Dissolve numpy array into binary array

Question

I have a numpy array

[0 0 0 0 0 0 0 1 1 2 2 2 2 2 1 1 0 0 0 0 0 0 0 0]

which I want to convert/dissolve into

[[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]]

my current approach is to first use a while loop to split the array into just 1s and then create an array based on np.where(x>0). I however believe that this is not the most efficient and elegant numpy solution. any ideas on how to improve this?

source = np.array([0., 0., 0., 0., 0., 0., 0., 1., 1., 2., 2., 2., 2.,
                   2., 1., 1., 0., 0., 0., 0., 0., 0., 0., 0.], dtype=np.int)

diss = None
while np.any(source):
    row = np.greater(source, 0).astype(np.int)
    if diss is None:
        diss = row
    else:
        diss = np.vstack([diss, row])
    source -= row

idx = np.where(diss > 0)
result = np.zeros((0,source.shape[0]), dtype=np.int)
for x, y in zip(*idx):
    row = np.zeros(source.shape, dtype=np.int)
    row[y] = 1
    result = np.vstack([result, row])

No, the order is not critical. The important thing is that any value >1 gets broken down into ones. — HTTPeter
– HTTPeter, Commented Apr 7, 2014 at 11:33

Warren Weckesser · Accepted Answer · 2014-04-07 12:33:17Z

2

Here's one way:

In [38]: x = np.array([0,0,0,0,0,1,1,2,2,2,1,1,0,0])

In [39]: n = x.sum()

In [40]: rows = np.arange(n)

In [41]: positions = np.nonzero(x)[0]

In [42]: cols = np.repeat(positions, x[positions])

In [43]: result = np.zeros((n, len(x)), dtype=int)

In [44]: result[rows, cols] = 1

In [45]: result
Out[45]: 
array([[0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]])

edited Apr 7, 2014 at 12:33

answered Apr 7, 2014 at 12:21

Warren Weckesser

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1 Comment

HTTPeter Over a year ago

I've timed both possible answers, and this one is faster: 10000 loops, best of 3: 21.8 µs per loop vs 10000 loops, best of 3: 36.2 µs per loop

askewchan · Accepted Answer · 2014-04-07 12:18:08Z

2

For your example, this is about 5 times faster, and it doesn't damage source.

source = np.array([0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 2, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], int)
m, h, w = source.max(), source.sum(), len(source)
i = np.concatenate([np.nonzero(source>i)[0] for i in xrange(m)])
result = np.zeros((h,w), int)
result[range(h), i] = 1

There is still a loop of length source.max() so if that is large (for the example it's only two) perhaps something better can be done.

edited Apr 7, 2014 at 12:18

answered Apr 7, 2014 at 12:13

askewchan

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2 Comments

user2357112 Over a year ago

source-i>0 seems like it'd be better written as source>i.

askewchan Over a year ago

Thanks @user2357112, it started as np.nonzero(source - i) before I realized it was counting negative values. That actually saves about 10% of the time :P

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