How to get top 3 count of names from list of dictionary in python

Question

Hello guys,

 resp = [{**"NR": "0"**,"Code": "4_RESOURCE","Cnt": "11"}, 
                {"NR": "10","Code": "10_humans","Cnt": "1"},
                {"NR": "1000","Code": "4_RESOURCE","Cnt": "120"}, 
                {**"NR": "0"**,"Code": "10_humans","Cnt": "12"},
                 {**"NR": "0"**,"Code": "4_RESOURCE","Cnt": "15"},
                 {**"NR": "0"**,"Code": "50_animals","Cnt": "20"}]

from it, if "NR" is "0" from unique Code from the above list of dictionary then need to take count and add that count against the unique code. like that need to take top three counts. it should be in dictionary as shown in output sample

output sample: count of Cnt {"4_RESOURCE": 26(15+11), "4_RESOURCE": 15, "10_humans":12} -------[Top three in dictionary]

I tired:

 def se_conc(response):
    cnt = 0
    list = []

    def key_func(k):
         return k['Code']
   INFO = sorted(resp, key=y_func)
   for k, v in groupby(INFO, y_func):
            print("codes:", k)
            print("v:", v)
            list.append(k)
            print("listcode", list)

            for key in list:
                if resp['NR'] == "0":
                    print("NR is 0:", k)

Thanks in advance

question from:https://stackoverflow.com/questions/65933417/how-to-get-top-3-count-of-names-from-list-of-dictionary-in-python

Answer 1

Use a collections.Counter and its most_commons method:

from collections import Counter

c = Counter()
for d in resp:
    if d["NR"] == "0":
        c[d["Code"]] += int(d["Cnt"])

dict(c.most_common(3))
# {'4_RESOURCE': 26, '50_animals': 20, '10_humans': 12}