Efficiently counting number of unique elements - NumPy / Python

Question

When running np.unique(), it first flattens the array, sorts the array, then finds the unique values. When I have arrays with shape (10, 3000, 3000), it takes about a second

Answer 1

Here's a method that works for an array with dtype np.uint8 that is faster than np.unique.

First, create an array to work with:

In [128]: a = np.random.randint(1, 128, size=(10, 3000, 3000)).astype(np.uint8)

For later comparison, find the unique values using np.unique:

In [129]: u = np.unique(a)

Here's the faster method; v will contain the result:

In [130]: q = np.zeros(256, dtype=int)

In [131]: q[a.ravel()] = 1

In [132]: v = np.nonzero(q)[0]

Verify that we got the same result:

In [133]: np.array_equal(u, v)
Out[133]: True

Timing:

In [134]: %timeit u = np.unique(a)
2.86 s ± 9.02 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [135]: %timeit q = np.zeros(256, dtype=int); q[a.ravel()] = 1; v = np.nonzero(q)
300 ms ± 5.52 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

So 2.86 seconds for np.unique(), and 0.3 seconds for the alternative method.