Comments on: Sorting Dictionaries by Value in Python (improved?)

By: Daniel Drucker

Daniel Drucker — Tue, 10 Nov 2009 15:09:29 +0000

There’s an error in sbv6; you say ‘reverse=True’ instead of ‘reverse=reverse’.

By: Gregg Lind

Gregg Lind — Tue, 30 Jun 2009 14:14:01 +0000

@Andy,

Python 3.1 ordered dictionaries won’t fix this problem unfortunately. All ordered dictionaries do is ensure that the dictionary always iterates in the same way, based on *insertion order*.

By: Andy

Andy — Tue, 30 Jun 2009 05:59:06 +0000

In a few years this will be just a little easier for everyone

Ordered dictionaries in 3.1:
http://docs.python.org/dev/py3k/whatsnew/3.1.html

By: Gregg Lind

Gregg Lind — Mon, 08 Jun 2009 13:40:03 +0000

This is maybe non-obvious, but if you convert it back to a dictionary, then you lose the ordering! That’s the whole point of sorting, it to end up with something sorted. Think about it a bit

To use:

for (k,v) in sbv6(my_dict): do_function(k,v) # example

By: Andy

Andy — Mon, 08 Jun 2009 13:32:04 +0000

>> It has to be list (or it wouldn’t be orderable), and the tuple makes sense since the data in each position has a meaning (KEY, X).

In that case, i think i will have to just use the list.

I already tried to convert it back to a dict, but that caused me to end up with my original input.

Thanks for your answer.

By: Andy

Andy — Mon, 08 Jun 2009 13:29:16 +0000

I seem to have hit submit too quick. The input was {'127.0.0.1': ['yes', '9999'], 'remote_host': ['no', '8888'], 'localhost': ['yes', '9999']} and the output is [('remote_host', ['no', '8888']), ('127.0.0.1', ['yes', '9999']), ('localhost', ['yes', '9999'])]

By: Gregg Lind

Gregg Lind — Mon, 08 Jun 2009 13:24:53 +0000

So, in your example, you have a dict of lists, or a dict of “X”s. The general pattern for what you get back is:

_LIST_ of _TUPLES_, where each TUPLE has 2 elements: (KEY, X).

It has to be list (or it wouldn’t be orderable), and the tuple makes sense since the data in each position has a meaning (KEY, X).

I’m glad you found this helpful!

By: Andy

Andy — Mon, 08 Jun 2009 13:18:12 +0000

Great post. I ran into the same requirement and sv6 seems to address it really well. Yet, as also mentioned in PEP 265, this is not really an approach obvious to newbies. Can you give me -and doubtlessly some other newbies- an example of how to get the output back in the dict structure which was passed to sv6? Here is an example. Having passed the following data: {'127.0.0.1': ['no', '9999'], 'remote_host': ['no', '888'], 'localhost': ['yes', '9999']} i get this back [('localhost', ['yes', '9999']), ('remote_host', ['no', '8888']), ('127.0.0.1', ['no', '9999'])] From what i have learned so far, it looks like a list with tuples and lists inside those tuples. Thanks in advance.

By: digital sanitation engineering: Sorting Python Dictionary By Value « 藏袍逢毅

digital sanitation engineering: Sorting Python Dictionary By Value « 藏袍逢毅 — Thu, 09 Apr 2009 04:55:08 +0000

[...] Gregg Lind (aka “write-only”) replied to the article with a comment pointing to his performance notes. PEP 0265 has the “best answer” that is at least 2x [...]

By: writeonly

writeonly — Fri, 12 Dec 2008 00:13:33 +0000

Paul, thanks for the tips. Looking through the code for nlargest, there’s a lot more going on there. Maybe this would be a good place to do a patch

_nlargest = nlargest
def nlargest(n, iterable, key=None):
“”"Find the n largest elements in a dataset.

Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
“”"
in1, in2 = tee(iterable)
keys = in1 if key is None else map(key, in1)
it = zip(keys, map(neg, count()), in2) # decorate
result = _nlargest(n, it)
return list(map(itemgetter(2), result)) # undecorate