sort by file/directory

[Mike Zupan]

I have a list that includes files and directories

ie: list = ['file', 'file2', 'dir1','file3','dir2' ]

I want to sort it so it looks like this

['dir1', 'dir2', 'file1','file2','file3' ]
I'm just wondering if there is an easy way to do this

Thanks
Mike

[Peter Hansen]

Mike Zupan wrote:

I have a list that includes files and directories

ie: list = ['file', 'file2', 'dir1','file3','dir2' ]

I want to sort it so it looks like this

['dir1', 'dir2', 'file1','file2','file3' ]

I'm just wondering if there is an easy way to do this

For the above list, executing .sort() would give the correct
result, but presumably you meant for this to work in a more
general manner.

Here's one way to do it, using the Decorate-Sort-Undecorate
pattern that should be familiar to all Python programmers:

(Untested code... likely to have various small errors, left
as an exercise for the reader. :-)

# best to avoid using the name "list" as it conflicts with
# a name in the built-in module
names = [ fill in your own data here ]

import os
sortme = []
for name in names:
# only works if names are absolute, or in current directory
# you probably have some other way to tell whether
# something is a folder, but the key thing is to
# form a tuple with 0 for directories, 1 for files, then
# the name
sortme.append((not os.path.isdir(name), name))

sortme.sort() # in-place sort, of course!

# remove "decoration" which forced the sort to put directories
# first, leaving just the names
names = [item[1] for item in sortme]

-Peter

[Jeremy Jones]

* Mike Zupan (mz****@meso.com) wrote:

I have a list that includes files and directories

ie: list = ['file', 'file2', 'dir1','file3','dir2' ]

I want to sort it so it looks like this

['dir1', 'dir2', 'file1','file2','file3' ]
I'm just wondering if there is an easy way to do this

Is there anything in the name of the files/directories that specified that
it is a file or a directory? If not, then I wouldn't think so. When you
are gathering the file and directory listing (maybe you're using os.walk?),
then you could at that time do something like (given a 'dir_list' and a
'file_list'):

if os.path.isdir(path_to_file): .... dir_list.append(os.path.basename(path_to_file))
.... else:
.... file_list.append(os.path.basename(path_to_file))

Then sort the lists and concatenate them together:
dir_list.sort()
file_list.sort()
new_list = dir_list + file_list
new_list

['bin', 'etc', 'sbin', 'tmp', '.muttrc', 'htpasswd', 'httpd.conf',
'passwd']
Jeremy Jones

[John Hunter]

>>>>> "Mike" == Mike Zupan <mz****@meso.com> writes:

Mike> I have a list that includes files and directories ie: list =
Mike> ['file', 'file2', 'dir1','file3','dir2' ]

Mike> I want to sort it so it looks like this

Mike> ['dir1', 'dir2', 'file1','file2','file3' ]
Mike> I'm just wondering if there is an easy way to do this
This is a good example of when you should use the decorate, sort,
undecorate (DSU) pattern. When python sorts tuples, it compares first
on the first element of the tuple and then on the next element, and
then so on. If you create tuples where the first element of dirs
compares less than the first element of files, and the second element
is just the dir/file name, then you'll get the sort you want. All you
have have to do is undecorate, ie, remove the unwanted information
from the tuples. Here is the approach

import os
fnames = ('emacs', '.emacs', 'tex','.tcshrc','diary' )

tmp = [ (not os.path.isdir(fname), fname) for fname in fnames] # decorate
tmp.sort() # sort
fnamesSorted = [fname for isdir, fname in tmp] # undecorate

For more info on DSU, see
http://groups.google.com/groups?num=...=Google+Search
and the Python Cookbook, which has a chapter on sorting mainly devoted
to DSU.

Mike> list = ['file', 'file2', 'dir1','file3','dir2' ]

You may want to avoid the name list because it is a built-in. For
example, to convert a string to a list of characters, you would do

seq = list('John Hunter')

After creating a variable named list, you can no longer do this.
Cheers,
John Hunter