Posted in Python onJune 04, 2020
我就废话不多说了,大家还是直接看代码吧!
import PIL.Image import numpy import os import shutil def sum_right(path): img = PIL.Image.open(path) array = numpy.array(img) num = array.sum(axis=0) print(type(num)) res_left = 0 res_right = 0 for i in range(256,512): res_right += num[i] print(res_right) if __name__ == '__main__': dir2 = r"C:\Users\Howsome\Desktop\tst" dir1 = r"C:\Users\Howsome\Desktop\AB" names = os.listdir(dir1) n = len(names) print("文件数量",n) res = 0 average_5 = 25565356 average_25 = 26409377 average_5_right = 10006019 #average_tmp = (average_25+average_5)//2 count = 0 #show(os.path.join(dir1, "uni4F6C.png")) for i in range(n): #取图片 img = PIL.Image.open(os.path.join(dir1,names[i])) file = os.path.join(dir1,names[i]) rmfile = os.path.join(dir2,names[i]) array = numpy.array(img) num = array.sum(axis=0) res_right = 0 for i in range(256, 512): res_right += num[i] average_5_right += res_right/n if res_right > average_5_right: shutil.copyfile(file, rmfile) os.remove(file) count += 1 print(average_5_right) print(count)
补充知识:python遍历灰度图像像素方法总结
啥也不说了,看代码吧!
import numpy as np import matplotlib.pyplot as plt import cv2 import time img = cv2.imread('lena.jpg',0) # 以遍历每个像素取反为例 # 方法1 t1 = time.time() img1 = np.copy(img) rows,cols = img1.shape[:2] for row in range(rows): for col in range(cols): img[row,col] = 255 - img[row,col] t2 = time.time() print('方法1所需时间:',t2-t1) # 方法2 t3 = time.time() img2 = np.copy(img) rows,cols = img2.shape[:2] img2 = img2.reshape(rows*cols) # print(img2) for i in range(rows*cols): img2[i] = 255-img2[i] img2 = img2.reshape(rows,cols) # print(img2) t4 = time.time() print('方法2所需时间:',t4-t3) # 方法3 t5 = time.time() img3 = np.copy(img) # 使用多维迭代生成器 for (x,y), pixel in np.ndenumerate(img3): img3[x,y] = 255-pixel t6 = time.time() print('方法3所需时间:',t6-t5)
测试结果:
方法1所需时间: 0.14431977272033691 方法2所需时间: 0.13863205909729004 方法3所需时间: 0.24196243286132812
以上这篇用python按照图像灰度值统计并筛选图片的操作(PIL,shutil,os)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持三水点靠木。
用python按照图像灰度值统计并筛选图片的操作(PIL,shutil,os)
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