python 图片批量指定局部区域打码处理

浅浅的花香味﹌ 2022-03-16 09:44 608阅读 0赞

### (一)法. 推荐使用 ###

### 1.看明白如下重复的for循环，就可以实现对同一张图片做任意位置的打码与任意数量的扩充了。 ###

import os
    from PIL import Image
    import glob
    import numpy as np
    
    outdir = r'C:\Users\admin\Desktop\c1\dama'
    count = 0
    for i in glob.glob(r'C:\Users\admin\Desktop\c1\resize/*.jpg'):
        im1 = Image.open(i)
        arr = np.array(im1)
        h = arr.shape[0]
        w = arr.shape[1]
    
        a = 7
        for m in range(a): #每张图可以生成a张打码图
            im2 = im1.copy() #同一张每次取出都复制，以防上次操作影响到下次
    
            for j in range(int((m/a) * w),int(((m+1)/a) * w),1):#每个m对应一段x
    
                for k in range(int(h-(m/a)*h - 1),int(h-((m+1)/a)*h - 1),-1):
    
                    im2.putpixel((j, k), (255,0,0))  # 这些像素点的颜色改成红色,改了图本身
            if m>0 and m<a-1: #我是按图的对角线位置依稀打码红色正方块的，此if舍弃了首尾两张(没挡住人脸，无意义)
                count += 1
                print(count)
                im2.save(os.path.join(outdir, '{}.jpg'.format(count)))
    
            im3 = im1.copy()
            for c in range(int((m / a) * w), int(((m + 1) / a) * w), 1):
    
                for d in range(int((m / a) * h), int(((m + 1) / a) * h), 1):
                    im3.putpixel((c, d), (255, 0, 0))
            if m > 0 and m < a - 1:
                count += 1
                print(count)
                im3.save(os.path.join(outdir, '{}.jpg'.format(count)))
    
            im4 = im1.copy()
            for e in range(int((m / a) * w), int(((m + 1) / a) * w), 1):
    
                for f in range(int(h/2 - 1/2 * (1 /a) * w), int(h/2 + 1/2 * (1 /a) * w), 1):
                    im4.putpixel((e, f), (255, 0, 0))
            if m > 0 and m < a - 1:
                count += 1
                print(count)
                im4.save(os.path.join(outdir, '{}.jpg'.format(count)))

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70][]

### 2.按对应顺序生成打码图 ###

import os
    from PIL import Image
    import numpy as np
    
    outdir = r'C:\Users\admin\Desktop\c1\trainB'
    count = 0
    path = r'C:\Users\admin\Desktop\c1\trainA'
    x_names = os.listdir(path)
    x_names.sort(key=lambda i:int(i[:-4]))
    
    for i in x_names:
        im1 = Image.open(os.path.join(path,i))
        arr = np.array(im1)
        h = arr.shape[0]
        w = arr.shape[1]
    
        for j in range(int((1 / 3) * w), int((2 / 3) * w), 1):
            for k in range(int((1 / 3) * h), int((2 / 3) * h), 1):
                im1.putpixel((j, k), (255, 0, 0))  
        count += 1
        print(count)
        im1.save(os.path.join(outdir, '{}.jpg'.format(count)))

### ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 1][] ###

### (二)法. ###

### 1.只用PIL ###

import os
    from PIL import Image
    import glob
    import numpy as np
    
    outdir = r'C:\Users\admin\Desktop\c1\dama'
    
    count = 0
    for i in glob.glob(r'C:\Users\admin\Desktop\c1\img/*.jpg'):
        im1 = Image.open(i)
        arr = np.array(im1)
        '在此也能体会到：若是要读取图片的尺寸等属性，最好直接用cv2，而不是PIL或matplotlib'
        # h = im1.shape[0]#错,im1作为图片，本身没有shape等np矩阵属性,故需先图转矩阵
        # w = im1.shape[1]
        h = arr.shape[0]
        w = arr.shape[1]
        a = 9
        for j in range(int(1 / 4 * w), int(3 / 4 * w), a):
            for k in range(int(1 / 4 * w), int(3 / 4 * w), a):
                arr[j:j + a, k:k + a] = arr[j + (a // 2)][k + (a // 2)]
    
        count += 1
        print(count)
        im2 = Image.fromarray(arr)
        im2.save(os.path.join(outdir,os.path.basename(i)))

效果如下(顺序因其他操作给打乱了)![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 2][]

注：我是直接对celeba数据集人脸做了打码处理。其中第18、19行的系数和depts值，要根据具体图片尺寸以及你想要打码的区域，自行作出改动。

### 2. 结合cv2 ###

1.cv2按BGR读取，以下是未做转RGB操作的效果

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 3][]

2.正确操作，效果如下

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 4][]

## ##

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[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 1]: /images/20220316/16d6d98f83ac4c008c719df3bb8caf35.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 2]: /images/20220316/f3bedd78b4074b4fa459e1f2991eb96f.png
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[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM5OTM4NjY2_size_16_color_FFFFFF_t_70 4]: /images/20220316/b7c5aa52b7bc453d89646384b36d6481.png