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Sorting dictionaries by value in Python 2.4 (Python recipe) by Nick Coghlan
ActiveState Code (http://code.activestate.com/recipes/304440/)

5

Python 2.4 adds a new builtin function sorted(), which can make obtaining the items of a dictionary sorted by key or value a single line operation.

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# Example from PEP 265 - Sorting Dictionaries By Value
#    Counting occurences of letters

d = {'a':2, 'b':23, 'c':5, 'd':17, 'e':1}

# operator.itemgetter is new in Python 2.4
#  `itemgetter(index)(container)` is equivalent to `container[index]`
from operator import itemgetter

# Items sorted by key
#   The new builtin `sorted()` will return a sorted copy of the input iterable.
print sorted(d.items())

# Items sorted by key, in reverse order
#   The keyword argument `reverse` operates as one might expect
print sorted(d.items(), reverse=True)

# Items sorted by value
#    The keyword argument `key` allows easy selection of sorting criteria
print sorted(d.items(), key=itemgetter(1))

# In-place sort still works, and also has the same new features as sorted
items = d.items()
items.sort(key = itemgetter(1), reverse=True)
print items

The new 'sorted' builtin is a convenient expression replacement for the old statement based approach to obtaining a sorted copy.

The new keyword arguments to the sort operations are extremely handy for tasks such as sorting a dictionary by value (see the "Sort names and separate by last initial" recipe for a more involved use of the new features)

Tags: search

3 comments

stewart midwinter 6 years, 4 months ago  # | flag

not a dictionary. note that the example creates a list, not a dictionary.

C T 5 years, 12 months ago  # | flag

I'm not sure how new this feature is, but in Python 2.5 i use this:

di = d.items()
if b_sorted:
    di.sort()  # by first item

if b_ranked:
    di.sort(key=lambda x: x[1])  # by second item
steve charlesworth 5 years, 3 months ago  # | flag

Uh, yeah... Since the goal was to get something that's ordered, a list makes sense.

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