Efficient way to count every pixels of an image of a particular value in Python?

Question

I have an RGB image composed of 7 different possible colors. I want to count how many of each pixel type is present in the image, in an efficient way. So not a loop on every pixels if possible, at least not manually (numpy operation is ok beacause it's way faster)

I tried loading it into a numpy array, which gives me a N*M*3 array, but I can't figure out a way to count the pixels of a particular value... Any ideas?

Thank you !

Possible duplicate of Count the number of pixels by color from an image loaded into a numpy array — Brenlla
– Brenlla, Commented Oct 4, 2018 at 11:22
Are the seven colors fixed? Do you need to know all color channels to decide which color? Because if for example the red (or any of the other two) channel were enough to discriminate between all seven colors, that would make things much easier. — Paul Panzer
– Paul Panzer, Commented Oct 4, 2018 at 11:24
@Brenlla I think it's not a direct dupe, because the small number of colors used may allow for some optimizations. — Paul Panzer
– Paul Panzer, Commented Oct 4, 2018 at 11:31

Daniel F · Accepted Answer · 2018-10-04 11:49:10Z

8

Just partially flatten and use np.unique with return_counts = True and axis = 0

flat_image = image.reshape(-1, 3)  # makes one long line of pixels
colors, counts = np.unique(flat_image, return_counts = True, axis = 0)

Or as one line:

colors, counts = np.unique(image.reshape(-1, 3), 
                           return_counts = True, 
                           axis = 0)

answered Oct 4, 2018 at 11:49

Daniel F

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Adrien Nivaggioli Over a year ago

Thank you! I've come up with a solution a bit similar but more complicated... This is perfect !

Paul Panzer · Accepted Answer · 2018-10-04 14:57:58Z

Since there are only seven colors simple masking will under reasonable assumptions be quite competitive. Timings below are for 100x100x3 @ 8bit random images:

timings
np.unique 6.510251379047986
masking   0.2401340039796196

Note that much but not all of the speedup is due to merging the channels into a single one.

Code:

import numpy as np

def create(M, N, k=7):
    while True:
        colors = np.random.randint(0, 256**3, (k,), dtype=np.int32)
        if np.unique(colors).size == k:
            break
    picture = colors[np.random.randint(0, k, (M, N))]
    RGB = np.s_[..., :-1] if picture.dtype.str.startswith('<') else np.s_[..., 1:]
    return picture.view(np.uint8).reshape(*picture.shape, 4)[RGB]

def f_df(image):
    return np.unique(image.reshape(-1, 3), 
                     return_counts = True, 
                     axis = 0)

def f_pp(image, nmax=50):
    iai32 = np.pad(image, ((0, 0), (0, 0), (0, 1)), mode='constant')
    iai32 = iai32.view(np.uint32).ravel()

    colors = np.empty((nmax+1,), np.uint32)
    counts = np.empty((nmax+1,), int)
    colors[0] = iai32[0]
    counts[0] = 0
    match = iai32 == colors[0]
    for i in range(1, nmax+1):
        counts[i] = np.count_nonzero(match)
        if counts[i] == iai32.size:
            return colors.view(np.uint8).reshape(-1, 4)[:i, :-1], np.diff(counts[:i+1])
        colors[i] = iai32[match.argmin()]
        match |= iai32 == colors[i]
    raise ValueError('Too many colors')



image = create(100, 100, 7)

col_df, cnt_df = f_df(image)
col_pp, cnt_pp = f_pp(image)
#print(col_df)
#print(cnt_df)
#print(col_pp)
#print(cnt_pp)
idx_df = np.lexsort(col_df.T)
idx_pp = np.lexsort(col_pp.T)

assert np.all(cnt_df[idx_df] == cnt_pp[idx_pp])

from timeit import timeit
print('timings')
print('np.unique', timeit(lambda: f_df(image), number=1000))
print('masking  ', timeit(lambda: f_pp(image), number=1000))

Anecdotally, this code was a lot faster for me than np.unique. Thanks Paul!

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