How to convert all pixel values of an image to a specific range -python
I have an rgb image with 12 different colors, but I don't know the colors (pixel values) in advance. I want to convert all pixel values between 0 and 11, each symbolizing a unique color of the original rgb image.
eg. all [230, 100, 140] are converted to [0,0,0], all [130,90,100] are converted to [0,0,1] and so on ... all [210,80,50] are converted to [ 0,0, 11].
source to share
A quick and dirty app. Much could be improved, especially if you go through the whole pixel of the image pixel by pixel, it is not very much, but very difficult, but I was too lazy to remember the threshold exactly and replace the RGB pixels.
import cv2
import numpy as np
#finding unique rows
#comes from this answer : http://stackoverflow.com/questions/8560440/removing-duplicate-columns-and-rows-from-a-numpy-2d-array
def unique_rows(a):
a = np.ascontiguousarray(a)
unique_a = np.unique(a.view([('', a.dtype)]*a.shape[1]))
return unique_a.view(a.dtype).reshape((unique_a.shape[0], a.shape[1]))
img=cv2.imread(your_image)
#listing all pixels
pixels=[]
for p in img:
for k in p:
pixels.append(k)
#finding all different colors
colors=unique_rows(pixels)
#comparing each color to every pixel
res=np.zeros(img.shape)
cpt=0
for color in colors:
for i in range(img.shape[0]):
for j in range(img.shape[1]):
if (img[i,j,:]==color).all(): #if pixel is this color
res[i,j,:]=[0,0,cpt] #set the pixel to [0,0,counter]
cpt+=1
source to share
You can use np.unique
with a bit of trickery:
import numpy as np
def safe_method(image, k):
# a bit of black magic to make np.unique handle triplets
out = np.zeros(image.shape[:-1], dtype=np.int32)
out8 = out.view(np.int8)
# should really check endianness here
out8.reshape(image.shape[:-1] + (4,))[..., 1:] = image
uniq, map_ = np.unique(out, return_inverse=True)
assert uniq.size == k
map_.shape = image.shape[:-1]
# map_ contains the desired result. However, order of colours is most
# probably different from original
colours = uniq.view(np.uint8).reshape(-1, 4)[:, 1:]
return colours, map_
However, if the number of pixels is much larger than the number of colors, the following heuristic algorithm can provide huge speedups. It tries to find a cheap hash function (for example, just looking at the red channel), and if it does, it uses that to create a lookup table. If it doesn't fall back to the above safe method.
CHEAP_HASHES = [lambda x: x[..., 0], lambda x: x[..., 1], lambda x: x[..., 2]]
def fast_method(image, k):
# find all colours
chunk = int(4 * k * np.log(k)) + 1
colours = set()
for chunk_start in range(0, image.size // 3, chunk):
colours |= set(
map(tuple, image.reshape(-1,3)[chunk_start:chunk_start+chunk]))
if len(colours) == k:
break
colours = np.array(sorted(colours))
# find hash method
for method in CHEAP_HASHES:
if len(set(method(colours))) == k:
break
else:
safe_method(image, k)
# create lookup table
hashed = method(colours)
# should really provide for unexpected colours here
lookup = np.empty((hashed.max() + 1,), int)
lookup[hashed] = np.arange(k)
return colours, lookup[method(image)]
Testing and timings:
from timeit import timeit
def create_image(k, M, N):
colours = np.random.randint(0, 256, (k, 3)).astype(np.uint8)
map_ = np.random.randint(0, k, (M, N))
image = colours[map_, :]
return colours, map_, image
k, M, N = 12, 1000, 1000
colours, map_, image = create_image(k, M, N)
for f in fast_method, safe_method:
print('{:16s} {:10.6f} ms'.format(f.__name__, timeit(
lambda: f(image, k), number=10)*100))
rec_colours, rec_map_ = f(image, k)
print('solution correct:', np.all(rec_colours[rec_map_, :] == image))
Sample output (12 colors, 1000x1000 pixels):
fast_method 3.425885 ms
solution correct: True
safe_method 73.622813 ms
solution correct: True
source to share