Conversion of a set of position vectors into a matrix of ones

Here is the input that I have

{(0, 1), (2, 1), (0, 0), (1, 1), (1, 0)}

These are the positions of the ones in a matrix of size, N=5.
My desired output is

[[1,1,0,0,0],
 [1,1,0,0,0],
 [0,1,0,0,0],
 [0,0,0,0,0],
 [0,0,0,0,0]]

Is there an efficient way to do the conversion? Perhaps a builtin numpy function? Creating a matrix of zeroes then cycling through the position vectors to push in the ones doesn't seem very efficient to me.

My ultimate goal with the output is to feed it to matplotlib's matshow() function.

@Ansagara's answer does the job, but requires a for loop. It can also be done without for loops as given below. This scales better for larger positions arrays.

import numpy as np

positions = {(0, 1), (2, 1), (0, 0), (1, 1), (1, 0)}
N = 5

# Create a matrix of zeros
matrix = np.zeros((N, N), dtype=int)

# Assign 1 to the positions based on the position vectors
positions = np.asarray(list(positions))
matrix[positions[:,0], positions[:,1]] = 1

print(matrix)

Which gives

[[1 1 0 0 0]
 [1 1 0 0 0]
 [0 1 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]]

You can utilize the numpy.zeros() function to create a matrix of zeros and then assign the value of 1 to the corresponding positions based on the position vectors.

Here's an example code that demonstrates this approach:

import numpy as np

positions = {(0, 1), (2, 1), (0, 0), (1, 1), (1, 0)}
N = 5

# Create a matrix of zeros
matrix = np.zeros((N, N), dtype=int)

# Assign 1 to the positions based on the position vectors
for pos in positions:
    matrix[pos] = 1

print(matrix)

Output:

[[1 1 0 0 0]
 [1 1 0 0 0]
 [0 1 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]]

using the numpy broadcast to set value.it may be better than the for loop.<br>
firstly, let us get the each row collection of columns, it is "dic".<br>
0 2 1 is the row, 0: [1,0] is the column index of the first row index.<br><br>
2: [1] is the column index of the third row index.<br>
1: [1, 0] is the column index of the second row index.<br>
with dic, you can using the boardcast to set value in the matrix<br>

dic: {0: [1, 0], 2: [1], 1: [1, 0]}

import numpy as np

kk = [(0, 1), (2, 1), (0, 0), (1, 1), (1, 0)]
# kk = sorted(kk.__iter__(), key=lambda j:j[0])
dic = {}
for i in kk:
    if i[0] not in dic.keys():
        dic[i[0]] = [i[1]]
    else:
        dic[i[0]].append(i[1])

zer = np.zeros((6,6))
for key, val in dic.items():
    zer[key, val] = 1
zer

array([[1., 1., 0., 0., 0., 0.],
       [1., 1., 0., 0., 0., 0.],
       [0., 1., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.]])