Pass shape tuple to Numpy `random.rand`

Question

I am using np.random.rand to create a matrix/tensor of the desired shape. However, this shape argument (generated in runtime) is a tuple such as: (2, 3, 4). How can we use this shape in np.random.rand?

Doing np.random.rand(shape) doesn't work and would give the following error:

TypeError: 'tuple' object cannot be interpreted as an index

Answer 1

You can unpack your shape tuple using * for example

>>> shape = (2,3,4)
>>> np.random.rand(*shape)
array([[[ 0.20116981,  0.74217953,  0.52646679,  0.11531305],
        [ 0.03015026,  0.3853678 ,  0.60093178,  0.20432243],
        [ 0.66351518,  0.45499515,  0.7978615 ,  0.92803441]],

       [[ 0.92058567,  0.27187654,  0.84221945,  0.19088589],
        [ 0.83938788,  0.53997298,  0.45754298,  0.36799766],
        [ 0.35040683,  0.62268483,  0.66754818,  0.34045979]]])

Answer 2

You can also use np.random.random_sample() which accepts the shape as a tuple and also draws from the same half-open interval [0.0, 1.0) of uniform distribution.

In [458]: shape = (2,3,4)

In [459]: np.random.random_sample(shape)
Out[459]: 
array([[[ 0.94734999,  0.33773542,  0.58815246,  0.97300734],
        [ 0.36936276,  0.03852621,  0.46652389,  0.01034777],
        [ 0.81489707,  0.1233162 ,  0.94959208,  0.80185651]],

       [[ 0.08508461,  0.1331979 ,  0.03519763,  0.529272  ],
        [ 0.89670103,  0.7133721 ,  0.93304961,  0.58961471],
        [ 0.27882714,  0.39493349,  0.73535478,  0.65071109]]])

---

In fact, if you see the NumPy notes about np.random.rand, it states :

This is a convenience function. If you want an interface that takes a shape-tuple as the first argument, refer to np.random.random_sample .

Answer 3

Realised that this was a case of argument unpacking and thus would need to use * operator. Here is the minimal working example.

import numpy as np
shape = (2, 3, 4)
H = np.random.rand(*shape)

The following StackOverflow answer has more details on the working of the star operator.