CODEWITHSUNDEEP
pythonlistalgorithmtuples
Ariel Marcelo Pardo
Ariel Marcelo Pardo

Reputation: 156

Returning a list of tuples of frequency, where did I do wrong?

I have to do an exercise that ask me to do a function that find the frequency of consecutive values on a list and return a list of tuples where I have to show how many times it was the repetition.

Input:

[1, 1, 1, 3, 5, 1, 1, 3, 3]

Output:

[(1,3) , (3,1) , (5,1) , (1,2) , (3,2)]

Code:

def empaquetar(l):

lista = []
for i in range(len(l)-1):
    cont = 0
    j = i + 1
    while l[i] == l[j] and j<len(l)-1:
        cont += 1
        i += 1
        j += 1
    if cont > 0:
        lista.append((l[i], cont + 1))
    else:
        lista.append((l[i],1))
return lista

Well, what I wrote does not quite exactly return what I want and I can't find a solution but I search a lot, for example the tool Counter from collection or something like. I need help, any suggestion?

Upvotes: 1

Views: 69

Answers (4)

Subhashis Pandey
Subhashis Pandey

Reputation: 1538

Use a second for loop to check how many times the number is there consecutively.

def empaquetar(l):

lista = []
for i in l:
    cont = 1
    for j in range(i+1, len(l)):
      If l[j]==i:
        cont += 1
      else:
        break
  lista.append((i, cont))

return lista

Upvotes: 1

Troll
Troll

Reputation: 1925

I have developed an algorithm with a time complexity of O(n) for this.

def consecutive(lst):
    lst.append(0)

    count = 1
    result = []

    for i, value in enumerate(lst):
        if value == lst[min(len(lst) - 1, i + 1)]:
            count += 1
        else:
            result.append((value, count))
            count = 1

    return result

Results

Let's try a few test cases for this function.

>>> lst = [1, 1, 1, 3, 5, 1, 1, 3, 3]
>>> consecutive(lst)

[(1, 3), (3, 1), (5, 1), (1, 2), (3, 2)]

Here is another test case to verify the code.

>>> lst = [1, 2, 5, 5, 3, 2, 2, 1]
>>> consecutive(lst)

[(1, 1), (2, 1), (5, 2), (3, 1), (2, 2), (1, 1)]

Upvotes: 1

nadapez
nadapez

Reputation: 2727

from itertools import groupby
lista = [group[0], len(list(group[1])) for group in groupby(l)]

Upvotes: 1

madmatrix
madmatrix

Reputation: 255

Not thoroughly tested, but here is a try

prev_val = None
counter = 1
for curr_val in input_list:
    if prev_val is not None:
        if prev_val != curr_val:
            output_list.append((prev_val, counter))
            counter = 1
        else:
            counter += 1
            
    prev_val = curr_val
            
output_list.append((prev_val, counter))
print (output_list)

Upvotes: 1

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