Question

"I have a number of data files with incomprehensible file names. In order to keep track of these, I'd like to prefix their file names or perhaps even rename them to some prefix + constant. What would be the easiest way to do so?"

SPSS Python Syntax Example

*1. Create test data files.

begin program.
rdir = 'd:/temp/' #Please specify an empty test folder.
import spss,random
while len(os.listdir(rdir))<10:
    fnlen = random.randrange(3,8)
    fn = ''
    for char in range(fnlen):
        fn += chr(random.randrange(97,123))
    if not rdir + fn + '.sav' in os.listdir(rdir):
        spss.Submit('data list free/id.\nbegin data\n1\nend data.\nsav out "%(rdir)s%(fn)s.sav".'%locals())
spss.Submit('new file.')
end program.

*2. Prefix names of all .sav files in rdir.

begin program.
rdir = 'd:/temp/' #Please specify folder containing relevant data files.
for num,fil in enumerate([fil for fil in os.listdir(rdir) if fil.endswith('.sav')]):
    os.rename(rdir + fil,rdir + str(num+1).zfill(3) + '_' + fil)
end program.

Description

Note that this syntax uses Python so make sure you have the SPSS Python Essentials installed.
It will first generate ten data files with incomprehensible file names.
Some empty folder is speficied in which the test files can be created.
In order to apply some structure, the second program block prefixes the names of all files with a (unique) file count number preceded by an underscore (_).

Alternatively, Discarding Old File Names

"Actually, the file names don't make any sense whatsoever. Can't I just rename them to some root name (say, "file_") with a suffix?"

SPSS Python Syntax Example

begin program.
rdir = 'd:/temp/' # Specify the folder containing all relevant files.
rootname = 'file_' # Specify a root for the new file names.
for num,fil in enumerate([fil for fil in os.listdir(rdir) if fil.endswith('.sav')]):
os.rename(rdir + fil,rdir + rootname + str(num + 1).zfill(3) + '.sav')
end program.