CODEWITHSUNDEEP
pythonnumpyopencvpython-imaging-library
Adrien Nivaggioli
Adrien Nivaggioli

Reputation: 365

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

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 !

Upvotes: 4

Views: 6358

Answers (2)

Paul Panzer
Paul Panzer

Reputation: 53089

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))

Upvotes: 3

Daniel F
Daniel F

Reputation: 14399

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)

Upvotes: 8

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