Comparing dictionary values() with list values

Question

I have the following dictionary and a list; from the dictionary I extracted the top 10 values. My question is this: I need to compare the keys of the respective values extracted from the dictMaior with the list. For example, Keys() do value() 72 is 19, and 19 is on the list? I need a code that returns me a list containing all the values of the dictMaior that is present in the list. I tried with items(), but it didn’t work ex. dictMaior = heapq.nlargest(10, dicionario.items())

import heapq

dicionario = {1: 68, 2: 53, 3: 57, 4: 56, 5: 60, 6: 58, 7: 62, 8: 53, 9: 57, 10: 64, 11: 53, 12: 54, 13: 64, 14: 58, 15: 63, 16: 61, 17: 56, 18: 64, 19: 72, 20: 65, 21: 63, 22: 55, 23: 57, 24: 62, 25: 65}

lista = [21, 20, 23, 8, 10, 18, 19, 2, 12, 14, 4, 9, 22, 16, 5]

# Extraindo os 10 maiores itens do dicionário através do values()
dictMaior = heapq.nlargest(10, dicionario.values())
# Saida do dictMaior:
# [72, 68, 65, 65, 64, 64, 64, 63, 63, 62]

Answer 1

Grateful to all who collaborated with me, but after eating a lot of documentation I arrived at the following code below that satisfied my need. I thought it was better to use "Collections" and it worked really well... Note: The list and the dictionary is just an adaptation of a csv of 1831 lines, just simplified to be able to post here... Thank you for all your efforts!

from collections import Counter

lista = [21, 20, 23, 8, 10, 18, 19, 2, 12, 14, 4, 9, 22, 16, 5]
dicionario = {1: 68, 2: 53, 3: 57, 4: 56, 5: 60, 6: 58, 7: 62, 8: 53, 9: 57, 10: 64, 11: 53, 12: 54, 13: 64, 14: 58, 15: 63, 16: 61, 17: 56, 18: 64, 19: 72, 20: 65, 21: 63, 22: 55, 23: 57, 24: 62, 25: 65}
numeros_presentes = []

cnt = Counter(dicionario)
ordenado = cnt.most_common()
ordenado = ordenado[0:10]

for num in ordenado:
    if num[0] in lista:
        numeros_presentes.append(num[0])

print(numeros_presentes)

# Saída do código:
# [19, 20, 10, 18, 21]

Answer 2

Well got a little confused even more because the return I got is an empty list but come on:

You have a list called: lista = [21, 20, 23, 8, 10, 18, 19, 2, 12, 14, 4, 9, 22, 16, 5]

and another list called: dictMaior = [72, 68, 65, 65, 64, 64, 64, 63, 63, 62]

What you need to check is if some dictMaior number is in the list then: numeros_presentes = [num for num in dictMaior if num in lista]

In other words num will be the name assigned to each value in the dictMaior then if num is present in lista it will be returned within numeros_presentes.

Answer 3

Only one variant of the @Claudinei version using lists in comprehension

from collections import Counter
lista = [ 21 ......
dic = { 1:68 ......
print([k for k,c in Counter(dic).most_common(10) if k in lista])

Answer 4

dicionario = {1: 68, 2: 53, 3: 57, 4: 56, 5: 60, 6: 58, 7: 62, 8: 53, 9: 57, 10: 64, 11: 53, 12: 54, 13: 64, 14: 58, 15: 63, 16: 61, 17: 56, 18: 64, 19: 72, 20: 65, 21: 63, 22: 55, 23: 57, 24: 62, 25: 65}
lista = [21, 20, 23, 8, 10, 18, 19, 2, 12, 14, 4, 9, 22, 16, 5]
lista2 = []
a = 0
for x in lista:
   lista2.append(dicionario[x])
   a += 1
lista2.sort(reverse=True)
print(lista2)