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.

Asked By: imsc

||

Source

Answer 1

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)

Answered By: Gareth Latty

Answer 2

>>> 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

Answered By: jamylak

Answer 3

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

>>> 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)]

Answered By: fedorqui

Answer 4

Late answer, and not a direct answer to the question, but if you end-up here from a "Sorting a defaultdict by value in python" google search, this is how I sort ( normal python dictionaries cannot be sorted, but they can be printed sorted) a defaultdict by its values:

orders = {
    'cappuccino': 54,
    'latte': 56,
    'espresso': 72,
    'americano': 48,
    'cortado': 41
}

sort_orders = sorted(orders.items(), key=lambda x: x[1], reverse=True)

for i in sort_orders:
    print(i[0], i[1])

Demo

Answered By: Pedro Lobito

Sorting a defaultdict by value in python

Question:

Answers: