how can I list only file names in a directory without directory info in the result? I tried
for file in glob.glob(dir+filetype): print file give me result /path_name/1.log,/path_name/2.log,.... but what I do need is file name only: 1.log, 2.log, etc. I do not need the directory info in the result. is there a simple way to get rid of the path info? I don't want to some substr on the result. Thank you!
Return the base name of pathname
path. This is the second element of the pair returned by passingpathto the functionsplit(). Note that the result of this function is different from the Unixbasenameprogram; wherebasenamefor '/foo/bar/' returns 'bar', thebasename()function returns an empty string ('').
So:
>>> os.path.basename('/path_name/1.log,/path_name/2.log') '2.log' 
glob.glob and os.path.basename better than the simple os.listdir? os.listdir strikes me as more flexible, especially since it'll give you a list right away, which is probably what you want in most cases. Maybe most people don't like list comprehension?
listdir plus fnmatch, or rethink his code (or possibly even his user interface) so he has some other rule instead of a glob-style wildcard expression… but why? That's exactly what glob is for.import os # Do not use 'dir' as a variable name, as it's a built-in function directory = "path" filetype = "*.log" # ['foo.log', 'bar.log'] [f for f in os.listdir(directory) if f.endswith(filetype[1:])] filetype="*.log", so you need to do file.endswith(filetype[1:]) or something… and hope that's the only kinds of glob pattern you ever get. (Also, you have to spell the methods right.)file with f to not overwrite class file(object)help is written separately from the code. In CPython 2.x, a file is a wrapper around a PyFile * (as defined here). There's no actual inheritance… but since object has no non-default method slots, the help system adds the class foo(object) for any builtin type that doesn't say otherwise, and isinstance and subclass treat any type as if it were a subclass of object, I guess it is correct as far as any reasonable attempt to detect it from within Python. Is that good enough?you can do
import os [file.rsplit(os.path.sep, 1)[1] for file in glob.glob(dir+filetype)] 
[-1] instead?
rsplit which I'm not familiar with.os.path is that you shouldn't try to do pathname manipulation in terms of string manipulation, because it's harder to read, easier to get wrong, and more likely to confuse other readers (as happened with Mark Ransom). Also, this is presumably the kind of thing the OP meant by "I don't want to some substr on the result".
diris a built-in function,dir(). You should choose another variable name in case any of your modules use the function, as it could lead to confusing bugs.fileeither, as it's the name of a builtin type (and therefore also callable as a function, which works just likeopen).