labelme批量json_to_dataset转换并解决无生成yaml文件问题

本文主要是介绍labelme批量json_to_dataset转换并解决无生成yaml文件问题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

关键代码

# 最前面加入导包
import yaml# 中间是代码的主体部分就不贴了
# 然后在最下面main函数之前加上这一部分：logger.warning('info.yaml is being replaced by label_names.txt')info = dict(label_names=label_names)with open(osp.join(out_dir, 'info.yaml'), 'w') as f:yaml.safe_dump(info, f, default_flow_style=False)logger.info('Saved to: {}'.format(out_dir))# 下面就是main函数了
if __name__ == '__main__':main()

count = os.listdir(json_dir)for i in range(0, len(count)):path = os.path.join(json_dir, count[i])if os.path.isfile(path):

完整代码
把整个文件代码替换就行

然后执行

labelme_json_to_dataset dir(你自己的文件夹路径)

import argparse
import base64
import json
import os
import os.path as ospimport imgviz
import PIL.Imagefrom labelme.logger import logger
from labelme import utils
import yamldef main():logger.warning("This script is aimed to demonstrate how to convert the ""JSON file to a single image dataset.")logger.warning("It won't handle multiple JSON files to generate a ""real-use dataset.")parser = argparse.ArgumentParser()parser.add_argument("json_dir")parser.add_argument("-o", "--out", default=None)args = parser.parse_args()json_dir = args.json_dircount = os.listdir(json_dir)for i in range(0, len(count)):path = os.path.join(json_dir, count[i])if os.path.isfile(path):if args.out is None:out_dir = osp.basename(path).replace(".", "_")out_dir = osp.join(osp.dirname(path), out_dir)else:out_dir = args.outif not osp.exists(out_dir):os.mkdir(out_dir)data = json.load(open(path))imageData = data.get("imageData")if not imageData:imagePath = os.path.join(os.path.dirname(json_dir), data["imagePath"])with open(imagePath, "rb") as f:imageData = f.read()imageData = base64.b64encode(imageData).decode("utf-8")img = utils.img_b64_to_arr(imageData)label_name_to_value = {"_background_": 0}for shape in sorted(data["shapes"], key=lambda x: x["label"]):label_name = shape["label"]if label_name in label_name_to_value:label_value = label_name_to_value[label_name]else:label_value = len(label_name_to_value)label_name_to_value[label_name] = label_valuelbl, _ = utils.shapes_to_label(img.shape, data["shapes"], label_name_to_value)label_names = [None] * (max(label_name_to_value.values()) + 1)for name, value in label_name_to_value.items():label_names[value] = namelbl_viz = imgviz.label2rgb(label=lbl, img=imgviz.asgray(img), label_names=label_names, loc="rb")PIL.Image.fromarray(img).save(osp.join(out_dir, "img.png"))utils.lblsave(osp.join(out_dir, "label.png"), lbl)PIL.Image.fromarray(lbl_viz).save(osp.join(out_dir, "label_viz.png"))with open(osp.join(out_dir, "label_names.txt"), "w") as f:for lbl_name in label_names:f.write(lbl_name + "\n")logger.info("Saved to: {}".format(out_dir))logger.warning('info.yaml is being replaced by label_names.txt')info = dict(label_names=label_names)with open(osp.join(out_dir, 'info.yaml'), 'w') as f:yaml.safe_dump(info, f, default_flow_style=False)logger.info('Saved to: {}'.format(out_dir))if __name__ == "__main__":main()

这篇关于labelme批量json_to_dataset转换并解决无生成yaml文件问题的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---