How do I check which items in List2 are in List1, then their count and order

Question

I am a beginner in Python and I would like to count on your help for the problems below. I have worked on two lists as the following objectives: 1.Check which items in List2 are in List1 2.Count each List2 item in List1 and sort it in Orderly List2.

Lista1 = ['a', 'k', 'j', 'd', 'p', 'u', 'j', 'h', 'k', 'k', 'c', 'p', 'e', 'u', 'z', 'z', 'f', 's', 'u', 'k', 'p', 'u', 'j', 'h', 'b', 'k', 'j', 'd', 'e', 'u', 'j', 'z']

Lista2 = ['j', 'u', 'z', 'k', 'u', 'p']

Lista_ordenada = [ ]


for i in  Lista1:
  if i == 'j' :
    break
Lista_ordenada.append('j')
.
.
.
for i in  Lista1:
  if i == 'p' :
    break
Lista_ordenada.append('p')

Of problems:

1. Although the code works, I see that it is very repetitive and slow;
2. Although I managed to put them in the ordered list (in a disorderly way), I could not do the count, nor order them.

Any help will be welcome.

Answer 1

I believe the solution below meets your need:

Input:

lista1 = ['a', 'k', 'j', 'd', 'p', 'u', 'j', 'h', 'k', 'k', 'c', 'p', 'e', 'u', 'z', 'z', 'f', 's', 'u', 'k', 'p', 'u', 'j', 'h', 'b', 'k', 'j', 'd', 'e', 'u', 'j', 'z']
lista2 = ['j', 'u', 'z', 'k', 'u', 'p']
lista_ordenada = []

for i in  lista2:
  if i in lista1:
    item_tuple = (i, lista1.count(i))
    lista_ordenada.append(item_tuple)
 
lista_ordenada = sorted(lista_ordenada, key=lambda tup: tup[1], reverse = True)
print(lista_ordenada)

Output:

[('j', 5), ('u', 5), ('k', 5), ('u', 5), ('z', 3), ('p', 3)]

The For loop checks if each item in List2 is in List1 and, if so, creates a tuple (item_tuple) of the item itself (i) with the number of occurrences (list1.Count(i)).

After creating the tuple, it adds this set in the sorted list via the . append(item_tuple).

Finally, the function Sorted() sorts the list_ordered by the second tuple item, which is the number of occurrences. Reverse = True was used to sort decreasingly, since the default function is the ascending sort.

Answer 2

The approach you used is not the most efficient because for each element of Lista2 you had to write a specific code to perform the comparison and addition of the element to Lista_ordenada, case Lista2 its code would no longer fulfill its purpose and would have to be written a new comparison loop for the new element. Another clear thing is the question of performance checking Lista1 is repeated making the algorithm inefficient.

Re-examined the question, there are three problems:

1. Check which items from Lista2 are in the Lista1.
2. Count each item of the Lista2 in Lista1.
3. Sort items in the Lista_ordenada.

To check which items of Lista2 are in the Lista1 transform them into ensembles and make the intercession with the operator &.

To count each item of the Lista2 in Lista1 brood accountants and use only those elements that are in Lista_ordenada to filter the result (because its elements are common the two lists).

To sort items in Lista_ordenada use the built-in function sorted().

from collections import Counter

Lista1 = ['a', 'k', 'j', 'd', 'p', 'u', 'j', 'h', 'k', 'k', 'c', 'p', 'e', 'u', 'z', 'z', 'f', 's', 'u', 'k', 'p', 'u', 'j', 'h', 'b', 'k', 'j', 'd', 'e', 'u', 'j', 'z']
Lista2 = ['j', 'u', 'z', 'k', 'u', 'p']

#Faz a intercessão de Lista1 e Lista2 e ordena o resultado.
Lista_ordenada = sorted(set(Lista1) & set(Lista2))

#Conta a frequencia de cada elemento de Lista1 e Lista2.
Contagem1 = Counter(Lista1)
Contagem2 = Counter(Lista2)

print(f"Elementos comuns as duas listas {Lista_ordenada}")

for k in Lista_ordenada:
  print(f"O elemento '{k}' aparece {Contagem1[k]} vez{'es' if Contagem1[k]>1 else '' } na lista1.")
  print(f"O elemento '{k}' aparece {Contagem2[k]} vez{'es' if Contagem2[k]>1 else '' } na lista2.")
  print(f"Num total de {Contagem1[k] + Contagem2[k]} aparições.")

#Elementos comuns as duas listas ['j', 'k', 'p', 'u', 'z']
#O elemento 'j' aparece 5 vezes na lista1.
#O elemento 'j' aparece 1 vez na lista2.
#Num total de 6 aparições
#O elemento 'k' aparece 5 vezes na lista1.
#O elemento 'k' aparece 1 vez na lista2.
#Num total de 6 aparições
#O elemento 'p' aparece 3 vezes na lista1.
#O elemento 'p' aparece 1 vez na lista2.
#Num total de 4 aparições
#O elemento 'u' aparece 5 vezes na lista1.
#O elemento 'u' aparece 2 vezes na lista2.
#Num total de 7 aparições
#O elemento 'z' aparece 3 vezes na lista1.
#O elemento 'z' aparece 1 vez na lista2.
#Num total de 4 aparições

^{Test the code on Ideone}