处理idl文件

Posted 2020-08-22 活在当下

将如下格式的数据：

{“60091.jpg”: [[214.1664, 192.99996, 261.00032, 224.83332000000001, 1], [74.83328, 160.49988, 224.0, 267.4998, 1]]}
{“60092.jpg”: [[15.166656, 197.49996, 80.16640000000001, 233.49996000000002, 1], [52.250048, 215.49996, 85.49951999999999, 254.5002, 3]]}
{“60093.jpg”: []}
{“60094.jpg”: [[171.33312, 188.49996000000002, 243.8336, 240.66647999999998, 1]]}
{“60095.jpg”: []}
{“60096.jpg”: []}
{“60097.jpg”: []}
转换为
000002.jpg dog 44 28 132 121
000002.jpg cat 46 18 112 123
000003.jpg ggg 54 19 243 178
进而实用工具转换为xml格式，实现可视化

1. import pandas as pd
2. import csv
5. label=open(‘label.idl‘,‘r‘)
6. label_list=label.readlines()
8. #fo = open("foo.txt", "wb")
10. output_file =‘./foo.txt‘
11. output = open(output_file,‘wb‘)
12. writer = csv.writer(output)
14. temp=[]#pre define tuple or list is ok
15. for content in label_list:
16. obj=pd.read_json(content,orient=‘index‘)
17. (row,col)=obj.shape
18. for i in range(col):
19. temp=[obj.ix[0].name,obj.ix[0][i][4],obj.ix[0][i][0],obj.ix[0][i][1],obj.ix[0][i][2],obj.ix[0][i][3]]
20. writer.writerow(temp)
21. # writer.writerow(obj)
22. # writer.writerow([obj.index,obj[i][0]])
25. # writer.writerow(obj[i][0])
26. # output.writelines(str(obj[i]))
27. # output.write(str(obj[i]))
29. # fo.write(str(obj[1][0][0]))
30. # fo.write(str(obj[1][0][0]);
31. #
32. # obj[1][0][0]
33. # obj[1][0][1]
34. # obj[1][0][2]
35. # obj[1][0][3]
36. # obj[1][0][4]
37. # i++
39. #obj[]
40. #print obj[0][]
41. #(obj[0].to_str()).split(‘,‘)[4]
43. output.close()
44. label.close()