CODEWITHSUNDEEP
pythonlistpython-2.7ioglob
newWithPython
newWithPython

Reputation: 883

How to carry in a list the name each file from a directory with python?

I am applying some regex to a folder full of .txt files in order to extract some specific patterns like this:

def retrive(directory, a_regex):
    for filename in glob.glob(os.path.join(directory, '*.txt')):
        with open(filename, 'r') as file:
            important_stuff = re.findall(a_regex, file.read(), re.S)
            my_list = [tuple([j.split()[0] for j in i]) for i in important_stuff]
            print my_list


lists_per_file = retrive(directory,regex_)

And the output is the desired content of all the files in a list:

[interesting stuff 1]
[interesting stuff 2]
[interesting stuff 3]
...
[interesting stuff n]
[interesting stuff n-1]

How can I carry or bind to the list the name of each document file, i.e. something like this:

[interesting stuff 1], name_of_document_1
[interesting stuff 2], name_of_document_2
[interesting stuff 3],name_of_document_3
...
[interesting stuff n], name_of_document_n
[interesting stuff n-1], name_of_document_n-1

Thanks in advance guys.

Upvotes: 0

Views: 46

Answers (1)

David
David

Reputation: 696

If you want to print the list and then the filename without a newline between the two, you will first have to turn the list into a string, then strip off the brackets from around the list. After that you can grab the filename from the filepath that you have, and put the two together.

See code below;

def retrive(directory, a_regex):
    for filename in glob.glob(os.path.join(directory, '*.txt')):
        with open(filename, 'r') as file:
            important_stuff = re.findall(a_regex, file.read(), re.S)
            my_list = [tuple([j.split()[0] for j in i]) for i in important_stuff]
            # print my_list # old line
            print str(my_list).strip('[]') + filename.split('/')[-1]


lists_per_file = retrive(directory,regex_)

Upvotes: 1

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