Hello everyone, I have a directory with 40 files. File names are divided into prefixes. I need to rename the prefix k3241 according to the name in the...

[weldermartins]

Hello everyone, I have a directory with 40 files.

File names are divided into prefixes. I need to rename the prefix k3241 according to the name in the last prefix.

I even managed to insert the csv extension at the end of the file. but renaming files based on various prefixes is the tricky one. Can anybody help me. I will be posting the image of the code that was created.

Thanks.

/*
This code inserts the csv extension in the filename.
*/
path ="/dbfs/tmp/data/dadosjuridicos/descompactados"
for i in os.listdir(path):
    files = os.path.join(path,i)
    split= os.path.splitext(files)
    if split[1]=='.COMPANY':
       os.rename(files,split[0]+'COMPANY.csv')
    if split[1]=='.ESTABLISHMENT':
       os.rename(files,split[0]+'ESTABLISHMENT.csv')
    if split[1]=='.PARTNER':
       os.rename(files,split[0]+'PARTNER.csv')

[Anonymous]

Hey there! Thanks for your question. I'm sorry to hear you're having difficulty. Let's see what your peers have to say.

[weldermartins]

Similar to the result of this code, but to change files.

path = 'k3250.seqy0.company.csv'
 
path1 = path.replace("k3250", "company")
 
print(path1)

[weldermartins]

I managed to solve it using the code below.

thank's

if split[1]=='.COMPANYCSV':
        os.rename(files,files.replace("K3241", "COMPANY")) 

[Anonymous]

@welder martins​ - Thank you for sharing all this. You are fantastic!

[Anonymous]

Hi @welder martins​ 

How are you doing?

Thank you for posting that question. We are glad you could resolve the issue. Would you want to mark an answer as the best solution?

Cheers