one dimension tuple to two dimension numpy array

Question

Convert a tuple into a Numpy matrix with the below conditions:

The shape of the array should be len(tuple) x len(tuple), ie a square matrix.
Elements in array at the location specified by (index of the element in the tuple, the value of the element in the tuple) should be one.

For example, I have a random tuple, like below:

# index means row ,value means col
(2,0,1)

I use two loop to change this tuple into Numpy array:

def get_np_represent(result):
    two_D = []
    for row in range(len(result)):
        one_D = []
        for col in range(len(result)):
            if result[row] == col:
                one_D.append(1)
            else:
                one_D.append(0)
        two_D.append(one_D)
    return np.array(two_D)

output:

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

But I have 10,000,000 such tuple, Is there a faster way?

Ke Yang · Accepted Answer · 2019-07-12 02:23:47Z

3

Something like this? Manipulate the matrix is quite faster than for loop.

import numpy as np

t = (2, 0, 1)
x = np.zeros([len(t),len(t)])

for i,v in enumerate(t):
    x[i, v] = 1

print(x)

outputs:

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

answered Jul 12, 2019 at 2:23

Ke Yang

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1 Comment

james Over a year ago

more Intuitive, Thanks

BENY · Accepted Answer · 2019-07-12 02:27:29Z

3

For example (setting up from Ke)

t = (2, 0, 1)
x = np.zeros([len(t),len(t)])
x[np.arange(len(x)),t]=1
x
Out[145]: 
array([[0., 0., 1.],
       [1., 0., 0.],
       [0., 1., 0.]])

answered Jul 12, 2019 at 2:27

BENY

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1 Comment

Ke Yang Over a year ago

I like this one, more concise.

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