How to find position of similar numbers in a list

Question

Given a list of elements, with certain similar elements in it, how do I write a code in Python to find the position of all the similar elements in it? This must be done using a for loop and if conditions within the for loop.

list1 = [5, 90, 10, 5, 100, 5]

So in this case, since 5 is the element that has repeats or ties, the output will be 0,3,5.

Possible duplicate of [Finding the index of an item given a list containing it in Python](https://stackoverflow.com/questions/176918/finding-the-index-of-an-item-given-a-list-containing-it-in-python) — SuperKogito, Mar 13 '19 at 16:18
*"This must be done using a for loop and if conditions within the for loop."* This is what we would like to tell you as well. Try and later come back with a specific question if you're stuck. — Austin, Mar 13 '19 at 16:19

score 2 · Answer 1 · answered Mar 13 '19 at 16:21

You can try to use set to find the unique set of elements in the list and then check for indexes of the repeated elements.

Try the code below:

from collections import defaultdict

list1 = [5, 90, 10, 5, 100, 5]

set1 = set(list1)
res_dict = defaultdict(list)

for x in set1:
    for i, y in enumerate(list1):
        if x == y:
            res_dict[x].append(i)

print res_dict

Output:

{90: [1], 100: [4], 10: [2], 5: [0, 3, 5]}

A l w a y s S u n n y · Answer 2 · 2019-03-13T16:29:40.040

1

With enumerate inside list comprehension for 5,

>>> list1 = [5, 90, 10, 5, 100, 5]
>>> all_index = [i for i, j in enumerate(list1) if j == 5]
>>> all_index

Output:

[0, 3, 5]

With loop for all element,

list1 = [5, 90, 10, 5, 100, 5]
result = {}
for e in list1:
    result[e] = [i for i, j in enumerate(list1) if j == e]
print(result)

Output:

 {90: [1], 10: [2], 100: [4], 5: [0, 3, 5]}

edited Mar 13 '19 at 16:29

answered Mar 13 '19 at 16:19

A l w a y s S u n n y

36,497
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score 0 · Answer 3 · answered Mar 13 '19 at 16:24

Using np.where will get you what you want:

import numpy as np

list1 = [5, 90, 10, 5, 100, 5]

data = np.array(list1)
unique = sorted(list(set(data)))

match = {}
for i in range(len(unique)):
    match[unique[i]] = np.where(data == unique[i])[0]

print(match)
>>> {5: array([0, 3, 5]), 10: array([2]), 90: array([1]), 100: array([4])}

score 0 · Answer 4 · answered Mar 13 '19 at 16:50

Using only a for loop and if conditions:

input_list = [5, 90, 100, 5, 100, 5]

elements = {}
result = []

for i, e in enumerate(input_list):
    if e in elements:
        if not elements[e][1]:
            result.append(elements[e][0])
            elements[e][1] = True
        result.append(i)
    else:
        elements[e] = [i, False]

print(sorted(result))

How to find position of similar numbers in a list

4 Answers4