python dictionary values sorting

Question

I have 2 dictionaries, dict1 and dict2 which contain the same keys, but different values for the keys. What I want to do is for each dictionary, sort the values from largest to smallest, and then give each value a rank 1-N, 1 being the largest value. From here, I want to get the difference of the ranks for the values in each dictionary for the same key. For example:

dict1 = {a:0.6, b:0.3, c:0.9, d:1.2, e:0.2}
dict2 = {a:1.4, b:7.7, c:9.0, d:2.5, e:2.0}

# sorting by values would look like this:
dict1 = {d:1.2, c:0.9, a:0.6, b:0.3, e:0.2}
dict2 = {c:9.0, b:7.7, d:2.5, e:2.0, a:1.4}

#ranking the values would produce this:
dict1 = {d:1, c:2, a:3, b:4, e:5}
dict2 = {c:1, b:2, d:3, e:4, a:5}

#computing the difference between ranks would be something like this:
diffs = {}
for x in dict1.keys():
    diffs[x] = (dict1[x] - dict2[x])

#diffs would look like this:
diffs[a] = -2
diffs[b] = 2
diffs[c] = 1
diffs[d] = -2
diffs[e] = 1

I know dictionaries are meant to be random and not sortable, but maybe there is a method to put the keys and values into a list? The main challenges I am facing are getting the keys and values sorted by value (largest to smallest) and then changing the value to its respective rank in the sorted list.

Answer 1

A simple solution for small dicts is

dict1 = {"a":0.6, "b":0.3, "c":0.9, "d":1.2, "e":0.2}
dict2 = {"a":1.4, "b":7.7, "c":9.0, "d":2.5, "e":2.0}
k1 = sorted(dict1, key=dict1.get)
k2 = sorted(dict2, key=dict2.get)
diffs = dict((k, k2.index(k) - k1.index(k)) for k in dict1)

A more efficient, less readable version for larger dicts:

ranks1 = dict(map(reversed, enumerate(sorted(dict1, key=dict1.get))))
ranks2 = dict(map(reversed, enumerate(sorted(dict2, key=dict2.get))))
diffs = dict((k, ranks2[k] - ranks1[k]) for k in dict1)

Answer 2

What version of python are you using? If 2.7, use OrderedDict.

Per the Python 2.7 docs:

OrderedDict(sorted(d.items(), key=d.get))

If you're using Python 2.4-2.6 you can still use OrderedDict by installing it from pypi here or if you have setuptools, run

easy_install ordereddict

Answer 3

You may be interested in collections.OrderedDict

Here's a sample, my initial thougth is you were also looking for dictionaries with keys ordered by values, things that od1 and od2 are.

d1 = {"a":0.6, "b":0.3, "c":0.9, "d":1.2, "e":0.2}
d2 = {"a":1.4, "b":7.7, "c":9.0, "d":2.5, "e":2.0}

od1 = OrderedDict(sorted(d1.items(), key=lambda t: t[1]))
od2 = OrderedDict(sorted(d2.items(), key=lambda t: t[1]))

k1 = od1.keys()
k2 = od2.keys()

diff = dict((k, n - k2.index(k)) for n, k in enumerate(k1))

If you don't need them then Sven solution is probably faster.

edit: not that faster honestly... (sven.py is his second, more efficient version):

$ cat /tmp/mine.py | time python -m timeit
10000000 loops, best of 3: 0.0842 usec per loop
real    0m 3.69s
user    0m 3.38s
sys 0m 0.03s
$ cat /tmp/sven.py | time python -m timeit
10000000 loops, best of 3: 0.085 usec per loop
real    0m 3.86s
user    0m 3.42s
sys 0m 0.03s

If someone wants to post formatted bigger dicts I'll test them too.