place smaller array in large array while ignoring zeros

Question

i have a large array and want to place a smaller array at some offset and at the same time ignore zeros in the smaller array. what's the best way to do that?

I tried masked arrays but somehow this is slow

mask = np.ma.masked_equal(pixels, 0, False)
output[offset_y:offset_y+tile_height,offset_x:offset_x+tile_width] = np.where(mask.mask, output[offset_y:offset_y+tile_height,offset_x:offset_x+tile_width], pixels)

then i tried this

np.place(output[offset_y:offset_y+tile_height,offset_x:offset_x+tile_width], pixels>0, pixels[pixels>0])

but it is very slow

what is the fastest way to do so?

Create a new empty array of the right size, and copy values to it from both the original and new arrays. I'd try to remove the zeros from the smaller array first. — hpaulj
– hpaulj, Commented Jan 26, 2015 at 22:44
Give us some idea of the size and shape of the arrays you are dealing with. Better yet, give us some small small arrays to test. — hpaulj
– hpaulj, Commented Jan 26, 2015 at 22:44
the size of output is 30000x50000 and the pixels are 16000x16000.. — haehn
– haehn, Commented Jan 26, 2015 at 22:45
i think it should be fast with masked arrays but i am not sure why it does not work — haehn
– haehn, Commented Jan 26, 2015 at 22:46

perimosocordiae · Accepted Answer · 2015-01-27 00:06:57Z

2

I suspect regular old logical indexing will be the most efficient here:

# make a view into output with the same dimensions as `pixels`
output_subarray = output[offset_y:offset_y+tile_height,offset_x:offset_x+tile_width]

mask = pixels != 0
output_subarray[mask] = pixels[mask]

answered Jan 27, 2015 at 0:06

perimosocordiae

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