speeden up function using numpy arrays

Question

from numba import jit

@jit
def interim_mk(x, unique_x):
    """

    :param x:
    :param unique_x:
    :return:
    """
    tp = np.zeros(unique_x.shape)

    for i in range(len(unique_x)):
        tp[i] = sum(x == unique_x[i])

    return tp

In the function above, I used jit to try and speeden it up. However, it does not seem to help. Both x and unique_x are numpy arrays, is there a way to speeden up this calculation (without using cython)

For more meaningful answers, it would be helpful to clarify the intent of your function. — cs95
– cs95, Commented Dec 11, 2017 at 0:21

Divakar · Accepted Answer · 2017-12-11 00:28:14Z

2

For case with positive numbers

You can use np.bincount -

count = np.bincount(x)
out = count[count!=0]

Using unique_x -

out = np.bincount(x)[unique_x]

For generic case

out = np.bincount(np.searchsorted(unique_x, x))

Of course, we could have gotten the counts directly from np.unique call, if that's how we had unique_x -

out = np.unique(x, return_counts=1)[1]

edited Dec 11, 2017 at 0:28

answered Dec 11, 2017 at 0:15

Divakar

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speeden up function using numpy arrays

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For case with positive numbers

For generic case

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