我试图计算NxN窗口中每个像素周围的对比度,并将结果保存在新图像中,其中新图像中的每个像素都是旧图像周围区域的对比度。从另一篇文章中我得到了这个:
1) Convert the image to say LAB and get the L channel
2) Compute the max for an NxN neighborhood around each pixel
3) Compute the min for an NxN neighborhood around each pixel
4) Compute the contrast from the equation above at each pixel.
5) Insert the contrast as a pixel value in new image.目前,我有以下几点:
def cmap(roi):
max = roi.reshape((roi.shape[0] * roi.shape[1], 3)).max(axis=0)
min = roi.reshape((roi.shape[0] * roi.shape[1], 3)).min(axis=0)
contrast = (max - min) / (max + min)
return contrast
def cm(img):
# convert to LAB color space
lab = cv2.cvtColor(img, cv2.COLOR_BGR2LAB)
# separate channels
L, A, B = cv2.split(lab)
img_shape = L.shape
size = 5
shape = (L.shape[0] - size + 1, L.shape[1] - size + 1, size, size)
strides = 2 * L.strides
patches = np.lib.stride_tricks.as_strided(L, shape=shape, strides=strides)
patches = patches.reshape(-1, size, size)
output_img = np.array([cmap(roi) for roi in patches])
cv2.imwrite("labtest.png", output_img)代码抱怨roi的大小。有什么更好的方法来做我想做的事吗?
发布于 2021-02-25 19:44:01
您可以使用扩张和侵蚀形态学操作来查找NxN邻域的最大值和最小值。
使用形态学操作使解决方案比“手动”将图像分割成小块要简单得多。
您可以使用以下阶段:
下面是一个完整的代码示例:
import numpy as np
import cv2
size_n = 5 # NxN neighborhood around each pixel
# Read input image
img = cv2.imread('chelsea.png')
# Convert to LAB color space
lab = cv2.cvtColor(img, cv2.COLOR_BGR2LAB)
# Get the L channel
L = lab[:, :, 0]
# Use "dilate" morphological operation (dilate is equivalent to finding maximum pixel in NxN neighborhood)
img_max = cv2.morphologyEx(L, cv2.MORPH_DILATE, np.ones((size_n, size_n)))
# Use "erode" morphological operation (dilate is equivalent to finding maximum pixel in NxN neighborhood)
img_min = cv2.morphologyEx(L, cv2.MORPH_ERODE, np.ones((size_n, size_n)))
# Convert to type float (required before using division operation)
img_max = img_max.astype(float)
img_min = img_min.astype(float)
# Compute contrast map (range of img_contrast is [0, 1])
img_contrast = (img_max - img_min) / (img_max + img_min)
# Convert contrast map to type uint8 with rounding - the conversion loosed accuracy, so I can't recommend it.
# Note: img_contrast_uint8 is scaled by 255 (scaled by 255 relative to the original formula).
img_contrast_uint8 = np.round(img_contrast*255).astype(np.uint8)
# Show img_contrast as output
cv2.imshow('img_contrast', img_contrast_uint8)
cv2.waitKey()
cv2.destroyAllWindows()输入图像:

L图像:

img_max

img_min

对比图img_contrast_uint8

https://stackoverflow.com/questions/66373003
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