python sort list of dicts by same key value count-CodePudding

I have list of dicts, key 'city' could be repetitive. I need to sort this list from most repetitive city to least.

My list of dicts:

data = [
    {'city': 'Zp', 'p': 8},
    {'city': 'Kyiv', 'p': 2},
    {'city': 'Lviv', 'p': 7},
    {'city': 'Kyiv', 'p': 3},
    {'city': 'Kyiv', 'p': 4},
    {'city': 'Brd', 'p': 1},
    {'city': 'Kyiv', 'p': 5},
    {'city': 'Zp', 'p': 9},
    {'city': 'Lviv', 'p': 6},
]

I tried to sort by key value count, but got same result:

data = sorted(data, key=lambda x: data.count(x['city']))
# >> data
#     [
#       {'city': 'Zp', 'p': 8},
#       {'city': 'Kyiv', 'p': 2},
#       {'city': 'Lviv', 'p': 7},
#       {'city': 'Kyiv', 'p': 3},
#       {'city': 'Kyiv', 'p': 4},
#       {'city': 'Brd', 'p': 1},
#       {'city': 'Kyiv', 'p': 5},
#       {'city': 'Zp', 'p': 9},
#       {'city': 'Lviv', 'p': 6},
#     ]

This would give me almost needed result, but there is 2 objects with 'Lviv' and 2 with 'Zp', and they are mixed in result (also I think this approach requires too much resources)

data = sorted(
    data,
    key=lambda x: len(
        [i for i in data if i['city'] == x['city']]
    ),
    reverse=True
)
# >> data
#  [
#    {'city': 'Kyiv', 'p': 2},
#    {'city': 'Kyiv', 'p': 3},
#    {'city': 'Kyiv', 'p': 4},
#    {'city': 'Kyiv', 'p': 5},
#    {'city': 'Zp', 'p': 8},
#    {'city': 'Lviv', 'p': 7},
#    {'city': 'Zp', 'p': 9},
#    {'city': 'Lviv', 'p': 6},
#    {'city': 'Brd', 'p': 1},
#]

What I want to get:

# >> data
#    [
#        {'city': 'Kyiv', 'p': 2},
#        {'city': 'Kyiv', 'p': 3},
#        {'city': 'Kyiv', 'p': 4},
#        {'city': 'Kyiv', 'p': 5},
#        {'city': 'Lviv', 'p': 6},
#        {'city': 'Lviv', 'p': 7},
#        {'city': 'Zp', 'p': 8},
#        {'city': 'Zp', 'p': 9},
#        {'city': 'Brd', 'p': 1},
#    ]

CodePudding user response：

from collections import Counter
from pprint import pprint

data = [
    {'city': 'Zp', 'p': 8},
    {'city': 'Kyiv', 'p': 2},
    {'city': 'Lviv', 'p': 7},
    {'city': 'Kyiv', 'p': 3},
    {'city': 'Kyiv', 'p': 4},
    {'city': 'Brd', 'p': 1},
    {'city': 'Kyiv', 'p': 5},
    {'city': 'Zp', 'p': 9},
    {'city': 'Lviv', 'p': 6},
]

# find the number of occurrences of each city
cities = map(lambda d: d['city'], data)
c = Counter(cities)

# sort data according to 1) frequency of the city, 2) name of the city
sorted_data = sorted(data, key=lambda d: (c[d['city']], d['city']), reverse=True)

pprint(sorted_data)

CodePudding user response：

You need to sort by two keys here:

the number of a city record occurrences in data (you need to extract the list of city names only for this)
The city name for tie breaking

>>> sorted(
    data, 
    key=lambda x: (list(y['city'] for y in data).count(x['city']), x['city']), 
    reverse=True
)
[
    {'city': 'Kyiv', 'p': 2}, 
    {'city': 'Kyiv', 'p': 3}, 
    {'city': 'Kyiv', 'p': 4}, 
    {'city': 'Kyiv', 'p': 5}, 
    {'city': 'Zp', 'p': 8}, 
    {'city': 'Zp', 'p': 9}, 
    {'city': 'Lviv', 'p': 6}, 
    {'city': 'Lviv', 'p': 7}, 
    {'city': 'Brd', 'p': 1}
]