Question

OK, I think this will be fairly simple, but my numpy-fu is not quite strong enough. I’ve got a an array A of ints; it’s tiled N times. I want a running count of the number of times each element is used.

For example, the following (I’ve reshaped the array to make the repetition obvious):

[0, 1, 2, 0, 0, 1, 0] \
[0, 1, 2, 0, 0, 1, 0] ...

would become:

[0, 0, 0, 1, 2, 1, 3] \
[4, 2, 1, 5, 6, 3, 7]

This python code does it, albeit inelegantly and slowly:

def running_counts(ar):
    from collections import defaultdict
    counts = defaultdict(lambda: 0)
    def get_count(num):
        c = counts[num]
        counts[num] += 1
        return c
    return [get_count(num) for num in ar]

I can almost see a numpy trick to make this go, but not quite.

Update

Ok, I’ve made improvements, but still rely on the above running_counts method. The following speeds things up and feels right-track-ish to me:

def sample_counts(ar, repititions):
    tile_bins = np.histogram(ar, np.max(ar)+1)[0]
    tile_mult = tile_bins[ar]
    first_steps = running_counts(ar)
    tiled = np.tile(tile_mult, repititions).reshape(repititions, -1)
    multiplier = np.reshape(np.arange(repititions), (repititions, 1))
    tiled *= multiplier
    tiled += first_steps
    return tiled.ravel()

Any elegant thoughts to get rid of running_counts()? Speed is now OK; it just feels a little inelegant.

Answer 1

Here’s my take on it:

def countify2(ar):
    ar2 = np.ravel(ar)
    ar3 = np.empty(ar2.shape, dtype=np.int32)
    uniques = np.unique(ar2)
    myarange = np.arange(ar2.shape[0])
    for u in uniques:
        ar3[ar2 == u] = myarange
    return ar3

This method is most effective when there are many more elements than there are unique elements.

Yes, it is similar to Sven’s, but I really did write it up long before he posted. I just had to run somewhere.