Reputation: 2683
I have a python image processing function, that uses tries to get the dominant color of an image. I make use of a function I found here https://github.com/tarikd/python-kmeans-dominant-colors/blob/master/utils.py
It works, but unfortunately I don't quite understand what it does and I learned that np.histogram
is rather slow and I should use cv2.calcHist
since it's 40x faster according to this: https://docs.opencv.org/trunk/d1/db7/tutorial_py_histogram_begins.html
I'd like to understand how I have to update the code to use cv2.calcHist
, or better, which values I have to input.
My function
def centroid_histogram(clt):
# grab the number of different clusters and create a histogram
# based on the number of pixels assigned to each cluster
num_labels = np.arange(0, len(np.unique(clt.labels_)) + 1)
(hist, _) = np.histogram(clt.labels_, bins=num_labels)
# normalize the histogram, such that it sums to one
hist = hist.astype("float")
hist /= hist.sum()
# return the histogram
return hist
The pprint
of clt
is this, not sure if this helps
KMeans(algorithm='auto', copy_x=True, init='k-means++', max_iter=300,
n_clusters=1, n_init=10, n_jobs=1, precompute_distances='auto',
random_state=None, tol=0.0001, verbose=0)
My code can be found here: https://github.com/primus852/python-movie-barcode
I am a very beginner, so any help is highly appreciated.
As per request:
rgb(22,28,37)
0.021515369415283203s
Upvotes: 24
Views: 32844
Reputation: 536
Awesome answer from @Divakar
With bit shifting instead of multiplication, it might be even a little faster. It seems to have a bigger effect on large pictures with more colors
import numexpr as ne
def get_dom_color(x):
def get_color_back(code):
blue = int(code % 256)
green = int((code % (256 * 256) - blue) / 256)
red = int((code - blue - green * 256) / (256 * 256))
return red, green, blue
arr = x.reshape(-1,x.shape[-1])
outa=ne.evaluate('(c1 << 16) + (c2 << 8) + c3',global_dict={},local_dict={'c1':arr[...,0],'c2':arr[...,1],'c3':arr[...,2]})
return get_color_back(np.bincount(outa).argmax())
def bincount_numexpr_app(a):
a2D = a.reshape(-1,a.shape[-1])
col_range = (256, 256, 256) # generically : a2D.max(0)+1
eval_params = {'a0':a2D[:,0],'a1':a2D[:,1],'a2':a2D[:,2],
's0':col_range[0],'s1':col_range[1]}
a1D = ne.evaluate('a0*s0*s1+a1*s0+a2',eval_params)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
np.random.seed(0)
x = np.random.randint(0,255,(4000,4000,3))
%timeit bincount_numexpr_app(x)
%timeit get_dom_color(x)
427 ms ± 14.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
412 ms ± 7.91 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Upvotes: 0
Reputation: 107
Improved version of @Divakar's answer. Separate solution for RGBA image. Fully transparent pixels will not be counted for the RGBA image.
import numpy as np
def most_common_color_RGB(image: np.ndarray):
"""input image ndarray shape should be RGB shape, for example: (512, 512, 3)"""
a2D = image.reshape(-1, image.shape[-1])
col_range = (256, 256, 256) # generically : a2D.max(0)+1
a1D = np.ravel_multi_index(a2D.T, col_range)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
def most_common_color_RGBA(image_RGBA: np.ndarray):
"""input image ndarray shape should be RGBA shape, for example: (512, 512, 4)"""
RGB_pixels = image_RGBA.reshape(-1, 4)
# remove transparent pixels
just_non_alpha = RGB_pixels[RGB_pixels[:, 3] != 0]
if just_non_alpha.shape[0] == 0:
return False
# delete alpha channel
just_non_alpha = np.delete(just_non_alpha, 3, axis=1)
col_range = (256, 256, 256) # generically : a2D.max(0)+1
a1D = np.ravel_multi_index(just_non_alpha.T, col_range)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
Upvotes: 0
Reputation: 221714
Two approaches using np.unique
and np.bincount
to get the most dominant color could be suggested. Also, in the linked page, it talks about bincount
as a faster alternative, so that could be the way to go.
Approach #1
def unique_count_app(a):
colors, count = np.unique(a.reshape(-1,a.shape[-1]), axis=0, return_counts=True)
return colors[count.argmax()]
Approach #2
def bincount_app(a):
a2D = a.reshape(-1,a.shape[-1])
col_range = (256, 256, 256) # generically : a2D.max(0)+1
a1D = np.ravel_multi_index(a2D.T, col_range)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
Verification and timings on 1000 x 1000
color image in a dense range [0,9)
for reproducible results -
In [28]: np.random.seed(0)
...: a = np.random.randint(0,9,(1000,1000,3))
...:
...: print unique_count_app(a)
...: print bincount_app(a)
[4 7 2]
(4, 7, 2)
In [29]: %timeit unique_count_app(a)
1 loop, best of 3: 820 ms per loop
In [30]: %timeit bincount_app(a)
100 loops, best of 3: 11.7 ms per loop
Further boost
Further boost upon leveraging multi-core
with numexpr
module for large data -
import numexpr as ne
def bincount_numexpr_app(a):
a2D = a.reshape(-1,a.shape[-1])
col_range = (256, 256, 256) # generically : a2D.max(0)+1
eval_params = {'a0':a2D[:,0],'a1':a2D[:,1],'a2':a2D[:,2],
's0':col_range[0],'s1':col_range[1]}
a1D = ne.evaluate('a0*s0*s1+a1*s0+a2',eval_params)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
Timings -
In [90]: np.random.seed(0)
...: a = np.random.randint(0,9,(1000,1000,3))
In [91]: %timeit unique_count_app(a)
...: %timeit bincount_app(a)
...: %timeit bincount_numexpr_app(a)
1 loop, best of 3: 843 ms per loop
100 loops, best of 3: 12 ms per loop
100 loops, best of 3: 8.94 ms per loop
Upvotes: 24
Reputation: 9835
The equivalent code for cv2.calcHist()
is to replace:
(hist, _) = np.histogram(clt.labels_, bins=num_labels)
with
dmin, dmax, _, _ = cv2.minMaxLoc(clt.labels_)
if np.issubdtype(data.dtype, 'float'): dmax += np.finfo(data.dtype).eps
else: dmax += 1
hist = cv2.calcHist([clt.labels_], [0], None, [num_labels], [dmin, dmax]).flatten()
Note that cv2.calcHist
only accepts uint8
and float32
as element type.
It seems like opencv's and numpy's binning differs from each other as the histograms differ if the number of bins doesn't map the value range:
import numpy as np
from matplotlib import pyplot as plt
import cv2
#data = np.random.normal(128, 1, (100, 100)).astype('float32')
data = np.random.randint(0, 256, (100, 100), 'uint8')
BINS = 20
np_hist, _ = np.histogram(data, bins=BINS)
dmin, dmax, _, _ = cv2.minMaxLoc(data)
if np.issubdtype(data.dtype, 'float'): dmax += np.finfo(data.dtype).eps
else: dmax += 1
cv_hist = cv2.calcHist([data], [0], None, [BINS], [dmin, dmax]).flatten()
plt.plot(np_hist, '-', label='numpy')
plt.plot(cv_hist, '-', label='opencv')
plt.gcf().set_size_inches(15, 7)
plt.legend()
plt.show()
Upvotes: 4
Reputation: 6468
@Divakar has given a great answer. But if you want to port your own code to OpenCV, then:
img = cv2.imread('image.jpg',cv2.IMREAD_UNCHANGED)
data = np.reshape(img, (-1,3))
print(data.shape)
data = np.float32(data)
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
flags = cv2.KMEANS_RANDOM_CENTERS
compactness,labels,centers = cv2.kmeans(data,1,None,criteria,10,flags)
print('Dominant color is: bgr({})'.format(centers[0].astype(np.int32)))
Result for your image:
Dominant color is: bgr([41 31 23])
Time it took: 0.10798478126525879 secs
Upvotes: 13