使用原始文件名保存图像 [英] Use original file name to save image

问题描述

我想用原始文件名保存图像，例如，如果原始文件名是hero，则处理后的图像名应为hero_Zero.png.
我不知道如何在def run(dirs, img):
中传递文件名 下面的代码是正确的并已修改.

I want to save image with original file name for example if original file name is hero so the processed image name should be hero_Zero.png.
I dont know how to pass file name in def run(dirs, img):
Below code is correct and modified.

def run(dirs, img, file_):
  for (x, y) in labels:
        component = uf.find(labels[(x, y)])
        labels[(x, y)] = component   
        if labels[(x, y)]==0:
            Zero[y][x]=int(255)
            count=count+1
            if count<=43:
                continue
            elif count>43:
                Zeroth = Image.fromarray(Zero)
                Zeroth.save(os.path.join(dirs, file_+'Zero.png'), 'png')  
           return (labels, output_img)
def main():
    path='E:/Dataset/1/'
    for root, dirs, files in sorted(os.walk(path)):
        for file_ in files:
            #print (dirs)
           # print (files)
            full_file_path = os.path.join(root, file_)
            img = Image.open(full_file_path)
            (labels, output_img) = run(root, img, file_[:-4])

推荐答案

在函数定义中添加一个额外的参数.除了传递目录和图像，还传递文件名.结果将如下所示:

Add an extra argument to your function definition. Instead of just taking the directory and image, also pass the filename. The result will look like this:

def run(dirs, img, f_name):
  for (x, y) in labels:
        component = uf.find(labels[(x, y)])
        labels[(x, y)] = component   
        if labels[(x, y)]==0:
            Zero[y][x]=int(255)
            count=count+1
            if count<=43:
                continue
            elif count>43:
                Zeroth = Image.fromarray(Zero)
                Zeroth.save(os.path.join(dirs, f_name + '_Zero.png'), 'png')

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