Unique entries in columns of a 2D numpy array

Question

I have an array of integers:

import numpy as np

demo = np.array([[1, 2, 3],
                 [1, 5, 3],
                 [4, 5, 6],
                 [7, 8, 9],
                 [4, 2, 3],
                 [4, 2, 12],
                 [10, 11, 13]])

And I want an array of unique values in the columns, padded with something if necessary (e.g. nan):

[[1, 4, 7, 10, nan],
 [2, 5, 8, 11, nan],
 [3, 6, 9, 12,  13]]

It does work when I iterate over the transposed array and use a boolean_indexing solution from a previous question. But I was hoping there would be a built-in method:

solution = []
for row in np.unique(demo.T, axis=1):
    solution.append(np.unique(row))

def boolean_indexing(v, fillval=np.nan):
    lens = np.array([len(item) for item in v])
    mask = lens[:,None] > np.arange(lens.max())
    out = np.full(mask.shape,fillval)
    out[mask] = np.concatenate(v)
    return out

print(boolean_indexing(solution))

Jérôme Richard · Accepted Answer · 2021-04-25 20:27:50Z

1

AFAIK, there are no builtin solution for that. That being said, your solution seems a bit complex to me. You could create an array with initialized values and fill it with a simple loop (since you already use loops anyway).

solution = [np.unique(row) for row in np.unique(demo.T, axis=1)]

result = np.full((len(solution), max(map(len, solution))), np.nan)
for i,arr in enumerate(solution):
    result[i][:len(arr)] = arr

edited Apr 25, 2021 at 20:27

answered Apr 25, 2021 at 19:24

Jérôme Richard

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1 Comment

Alexander L. Hayes Over a year ago

Thank you, this is definitely cleaner than what I had. I hoped to avoid the loop, but numpy does tend to avoid cases where the output sizes have less-than-obvious layouts.

Kevin · Accepted Answer · 2021-04-25 23:26:29Z

If you want to avoid the loop you could do:

demo = demo.astype(np.float32) # nan only works on floats

sort = np.sort(demo, axis=0)
diff = np.diff(sort, axis=0)
np.place(sort[1:], diff == 0, np.nan)
sort.sort(axis=0)
edge = np.argmax(sort, axis=0).max()
result = sort[:edge]

print(result.T)

Output:

array([[ 1.,  4.,  7., 10., nan],
       [ 2.,  5.,  8., 11., nan],
       [ 3.,  6.,  9., 12., 13.]], dtype=float32)

Not sure if this is any faster than the solution given by Jérôme.

EDIT

A slightly better solution

demo = demo.astype(np.float32)

sort = np.sort(demo, axis=0)
mask = np.full(sort.shape, False, dtype=bool)
np.equal(sort[1:], sort[:-1], out=mask[1:])
np.place(sort, mask, np.nan)
edge = (~mask).sum(0).max()
result = np.sort(sort, axis=0)[:edge]

print(result.T)

Output:

array([[ 1.,  4.,  7., 10., nan],
       [ 2.,  5.,  8., 11., nan],
       [ 3.,  6.,  9., 12., 13.]], dtype=float32)

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