix(n,m):
zeros_a=np.zeros((n,m))
for i in range(n):
for j in range(m):
if (i+j)%2==0:
zeros_a[i,j]=1
return zeros_a
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