直方图均衡化Python(无Numpy和无打印)

这是一个经过一些修改的解决方案。它给出以下输出

原件:

均衡:

主要修改:

1. 这个 df() 和 cdf() 功能变得简单。执行时一定要打印他们的输出,以检查它是否与您期望的结果相匹配
2. 这个 equalize_image() 函数通过从正常像素范围(即 range(0,256) )累积分布函数

代码如下:

import cv2
img = cv2.imread(raw_input('Please enter the name of your image:'),0) #The ',0' makes it read the image as a grayscale image
row, col = img.shape[:2]


def df(img):  # to make a histogram (count distribution frequency)
    values = [0]*256
    for i in range(img.shape[0]):
        for j in range(img.shape[1]):
            values[img[i,j]]+=1
    return values


def cdf(hist):  # cumulative distribution frequency
    cdf = [0] * len(hist)   #len(hist) is 256
    cdf[0] = hist[0]
    for i in range(1, len(hist)):
        cdf[i]= cdf[i-1]+hist[i]
    # Now we normalize the histogram
    cdf = [ele*255/cdf[-1] for ele in cdf]      # What your function h was doing before
    return cdf

def equalize_image(image):
    my_cdf = cdf(df(img))
    # use linear interpolation of cdf to find new pixel values. Scipy alternative exists
    import numpy as np
    image_equalized = np.interp(image, range(0,256), my_cdf)
    return image_equalized

eq = equalize_image(img)
cv2.imwrite('equalized.png', eq)

如果您不知道,opencv提供了一个用于直方图均衡的内置函数,有文档记录 here .

关于您的代码:

分布频率(或直方图)计算不正确,因为您只计算图像中确实出现的颜色的频率。您应该计算所有颜色值的外观,即使它们没有出现。 此外,每次你的颜色再次出现时,你都会将该颜色的一个新元素添加到列表中,这没有多大意义。我不太确定+=3从哪里来。

我会这样做:

def df(img): #to make a histogram (count distribution frequency)
    values = [0] * 256
    for i in range(len(img)):
        for j in img[i]:
           values[j] += 1