迭代文件夹打开，逐个处理并保存图像

Question

我有一个文件夹“Images-2”，它有超过100个子文件夹，这些子文件夹由每个文件夹一个图像组成。 def main()打开每个图像，然后def run(img)拍摄图像并对其进行处理，但现在我无法将该图像保存在其子文件夹中。

例如def main C：/Images-2/1/1.png（1是文件夹的名称，所以我有100个文件夹中的图像-2）

如果条件将保存处理后的图像（零。 png）in folder Images-2/1/

它如何工作100个文件夹，每个文件夹1个图像？

def run(img): 
    data = img.load() 
    width, height = img.size 
    output_img = Image.new("RGB", (100, 100)) 
    Zero=np.zeros(shape=(100, 100),dtype=np.uint8) 

    for (x, y) in labels: 
      component = uf.find(labels[(x, y)]) 
      labels[(x, y)] = component 
      path='C:/Python27/cclabel/Images-2/' 
      if labels[(x, y)]==0: 
       Zero[y][x]=int(255) 
       Zeroth = Image.fromarray(Zero) 
       for root, dirs in os.walk(path): 
        print root 
        print dirs 
        Zeroth.save(path+'Zero'+'.png','png') 
def main(): 
    # Open the image 
    path="C:/Python27/cclabel/Images-2/" 
    for root, dirs, files in os.walk(path): 
     for file_ in files: 
      img = Image.open(os.path.join(root, file_)) 
      img = img.point(lambda p: p > 190 and 255) 
      img = img.convert('1') 
      (labels, output_img) = run(img) 

if __name__ == "__main__": main() 

Answer 1

您打电话给os.walk两次。那是你的问题。这就是我的意见在我的意见：

def run(dirname, img): 
    data = img.load() 
    width, height = img.size 
    output_img = Image.new("RGB", (100, 100)) 
    Zero=np.zeros(shape=(100, 100), dtype=np.uint8) 

    for (x, y) in labels: 
     component = uf.find(labels[(x, y)]) 
     labels[(x, y)] = component 
     path = 'C:/Python27/cclabel/Images-2/' 
     if labels[(x, y)] == 0: 
      Zero[y][x] = 255 
      Zeroth = Image.fromarray(Zero) 
      Zeroth.save(os.path.join(dirname, 'Zero.png'), 'png') 


def main(): 
    path = "C:/Python27/cclabel/Images-2/" 
    for root, dirs, files in os.walk(path): 
     for file_ in files: 
      img = Image.open(os.path.join(root, file_)) 
      img = img.point(lambda p: p > 190 and 255) 
      img = img.convert('1') 
      (labels, output_img) = run(root, img) 


if __name__ == "__main__": 
    main()