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Count all dictionary values by specific key and specific another dictionary value Python

November 18, 2021

I have this code:

base = [
    {'num': 10, 'bet': '100EUR'},
    {'num': 22, 'bet': '10EUR'},
    {'num': 10, 'bet': '210EUR'},
    {'num': 11, 'bet': '100EUR'},
    {'num': 8, 'bet': '50EUR'},
    {'num': 10, 'bet': '10EUR'},
    {'num': 8, 'bet': '30EUR'},
    {'num': 32, 'bet': '10EUR'}]

newbase = [{'num': b['num'], 'bet': int(b['bet'][:-3])}for b in base]
print(newbase)

There are duplicated values for dict ‘num’ in base. newbase = [{'num': b['num'], 'bet': int(b['bet'][:-3])}for b in base] deletes last 3 letter, so to be able to convert it to number. I have to make a program, that counts for each number how is sum of its money, and show 3 numbers with most money in them. For example output have to be like that:

10 = 320 EUR
11 = 100 EUR
8 = 80 EUR

How to do this?

>Solution :

I’m always looking to practice my Python, so here’s what I came up with.

import itertools

base = [
    {'num': 10, 'bet': '100EUR'},
    {'num': 22, 'bet': '10EUR'},
    {'num': 10, 'bet': '210EUR'},
    {'num': 11, 'bet': '100EUR'},
    {'num': 8, 'bet': '50EUR'},
    {'num': 10, 'bet': '10EUR'},
    {'num': 8, 'bet': '30EUR'},
    {'num': 32, 'bet': '10EUR'}]

# Sort by "num"
base.sort(key=lambda x:x['num'])

# Use groupby to create a list of tuples
# Where index 0 is "num" and index 1 is sum of bet
temp = []
for k,v in itertools.groupby(base, key=lambda x:x['num']):
    temp.append((k, sum(int(x['bet'][:-3]) for x in v)))

# Sort the tuples on bet
temp.sort(key=lambda x:x[1], reverse=True)

for i in range(3):
    print(f"{temp[i][0]} = {temp[i][1]} EUR")

Output: