Is there a difference between the input paramaters of numpy.random.choice and random.choice?
numpy, random
Solution
Well `np.random.choice` as noted in the docs, expects a 1D array and your input when expressed as an array would be `2D`. So, it won't work simply like that.
To make it work, we can feed in the length of the input and let it select one index, which when indexed into the input would be the equivalent one from `random.choice`, as shown below -
out = a[np.random.choice(len(a))] # a is input
Sample run -
In [74]: a = [(1,2),(4,3),(6,9)]
In [75]: a[np.random.choice(len(a))]
Out[75]: (6, 9)
In [76]: a[np.random.choice(len(a))]
Out[76]: (1, 2)
Alternatively, we can convert the input to a 1D array of object dtype and that would allow us to directly use `np.random.choice`, as shown below -
In [131]: a0 = np.empty(len(a),dtype=object)
In [132]: a0[:] = a
In [133]: a0.shape
Out[133]: (3,) # 1D array
In [134]: np.random.choice(a0)
Out[134]: (6, 9)
In [135]: np.random.choice(a0)
Out[135]: (4, 3)
Problem
Why does numpy.random.choice not work the same as random.choice? When I do this : ``` >>> random.choice([(1,2),(4,3)]) (1, 2) ``` It works. But when I do this: ``` >>> np.random.choice([(1,2), (3,4)]) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "mtrand.pyx", line 1393, in mtrand.RandomState.choice (numpy/random/mtrand/mtrand.c:15450) ValueError: a must be 1-dimensional ``` How do I achieve the same behavior as random.choice() in numpy.random.choice()?