Finding and copying files with python.

[gtb]

I wish to copy the highest version number of a file from directory \
\
\fileserver\D:\scripts to C:\scripts where the file names are of the
form

filename_MM.NN.SS.zip, where MM, NN, and SS can be one to three
digits.

Example directory:
other.zip
dx_ver_1.1.63.zip
dx_ver_1.2.01.zip
dx_ver_1.12.7.zip
temp.txt


Does python have string matching routines that would find the bottom
listed zip file and/or file copying routines?

A little pointer or two would be much appreciated.

Thanks,

jh

 

[oliver]

Just sort them and then select the bottom one from a list by using a
negative indices. I.e.:

list[-1]

Would return the bottom result out of a list

 

[kyosohma]

I wish to copy the highest version number of a file from directory \
\
\fileserver\D:\scripts to C:\scripts where the file names are of the
form

filename_MM.NN.SS.zip, where MM, NN, and SS can be one to three
digits.

Example directory:
other.zip
dx_ver_1.1.63.zip
dx_ver_1.2.01.zip
dx_ver_1.12.7.zip
temp.txt

Does python have string matching routines that would find the bottom
listed zip file and/or file copying routines?

A little pointer or two would be much appreciated.

Thanks,

jh

You could just use string slicing to cut off the first 7 characters
and have the numbers available to compare. There's also the os.stat
module to find the last modified date of the file. You might be able
to use the glob module to grab a list of the files and then sort the
list too.

Mike

 

[gtb]

You could just use string slicing to cut off the first 7 characters
and have the numbers available to compare. There's also the os.stat
module to find the last modified date of the file. You might be able
to use the glob module to grab a list of the files and then sort the
list too.

Mike

Thanks for posting folks. I didn't make my question clear. Before I
sort the files I need to ensure that I am only sorting the files that
match the profile of "filename_MM.NN.SS.zip", where MM, NN, and SS can
be one to three
digits.

Thanks again,

jh

 

[kyosohma]

Thanks for posting folks. I didn't make my question clear. Before I
sort the files I need to ensure that I am only sorting the files that
match the profile of "filename_MM.NN.SS.zip", where MM, NN, and SS can
be one to three
digits.

Thanks again,

jh

Then you probably need to use the glob module and the re module. You
may even be able to just use the glob module by doing something like:

filenames = glob.glob(r'pathtofiles\*.*.*.*.zip')

and then sort that.

Mike

 

[Steve Holden]

Thanks for posting folks. I didn't make my question clear. Before I
sort the files I need to ensure that I am only sorting the files that
match the profile of "filename_MM.NN.SS.zip", where MM, NN, and SS can
be one to three
digits.

Thanks again,

jh

OK, well look at the glob module to get a list of the filenames.

You are going to have to be careful doing the sort, however, because a
simple string comparison won't work for numbers of differing lengths.

One way would be to define a function that extracts the numeric
components from a filename and produces a three-element tuple or list.
Then pass this function to sort() as the "key" argument.

regards
Steve

 

[Daniele Varrazzo]

I wish to copy the highest version number of a file from directory \
Comparing the version strings is not enough: you have to convert the
parts into integers, because else:

True

Thanks for posting folks. I didn't make my question clear. Before I
sort the files I need to ensure that I am only sorting the files that
match the profile of "filename_MM.NN.SS.zip", where MM, NN, and SS can
be one to three
digits.

Match the file names against the pattern "dx_ver_(\d+).(\d+).(\d
+).zip". You may also use the glob function, but then you will have to
parse the version number from the file name anyway: with the regexp
you can use match.groups() to get the version number.

You can do:

import re
ver_re = re.compile(r"dx_ver_(\d+).(\d+).(\d+).zip")

def getVer(fn):
"""Return a *comparable* file version, None for bad file names"""
m = ver_re.match(fn)
return m and map(int, m.groups())

print sorted(os.listdir('/path/to/wherever'), key=getVer)[-1]

--Daniele

P.S. I guess in Obfuscated Python one would write something like:

print sorted((pair for pair in ((re.match(r"dx_ver_(\d+).(\d+).(\d+).zip", fn), fn) for fn in os.listdir('/path/to/wherever')) if pair[0]), key=lambda _: map(int, _[0].groups()))[-1][1]

dx_ver_1.12.7.zip

 

[gtb]

Comparing the version strings is not enough: you have to convert the
parts into integers, because else:

True

Thanks for posting folks. I didn't make my question clear. Before I
sort the files I need to ensure that I am only sorting the files that
match the profile of "filename_MM.NN.SS.zip", where MM, NN, and SS can
be one to three
digits.

Match the file names against the pattern "dx_ver_(\d+).(\d+).(\d
+).zip". You may also use the glob function, but then you will have to
parse the version number from the file name anyway: with the regexp
you can use match.groups() to get the version number.

You can do:

import re
ver_re = re.compile(r"dx_ver_(\d+).(\d+).(\d+).zip")

def getVer(fn):
"""Return a *comparable* file version, None for bad file names"""
m = ver_re.match(fn)
return m and map(int, m.groups())

print sorted(os.listdir('/path/to/wherever'), key=getVer)[-1]

--Daniele

P.S. I guess in Obfuscated Python one would write something like:

print sorted((pair for pair in ((re.match(r"dx_ver_(\d+).(\d+).(\d+).zip", fn), fn) for fn in os.listdir('/path/to/wherever')) if pair[0]), key=lambda _: map(int, _[0].groups()))[-1][1]

dx_ver_1.12.7.zip

Thanks all. Much to learn, but you have certainly helped me get
started.

jh