Python，字符串切片（从文件位置列表中获取文件名）

Question

I am trying to get the files names from a list of file locations. Thinking it involves string slicing.

The one I worked out is:

L = ['C:\\Design\dw\file4.doc',
'C:\\light\PDF\downloads\list.doc',
'C:\\Design\Dq\file4g.doc',
'C:\\Design\Dq\file4r.doc',
'C:\\Design\Dq\file4k.doc',
'C:\\Design\Dq\ole.doc',
'C:\\GE\easy\file\os_references(9).doc',
'C:\\mate\KLO\Market\BIZ\KP\who\Documents\REF.doc']

LL = []

for a in L:
    b = a.split('\')
    for c in b:
        if c.endswith('.doc'):
            c.replace('.doc', '')
            LL.append(c)

print LL

question 1: the output still contains '.doc'. why, and how can I have them removed?

question 2: what's the better way to get the file names?

Thanks.

Answer 1

The answer to the first question is that strings are immutable, .replace() doesn't modify the string in place, viz:

blaize@bolt ~ $ python 
>>> s = "foobar"
>>> s2 = s.replace("o", "x")
>>> print s
foobar
>>> print s2
fxxbar

My answer to the second question follows:

# I use ntpath because I'm running on Linux.
# This way is more robust if you know you'll be dealing with Windows paths.
# An alternative is to import from os.path then linux filenames will work 
# in Linux and Windows paths will work in Windows.
from ntpath import basename, splitext

# Use r"" strings as people rightly point out.
# "\n" does not do what you think it might.
# See here: https://docs.python.org/2.0/ref/strings.html.
docs = [r'C:\Design\dw\file4.doc',
        r'C:\light\PDF\downloads\list.doc',
        r'C:\Design\Dq\file4g.doc',
        r'C:\Design\Dq\file4r.doc',
        r'C:\Design\Dq\file4k.doc',
        r'C:\Design\Dq\ole.doc',
        r'C:\Design/Dq/test1.doc',  # test a corner case
        r'\\some_unc_machine\Design/Dq/test2.doc',  # test a corner case
        r'C:\GE\easy\file\os_references(9).doc',
        r'C:\mate\KLO\Market\BIZ\KP\who\Documents\REF.doc']

# Please use meaningful variable names:
basenames = []

for doc_path in docs:

    # Please don't reinvent the wheel.
    # Use the builtin path handling functions.
    # File naming has a lot of exceptions and weird cases 
    # (particularly on Windows).
    file_name = basename(doc_path)
    file_basename, extension = splitext(file_name)
    if extension == ".doc":
        basenames.append(file_basename)

print basenames

Best of luck mate. Python is an excellent language.

Answer 2

[file.split('\\')[-1].split('.')[0] for file in L]

You're actually not doing any slicing in your example. You are splitting and replacing. Since we know the file name and extension will always be the last part of a path we can use a negative index to access it after splitting.

Once we split again on the period the file name will always be the 0th element so we can just grab that and add it to a list.

EDIT: I just noticed that this method will have problems with paths that contain \\f since this is a special Python character.

Answer 3

try this if there is no space or other symbols in filename

[re.findall('\w+.doc$', L) for x in L]

Try to take a look at

ntpath module

Answer 4

First thing replace method returns the string with the replaced value. It does not changes the string. So you need to do

c = c.replace('.doc', '')

Answer 5

First answer: replace returns a copy of string, so you doesn't save your changes.
Second answer: You need to get the raw representation of several of the paths because combinations like '\\f' are interpretated as an utf-8 char.
So the tricky part is format the strings to its raw representation. For this i've used the raw() of this answer
Once we have this function, we can manipulate well the strings.
I've used re.split to accept unix and dos format paths

>>> L = [re.split(r'[\/\\]', raw(path)) for path in L]
>>> L
[['C:', 'Design', 'dw', 'file4.doc'], ['C:', 'light', 'PDF', 'downloads', 'list.doc'], ['C:', 'Design', 'Dq', 'file4g.doc'], ['C:', 'Design', 'Dq', 'file4r.doc'], ['C:', 'Design', 'Dq', 'file4k.doc'], ['C:', 'Design', 'Dq', 'ole.doc'], ['C:', 'GE', 'easy', 'file', 'os_references(9).doc'], ['C:', 'mate', 'KLO', 'Market', 'BIZ', 'KP', 'who', 'Documents', 'REF.doc']]

Now L contains a list of path parts, so you can access to file name and its extension getting the last element of every list

>>> L_names = [path_parts[-1] for path_parts in L if path_parts[-1].endswith('.doc')]
>>> L_names
['file4.doc', 'list.doc', 'file4g.doc', 'file4r.doc', 'file4k.doc', 'ole.doc', 'os_references(9).doc', 'REF.doc']

Answer 6

The first important point is that you should input your list with raw string ( r prefix):

L = [r'C:\\Design\dw\file4.doc',
     r'C:\\light\PDF\downloads\list.doc',
     …]

Otherwise, characters are interpolated, in your file names ( \\… is generally replaced by a single character).

Python 2 has a dedicated sub-module just for manipulating paths, which gives you the expected result:

from os.path import basename, splitext                                          
print [splitext(basename(path))[0] for path in L]

Note that the paths and this script must be run on systems that use the same path separator ( / or \\ ) convention (which should usually be the case, as paths generally make sense locally on a machine). You can make it work specifically for Windows path (on any operating system) by doing instead:

from ntpath import basename, splitext 

You then get, on any machine:

['file4', 'list', 'file4g', 'file4r', 'file4k', 'ole', 'os_references(9)', 'REF']