Group dictionary key values in python

Question

I have the following dictionary

mylist = [{\'tpdt\': \'0.00\', \'tmst\': 45.0, \'tmdt\': 45.0, \'pbc\': 30, \'remarks\': False, \'shift\': 1, \'ebct\': \'0.         


        

Answer 1

Here's a code that doesn't require mylist to be sorted by "mc_no" key:

from collections import defaultdict

sums = defaultdict(int)  # key -> sum
for d in mylist:
    sums[d["mc_no"]] += d["tmst"]

Answer 2

You can use itertools.groupby:

>>> for key, group in itertools.groupby(mylist, lambda item: item["mc_no"]):
...     print key, sum([item["tmst"] for item in group])
... 
KA20 90.0
KA23 110.0

Note that for groupby to work properly, mylist has to be sorted by the grouping key:

from operator import itemgetter

mylist.sort(key=itemgetter("mc_no"))

Answer 3

First you needs to sort that list of dictionaries..using following code.. Ex.

    animals = [{'name':'cow', 'size':'large'},{'name':'bird', 'size':'small'},{'name':'fish', 'size':'small'},{'name':'rabbit', 'size':'medium'},{'name':'pony', 'size':'large'},{'name':'squirrel', 'size':'medium'},{'name':'fox', 'size':'medium'}]

    import itertools
    from operator import itemgetter
    sorted_animals = sorted(animals, key=itemgetter('size'))

then you use following code

    for key, group in itertools.groupby(sorted_animals, key=lambda x:x['size']):
        print key,
        print list(group)

After you will get following result...

large [{'name': 'cow', 'size': 'large'}, {'name': 'pony', 'size':'large'}]
medium [{'name': 'rabbit', 'size': 'medium'}, {'name': 'squirrel', 'size':'medium'}, {'name': 'fox', 'size': 'medium'}]
small [{'name': 'bird', 'size': 'small'}, {'name': 'fish', 'size': 'small'}]