sorting by value

Let's say I want to build a map, where the key is the file name and the value is the size in bytes:

files["foo"] = 12345

Once it is built, I want to sort by the file size, largest first. In Perl, I could do this:

foreach my $key (sort { $files{$b} <=> $files{$a} } keys %files)

How do I do this in python??

I came up with a solution, but it may not be the best way. Opinions?

--------------
def sort_byval(dict, reverse=0):
if type(dict) is not type({}): return []
keys = dict.keys()

s = map(lambda k: (dict[k], k), keys)
s.sort()
if reverse: s.reverse()

return s
# end sort_byval

m = { 'a': 1000, 'b': 2000, 'c': 3000 }
v = sort_byval(m,1)

for v1,v2 in v: print v1

---------------------

prints:
3000
2000
1000

I can't think of a better way to do it really, but I'd like to propose a change to is so it returns a sorted dictionary instead of a list of (val, key) tuples.



v would now contain:


Which is the dictionary passed, sorted by value.

Sorry, ZeUs, but dictionaries are always unsorted. You cannot rebuild a dictionary in a different order and expect it to remain that way. In fact, after a call "v = sort_byval(m)" v and m will be identical copies of each other.

Either stick with vpopper's solution or, if you really must have a dictionary, look for sorted dictionary implementations on the Vaults of Pernassus.

Duh.... I knew that...

See what happens if you don't pay attention?