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Python 实现中值滤波、均值滤波的方法

作者:bllddee  发布时间:2022-05-09 08:58:38 

标签:Python,中值,滤波,均值

红包:

Python 实现中值滤波、均值滤波的方法

Lena椒盐噪声图片:

Python 实现中值滤波、均值滤波的方法


# -*- coding: utf-8 -*-
"""
Created on Sat Oct 14 22:16:47 2017

@author: Don
"""

from tkinter import *
from skimage import io
import numpy as np

im=io.imread('lena_sp.jpg', as_grey=True)
im_copy_med = io.imread('lena_sp.jpg', as_grey=True)
im_copy_mea = io.imread('lena_sp.jpg', as_grey=True)
#io.imshow(im)
for i in range(0,im.shape[0]):
for j in range(0,im.shape[1]):
 im_copy_med[i][j]=im[i][j]
 im_copy_mea[i][j]=im[i][j]
#ui
root = Tk()
root.title("lena")
root.geometry('300x200')

medL = Label(root, text="中值滤波:")
medL.pack()
med_text = StringVar()
med = Entry(root, textvariable = med_text)
med_text.set("")
med.pack()

meaL = Label(root, text="均值滤波:")
meaL.pack()
mea_text = StringVar()
mea = Entry(root, textvariable = mea_text)
mea_text.set("")
mea.pack()

def m_filter(x, y, step):
sum_s=[]
for k in range(-int(step/2),int(step/2)+1):
 for m in range(-int(step/2),int(step/2)+1):
  sum_s.append(im[x+k][y+m])
sum_s.sort()
return sum_s[(int(step*step/2)+1)]

def mean_filter(x, y, step):
sum_s = 0
for k in range(-int(step/2),int(step/2)+1):
 for m in range(-int(step/2),int(step/2)+1):
  sum_s += im[x+k][y+m] / (step*step)
return sum_s

def on_click():
if(med_text):
 medStep = int(med_text.get())
 for i in range(int(medStep/2),im.shape[0]-int(medStep/2)):
  for j in range(int(medStep/2),im.shape[1]-int(medStep/2)):
   im_copy_med[i][j] = m_filter(i, j, medStep)
if(mea_text):
 meaStep = int(mea_text.get())
 for i in range(int(meaStep/2),im.shape[0]-int(meaStep/2)):
  for j in range(int(meaStep/2),im.shape[1]-int(meaStep/2)):
   im_copy_mea[i][j] = mean_filter(i, j, meaStep)
io.imshow(im_copy_med)
io.imsave(str(medStep) + 'med.jpg', im_copy_med)
io.imshow(im_copy_mea)
io.imsave(str(meaStep) + 'mea.jpg', im_copy_mea)

Button(root, text="filterGo", command = on_click).pack()

root.mainloop()

运行结果截图:

Python 实现中值滤波、均值滤波的方法

来源:https://blog.csdn.net/Dooonald/article/details/78260299

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