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Jorge Ponti
Asked: 2020-07-04 10:58:21 +0800 CST 2020-07-04 10:58:21 +0800 CST

Show position of repeated elements of a Python list

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I am doing a little development, and I need to have the position of the repeated elements of a list.

What I have done for this is the following:

FDE = [1,2,3,4,4,3]
for k in range(6):
   m=[i for i,x in enumerate(FDE) if x==FDE[k]]
print(m)

The result that the screen shows me is:

m=[2.5]

In short, it only looks for the repeated one '3'but not the '4'one that is also repeated , ideally it would print :

m=[2,3,4,5]
python

2 Answers

  1. Best Answer

    There are several ways to do it, one of them is to use the method countto see if an element is found more than once in the list. It is not the most efficient but it is concise:

    FDE = [1,2,3,4,4,3]
m=[i for i, _ in enumerate(FDE) if FDE.count(x)>1]
print(m)

    Departure:

    [2. 3. 4. 5]

    Edition:

    As I have told you, one option to know the value associated with the indices is to use a dictionary that has each unique duplicate element as its key and a list with the indices where it is found as its value.

    Again there are several options, a relatively simple one is to use collections.defaultdict. We create a dictionary that stores the unique elements as keys and their indices as values, then we keep those pairs that have more than one index:

    from collections import defaultdict

FDE = [1,2,3,4,4,3]

aux = defaultdict(list)
for index, item in enumerate(FDE):
    aux[item].append(index)
result = {item: indexs for item, indexs in aux.items() if len(indexs) > 1}
print(result)

    Departure:

    {3: [2, 5], 4: [3, 4]}

    • 2

  2. Here is a modification of the great solution given by @FJSevilla that can run a bit faster, since it removes a loop forin exchange for creating an object Counterfrom the list, which gets a dictionary with the number of occurrences (values ) for each element of the list (keys).

    from collections import Counter
from collections import defaultdict


lista = [1,2,3,4,4,3]
counts_por_elem = Counter(lista)

indices_por_elem = defaultdict(list)
indices = []   

for indice, elem in enumerate(lista):
  if counts_por_elem[elem] > 1:
    indices.append(indice)  
    indices_por_elem[elem].append(indice)

print(indices_por_elem)
print(indices)

    As output you get both a dictionary with the indices per element, and a list with the indices of the repeated elements.

    defaultdict(<class 'list'>, {3: [2, 5], 4: [3, 4]})

    [2, 3, 4, 5]

    • 2