Sorting a defaultdict by value in python
defaultdict, dictionary, python, sorting
Solution
>>> 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
Problem
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.