Rename files with Python, adding numbering dynamically

[Analeea]

I have a directory with many pdf lists, and I want to rename the files dynamically.

Ex:

~/teste/
|__ProjMed-Química-Físico.Química-quimica_propriedades_coligativas_exercicios.pdf
|__ProjMed-Química-Físico.Química-quimica_termoquimica_entalpia_lei_de_hess_energia_de_ligacao_exercicios_gabarito.pdf

So far, my code is like this:

import os

os.chdir('/home/matheus/teste/')

for f in os.listdir():

f_name, f_ext = os.path.splitext(f)

f_id, f_subject, f_frente, f_topic = f_name.split('-')

f_name1 = '{}-{}-{}-{}{}'.format(f_id, f_subject, f_frente, f_topic, f_ext)

print(f_name1)

print(sorted(f_name1, key=str))

how one can notice in f_id, there will always be a repeat identifier. in the case of the example, 'ProjMed'.

I would like to count how many ProjMed appear (total files with this identifier) and rename them so that it gets something like this:

1-ProjMed

2-ProjMed

3-ProjMed

etc.

I am trying to make a script that surrounds every time a new file is added in a folder. (it would be activated with the ♪, from Linux) then, on day to day, there will be some files that already have a numbering and others that must still have the numbering added (always 1 to more than the highest value already present in another file). Logically, if there is already any file with numbering, it should be ignored in rename.

how could I modify my current script to achieve these goals?

• Count how many 'identifiers' appear by folder ('ProjMed', for example)
• ignore the rename of files that already have numbering
• rename files without numbering with n+1, being n the largest numbering present in a file of this folder

[jeanid]

Hi.

follows a solution proposal for your problem.

To facilitate mainly the appointment of the file was made the following class:

class fileDataModel:
    def __init__(self, id_name, subject, frente, topic, ext, id_num = 0):
        self.id_num = id_num
        self.id_name = id_name
        self.subject = subject
        self.frente = frente
        self.topic = topic
        self.ext = ext
def __repr__(self):
    return self.getFileName()

def getFileName(self):
    if self.id_num != 0:
        return "{}-{}-{}-{}-{}{}".format(self.id_num, self.id_name, self.subject, self.frente, self.topic, self.ext)
    return "{}-{}-{}-{}{}".format(self.id_name, self.subject, self.frente, self.topic, self.ext)

Note that the method getFileName returns different strings depending on the value of id_num. It was considered when id_num is equal to 0, the file is new and still has no numbering.

Now follows a proposal for automatic numbering of new files:

import os

os.chdir('/home/matheus/teste/')
files = dict()

for f in os.listdir():

f_name, f_ext = os.path.splitext(f)

f_name = f_name.split('-')
f_id_num = 0

if len(f_name) == 4: # Arquivo sem identificador
    f_id, f_subject, f_frente, f_topic = f_name
elif len(f_name) == 5: # Arquivo ja com identificador
    f_id_num, f_id, f_subject, f_frente, f_topic = f_name
    f_id_num = int(f_id_num)
else:
    print(f"Arquivo com nome fora do padrao: {f}")
    continue

newFileDataModel = fileDataModel(f_id, f_subject, f_frente, f_topic, f_ext, f_id_num)

if f_id not in files.keys():
    files[f_id] = {'last_id' : f_id_num, 'files' : [newFileDataModel]}
else:
    files[f_id]['files'].append(newFileDataModel)
    if f_id_num > files[f_id]['last_id']:
        files[f_id]['last_id'] = f_id_num

for key in files.keys():
last_id = files[key]['last_id']

for file in files[key]['files']:
    if file.id_num != 0:
        continue

    last_id += 1

    old_file_name = file.getFileName()
    file.id_num = last_id

    os.rename(old_file_name, file.getFileName())

It is noteworthy that the above code always considers the highest id present in the added files. That is, if the highest id is 4, a new file will have id equal to 5. In addition, if there is deletion of any file with id lower than the largest id, a new file added will not be contemplated with this missing id in the sequence and yes with a number more than the higher id present.