打印文件名称

Question

f = glob.glob('/fulldirectory/*.txt')

for index, files in enumerate(f, 1):
    r = open(files)
    reader = csv.DictReader(r)

So I am trying to print off the actual name of a file as part of my analysis.

Each file in the directory above is named with this convention: R1.txt, R2.txt, R3.txt, etc.

At the moment I am simply using the enumerate function to print off the number - but this only works under the assumption that no files are missing in the directory.

EDIT:

I tried this, but it's not giving me quite what I want:

p = [int(s) for s in files if files.isdigit()]

print p

>[0,1]
>[0,2]

Answer 1

You can just do a simple re.sub to substitute the .txt to an empty string.

import re, glob
f = glob.glob('/fulldirectory/*.txt')
for file in f:
    print(re.sub('\.txt$', '', file))
    r = open(file)
    reader = csv.DictReader(r)

In an ideal world I would print 'index', and on the first iteration, R01 would be printed. Then R02, etc.

If you like them to always be in order, do this instead to first sort the file names:

f = sorted(glob.glob('/fulldirectory/*.txt'))

If you only want to print the base name of the file, you can print this instead:

import os
print(re.sub('\.txt$', '', os.path.basename(file)))

Note: the other way suggested may not be very safe, because it's not recommended to use multiple splits on file names.

Here is a complete example with decent explanation that the OP requested:

import re, os, glob
file_list = glob.glob('/fulldirectory/*.txt') # get the list of file names that ends in .txt
f = sorted(file_list, key = lambda x: int(re.findall('\d+\.txt$',os.path.basename(x))[0]))
    # 1               2     3          8  4          5            6                   7
for file in f:
    print(re.sub('\.txt$', '', file))
          # 9
    # do your stuff....

1. The sorted() function is used to sort the list of file names then store it into f (f is the sorted version of file_list)

2. the key argument is a function that accepts an argument and outputs a sortable object(ie. str , int , list ...), it is used to define the key it's sorting with

3. lambda is an anonymous function that accepts argument 'x', this works the same way as a def NoName(x): return something

4. Use re.findall to find all substrings that matches the regex, in this case, there only should one one match [ie. 'abc123.txt' will return [123] ]

5. '\\d+\\.txt$' is a regex, \\d+ - any number repeating one or more times, \\. is a regular dot . theres a \\ in front because normally in regex, a . has the special meaning which it represents any character, the \\ escapes it, making it only a regular . , txt is a string to match at that given location, and $ is the symbol indicating it to match only at the end of the string.

6. os.path.basename() is used to retrieve the basename (the final part of the path [ie. 'abc123.txt' of '\\a\\b\\c\\abc123.txt' ])

7. since re.findall() always return a list so to retrieve the only match will involve using [0] (ie. ['123',][0] => '123' )

8. because the data retrieved is a string, have to use the int() to change it to an int for comparing. The int is what got passed to key in #2.

9. re.sub('\\.txt$', '', file) the first argument is a regex, the second argument is the string to replace it to, the third argument is the string that need replacing (ie. re.sub('a', '', 'banana') => bnn because it replaces every a with nothing) check #5 for more information about the regex

If you need any more clarification, tell me

Alternative method of importing is this:

import glob
from re import sub, findall
from os.path import basename
file_list = glob.glob('/fulldirectory/*.txt') # get the list of file names that ends in .txt
f = sorted(file_list, key = lambda x: int(findall('\d+\.txt$',basename(x))[0]))

for file in f:
    print(sub('\.txt$', '', file))

    # do your stuff....

helpful links:

https://docs.python.org/3/library/re.html

https://docs.python.org/3/library/os.path.html

https://docs.python.org/3/tutorial/

Answer 2

Just go like this:

f = glob.glob('/fulldirectory/*.txt')

for files in f:
    print files.split('\\')[-1].split('.')[0]