creating a binary matrix using list values-CodePudding

I am new to python and stack overflow. I need to create a binary matrix that the value takes 1 if the customers exist in a given list. For example: assume a list as [4,5,8]. i need a 10x10 matrix with all zero excepts matrix[4,5],matrix[5,4], matrix[5,8],matrix[8,5] and matrix[4,8], matrix[8,4] are 1. Is there an easy way to do this? Thank you.

CodePudding user response：

You have:

data = [4, 5, 8]

You calculate:

matrix = [[int(c in data and r in data and c != r) for c in range(10)] for r in range(10)]

You get:

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

The solution shown above uses zero-based indexing.

CodePudding user response：

You can do this with numpy as vectorize and without use of for-loop like below:

arr = [4,5,8]
idx = np.array(np.meshgrid(arr,arr)).T.reshape(-1,2)

#romove (4,4) , (5,5) , (8,8)
idx = idx[idx[:,0] != idx[:,1]]

mtx = np.zeros((10,10))

mtx[idx[:,0], idx[:,1]] = 1
print(mtx)

Output:

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