Rename and read multiple JSON files in python-CodePudding

I'm trying to rename my 6 JSON files present in logs folder using python. Say I have files like-

File1: results_HAT_6e0_57-graph_52.19_test_results_tests_storage.Full.up_ bargraph _results-chart

File2: results_HAT_6e0_55-graph_52.19_test_results_tests_storage.Full.down_ bargraph _results-chart

File3: results_HAT_6e0_58-graph_52.2_test_results_tests_storage.Full.up_ bargraph _results-chart

File4: results_HAT_6e0_56-graph_52.2_test_results_tests_storage.Full.down_ bargraph _results-chart

File5: results_HAT_6e0_59-graph_52.242_test_results_tests_storage.Full.up_ bargraph _results-chart

File6: results_HAT_6e0_54-graph_52.242_test_results_tests_storage.Full.down_bargraph_results-chart

and i want to make it more easy to understand , therefore want to rename it like:

File1: 52.19_up

File2: 52.19_down

File3: 52.2_up

File4: 52.2_down

File5: 52.242_up

File6: 52.242_down

PS- the values after dot(.) may vary. eg- 52.67, 52.9, 52.002...

I tried doing through this similar method-

import numpy as np
import os

os.chdir(r'C:\Users\me\desktop\json\logs')
print(os.getcwd())

for count, f in enumerate(os.listdir()):
    f_name, f_ext = os.path.splitext(f)
    f_name = "f"   str(count)

    new_name = f'{f_name}{f_ext}'
    os.rename(f, new_name)

with open('f0.json', 'r') as f1:              
    data1 = json.load(f1)

with open('f1.json', 'r') as f2:                      
    data2 = json.load(f2)

with open('f2.json', 'r') as f3:                    
    data3 = json.load(f3)

with open('f3.json', 'r') as f4:                   
    data4 = json.load(f4)

with open('f4.json', 'r') as f5:                       
    data5 = json.load(f5)

with open('f5.json', 'r') as f6:                     
    data6 = json.load(f6)

But through this method, files are renaming as f0...f5. which is still confusing and unclear. Any help would be appreciated.

CodePudding user response：

Well, Im not sure about the format of these filenames. But as the file names you provided. I can suggest this:

import re
import glob
import os 

files = glob.glob("/home/adam/*")
for file in files:
    prefix = re.findall(r"_(\d \.\d )_", file)
    postfix = re.findall(r"Full\.(\w )_", name)
    prefix = prefix[0] if prefix else ""
    postfix = postfix[0] if postfix else ""
    new_name = f"{prefix}_{postfix}"
    os.rename(file, new_name)

CodePudding user response：

You could do it this way:

for count, f in enumerate(os.listdir()):
    f_name, f_ext = os.path.splitext(f)
    f_name = "f"   str(count)
    f_name_split = f.split('_')
    new_name = f"{f_name_split[4]}_{f_name_split[8].split('.')[-1]}{f_ext}"
    os.rename(f, new_name)

Note: If the file name structure changes, then it will probably not work anymore.