Sorting a defaultdict by value in python

>>> sorted(cityPopulation.iteritems(),key=lambda (k,v): v[0],reverse=True) #1990[('C', [30, 10, 20]), ('B', [20, 30, 10]), ('A', [10, 20, 30])]>>> sorted(cityPopulation.iteritems(),key=lambda (k,v): v[2],reverse=True) #2010[('A', [10, 20, 30]), ('C', [30, 10, 20]), ('B', [20, 30, 10])]

Note in python 3 you can't automagically unpack lambda arguments so you would have to change the code

sorted(cityPopulation.items(), key=lambda k_v: k_v[1][2], reverse=True) #2010

If you want to sort based on the values, not in the keys, use data.items() and set the key with lambda kv: kv[1] so that it picks the value.

See an example with this defaultdict:

>>> from collections import defaultdict>>> data = defaultdict(int)>>> data['ciao'] = 17>>> data['bye'] = 14>>> data['hello'] = 23>>> datadefaultdict(<type 'int'>, {'ciao': 17, 'bye': 14, 'hello': 23})

Now, let's sort by value:

>>> sorted(data.items(), lambda kv: kv[1])[('bye', 14), ('ciao', 17), ('hello', 23)]

Finally use reverse=True if you want the bigger numbers to come first:

>>> sorted(data.items(), lambda kv: kv[1], reverse=True)[('hello', 23), ('ciao', 17), ('bye', 14)]

Note that key=lambda(k,v): v is a clearer (to me) way to say key=lambda(v): v[1], only that the later is the only way Python 3 allows it, since auto tuple unpacking in lambda is not available.

In Python 2 you could say:

>>> sorted(d.items(), key=lambda(k,v): v)[('bye', 14), ('ciao', 17), ('hello', 23)]

A defaultdict doesn't hold order. You might need to use a OrderedDict, or sort the keys each time as a list.

E.g:

  from operator import itemgetter  sorted_city_pop = OrderedDict(sorted(cityPopulation.items()))

Edit: If you just want to print the order, simply use the sorted builtin:

for key, value in sorted(cityPopulation.items()):    print(key, value)