OpenCV利用python来实现图像的直方图均衡化

1.直方图

直方图: (1) 图像中不同像素等级出现的次数 (2) 图像中具有不同等级的像素关于总像素数目的比值。

我们使用cv2.calcHist方法得到直方图

cv2.calcHist(images, channels, mask, histSize, ranges):

-img: 图像

-channels: 选取图像的哪个通道

-histSize: 直方图大小

-ranges: 直方图范围

cv2.minMaxLoc: 返回直方图的最大最小值,以及他们的索引

import cv2

import numpy as np

def ImageHist(image, type):

color = (255, 255,255)

windowName = 'Gray'

if type == 1: #判断通道颜色类型 B-G-R

color = (255, 0, 0)

windowName = 'B hist'

elif type == 2:

color = (0,255,0)

windowName = 'G hist'

else:

color = (0,0,255)

# 得到直方图

hist = cv2.calcHist([image],[0],None,[256],[0,255])

# 得到最大值和最小值

minV,maxV,minL,maxL = cv2.minMaxLoc(hist)

histImg = np.zeros([256,256,3],np.uint8)

#直方图归一化

for h in range(256):

interNormal = int(hist[h] / maxV * 256)

cv2.line(histImg, (h, 256), (h, 256 - interNormal), color)

cv2.imshow(windowName, histImg)

return histImg

img = cv2.imread('img.jpg', 1)

channels = cv2.split(img) # R-G-B

for i in range(3):

ImageHist(channels[i], 1 + i)

cv2.waitKey(0)


2.直方图均衡化

灰色图像直方图均衡化

这里我们直接使用cv2.equalizeHist方法来得到直方图均衡化之后的图像

import cv2

import numpy as np

img = cv2.imread('img.jpg', 1)

gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

dat = cv2.equalizeHist(gray)

cv2.imshow('gray', gray)a

cv2.imshow('dat', dat)

cv2.waitKey(0)

原图像:


直方图均衡化后的图像:

彩色图像直方图均衡化

彩色图像有3个通道,直方图是针对单通道上的像素统计,所以使用cv2.split方法分离图像的颜色通道,分别得到各个通道的直方图,最后使用cv2.merge()方法合并直方图,得到彩色图像的直方图均衡化

import cv2

import numpy as np

img = cv2.imread('img.jpg', 1)

cv2.imshow('img', img)

(b, g, r) = cv2.split(img)

bH = cv2.equalizeHist(b)

gH = cv2.equalizeHist(g)

rH = cv2.equalizeHist(r)

dat = cv2.merge((bH, gH, rH))

cv2.imshow('dat', dat)

cv2.waitKey(0)

D:\Anaconda\lib\site-packages\numpy\_distributor_init.py:32: UserWarning: loaded more than 1 DLL from .libs:

D:\Anaconda\lib\site-packages\numpy\.libs\libopenblas.NOIJJG62EMASZI6NYURL6JBKM4EVBGM7.gfortran-win_amd64.dll

D:\Anaconda\lib\site-packages\numpy\.libs\libopenblas.PYQHXLVVQ7VESDPUVUADXEVJOBGHJPAY.gfortran-win_amd64.dll

stacklevel=1)

原图像:


直方图均衡化之后的图像:

3.源代码实现直方图均衡化

下面我们用源代码来实现直方图

横坐标为像素等级,纵坐标为出现的概率

import cv2

import numpy as np

import matplotlib.pyplot as plt

img = cv2.imread('img.jpg', 1)

gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

count = np.zeros(256, np.float)

for i in range(img.shape[0]):

for j in range(img.shape[1]):

count[int(gray[i, j])] += 1 # 统计该像素出现的次数

count = count / (img.shape[0] * img.shape[1]) # 得到概率

x = np.linspace(0,255,256)

plt.bar(x, count,color = 'b')

plt.show()

# 计算累计概率

for i in range(1,256):

count[i] += count[i - 1]

# 映射

map1 = count * 255

for i in range(img.shape[0]):

for j in range(img.shape[1]):

p = gray[i, j]

gray[i, j] = map1[p]

cv2.imshow('gray', gray)

cv2.waitKey(0)

直方图:

直方图均衡化后的图像:

彩色直方图源码

import cv2

import numpy as np

import matplotlib.pyplot as plt

img = cv2.imread('img.jpg', 1)

# R-G-B三种染色直方图

countb = np.zeros(256, np.float32)

countg = np.zeros(256, np.float32)

countr = np.zeros(256, np.float32)

for i in range(img.shape[0]):

for j in range(img.shape[1]):

(b,g,r) = img[i,j]

b = int(b)

g = int(g)

r = int(r)

countb[b] += 1 # 统计该像素出现的次数

countg[g] += 1

countr[r] += 1

countb = countb / (img.shape[0] * img.shape[1]) # 得到概率

countg = countg / (img.shape[0] * img.shape[1])

countr = countr / (img.shape[0] * img.shape[1])

x = np.linspace(0,255,256)

plt.figure()

plt.bar(x, countb,color = 'b')

plt.figure()

plt.bar(x, countg,color = 'g')

plt.figure()

plt.bar(x, countr,color = 'r')

plt.show()

# 计算直方图累计概率

for i in range(1,256):

countb[i] += countb[i - 1]

countg[i] += countg[i - 1]

countr[i] += countr[i - 1]

#映射表

mapb = countb * 255

mapg = countg * 255

mapr = countr * 255

dat = np.zeros(img.shape, np.uint8)

for i in range(img.shape[0]):

for j in range(img.shape[1]):

(b,g,r) = img[i, j]

dat[i, j] = (mapb[b],mapg[g],mapr[r])

cv2.imshow('dat', dat)

cv2.waitKey(0)

R-G-B 3 种颜色通道的直方图如下:




图像均衡化之后的结果:

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