Sorting a defaultdict by value in python

Question

I have a data-structure which is something like this:

The population of three cities for different year are as follows.

Name  1990 2000 2010 A     10   20   30 B     20   30   10 C     30   10   20 

I am using a defaultdict to store the data.

from collections import defaultdict cityPopulation=defaultdict(list) cityPopulation['A']=[10,20,30] cityPopulation['B']=[20,30,10] cityPopulation['C']=[30,10,20] 

I want to sort the defaultdict based on a particular column of the list (the year). Say, sorting for 1990, should give C,B,A, while sorting for 2010 should give A,C,B.

Also, is this the best way to store the data? As I am changing the population values, I want it to be mutable.

Answer 1

>>> 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 

Answer 2

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.

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See an example with this defaultdict:

>>> from collections import defaultdict >>> data = defaultdict(int) >>> data['ciao'] = 17 >>> data['bye'] = 14 >>> data['hello'] = 23  >>> data defaultdict(<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)]