ansaurus

Question

Dynamic filenames

Answer 1

+1 A:

Look into python's glob module.

It uses the usual shell wildcard syntax, so ?? would match any two characters, while * would match anything. You could use either f??.txt or f*.txt, but the former is a more strict match, and wouldn't match something like fact_or_fiction.txt while the latter would.

E.g.:

import glob
for filename in glob.iglob('f??.txt'):
    infile = file(filename)
    # Do stuff...

Joe Kington 2010-08-14 17:25:15

`glob` is a great option. If, however, you only wanted to access a certain set (say, f0-99 existed, but you only wanted f0-19), then you could use string interpolation: `"f%02d" % 4` becomes `"f04"`.

Amber 2010-08-14 17:26:58

+1 to Amber. I would use glob only if I wanted all files matching a pattern. For a series of files whose names are known a priori, I would generate the names.

Noufal Ibrahim 2010-08-14 17:31:33

If you need a reference for interpolation: http://docs.python.org/library/stdtypes.html#string-formatting-operations

Amber 2010-08-14 17:33:04

@Amber - True, and it's worth noting that glob won't necessarily return things in sorted order (it depends on the underlying filesystem).

Joe Kington 2010-08-14 17:33:40

But `sorted()` can be used to fix the latter.

Amber 2010-08-14 17:35:11

in addition, glob will only return you **existing** matching files, which does not help you if you want to **create** them as the original poster does (and they did not exist before)

Andre Holzner 2010-08-14 17:56:10

Answer 2

+4 A:

The names are regular. You can create the list of filenames with a simple list comprehension.

["f%02d"%x for x in range(1,21)]

Noufal Ibrahim 2010-08-14 17:30:36

For more details on string interpolation: http://docs.python.org/library/stdtypes.html#string-formatting-operations

Amber 2010-08-14 17:33:29

Answer 3

A:

An example of writing t to all files:

for x in range(22): #Remember that the range function returns integers up to 22-1
    exec "f%02d = open('file%02d.txt', 'w')" % (x, x)

I use the exec statement but there's probably a better way. I hope you get the idea though.

NOTE: This method will give you the variable names fXX to work with later if needed. The last two lines are just examples. Not really needed if all you need is to assign fileXX.txt to fXX.

EDIT: Removed the other last two lines because it seemed that people just weren't too happy with me putting them there. Explanations for downvotes are always nice.

vlad003 2010-08-14 17:39:24

Why would you use `exec`? You can just reuse the variable storing the file handle, so there's no need to make ones with different names - the only interpolation you need to do is in the filename: `f = open('file%02d.txt' % x, 'w')`.

Amber 2010-08-14 17:43:37

Because the OP might need those variables in that format later as he says: *"i want the variables assigned to them to be fxx (f00,f01...)"*. Using just `f =` won't allow him to have those variable names. `exec` allows him to have the variable names he needs

vlad003 2010-08-14 17:50:12

You are simultaneously right and wrong =p.

katrielalex 2010-08-14 17:50:54

Well I said there's probably a better way than to use `exec` but this method does allow him to have what he wants, no?

vlad003 2010-08-14 17:53:03

Well, yes, you're right in that it works but wrong in that `exec` is not the right tool here. I didn't downvote you, by the way.

katrielalex 2010-08-14 18:25:05

No worries, I could tell you didn't :P.

vlad003 2010-08-14 18:41:41

yah, it works thanks guys. i know that some might think it's bad style, but i just need to get this done :P

pjehyun 2010-08-15 23:28:08

Answer 4

+1 A:

I think the point is the OP wants to register the variables programagically:

for i in range( 20 ):
    locals()[ "f%02d" % i ] = open( "file%02d.txt" % i )

and then e.g.

print( f01 )
...

katrielalex 2010-08-14 17:50:24

How would `locals()` be different from `globals()` and `vars()`? If it's all in a loop (and not in a function) you could use either one of them, or am I missing something?

vlad003 2010-08-14 17:55:01

@vlad003 - http://docs.python.org/tutorial/classes.html#python-scopes-and-namespaces Also, looking up a global variable is slower than looking up a local variable, so beyond the usual avoidance of global variables, there's a speed difference, as well.

Joe Kington 2010-08-14 18:04:10

@vlad003: I never said it was better than `globals()` (it is equivalent to `vars()` with no arguments), I said that it was better than `exec`. However, Joe is right in that it *is* better.

katrielalex 2010-08-14 18:26:31

Anything's probably better than `exec`; I only put it here b/c it *works, even though not the best*.@Joe: thanks for the link. I never read the tutorial. I usually learn by doing so I learn as I go.

vlad003 2010-08-14 18:45:36

This'd work but I don't like the idea of magically creating variables in my namespace. I'd much prefer that these names were put into a dictionary or something similar.

Noufal Ibrahim 2010-08-15 17:21:58

Answer 5

A:

files = [open("file%02d", "w") for x in xrange(0, 20+1)]
# now use files[0] to files[20], or loop

# example: write number to each file
for n, f in enumerate(files):
  files.write("%d\n" % n)

# reopen for next example
files = [open("file%02d", "r") for x in xrange(0, 20+1)]
# example: print first line of each file
for n, f in enumerate(files):
  print "%d: %s" % (n, f.readline())

Roger Pate 2010-10-20 18:34:19

ansaurus

tags:

views:

answers:

Dynamic filenames

related questions