在python中按值对defaultdict进行排序

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.

---

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)]

Answer 3

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

Eg:

  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)

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])