Index of Duplicates Items in a Python List

List indexes of duplicate values in a list with Python

a, seen, result = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5], set(), []
for idx, item in enumerate(a):
    if item not in seen:
        seen.add(item)          # First time seeing the element
    else:
        result.append(idx)      # Already seen, add the index to the result
print result
# [3, 4, 7, 8, 9]

Edit: You can just use list comprehension in that function, like this

def list_duplicates(seq):
    seen = set()
    seen_add = seen.add
    return [idx for idx,item in enumerate(seq) if item in seen or seen_add(item)]

print list_duplicates([1, 2, 3, 2, 1, 5, 6, 5, 5, 5])
# [3, 4, 7, 8, 9]

How to fix the index of duplicate in a list

"Return Value from List index()
The index() method returns the index of the given element in the list.
If the element is not found, a ValueError exception is raised.
Note: The index() method only returns the first occurrence of the matching element.(https://www.programiz.com/python-programming/methods/list/index)"

https://www.geeksforgeeks.org/enumerate-in-python/

import copy

a = [1,2,2,6,7]
b = copy.deepcopy(a)
b.sort(reverse = True)
position = []

arr=[]
for i in enumerate(b):# a[7, 6, 2, 2, 1]
    ''' 
    ind val
    (0, 7)
    (1, 6)
    (2, 2)
    (3, 2)
    (4, 1)
    '''
    arr.append(i[0])
position=arr[::-1]
'''
a=  [1,2,2,6,7]
ind=[0,1,2,3,4]
reverse ...
b=   [7,6,2,2,1]
ind= [0,1,2,3,4]
reverse ...
resul=[4,3,2,1]
'''

print(position)

Finding indices of duplicate items in Python

If you already have a numpy array, you can use np.unique and use the return_inverse flag. Use the inverse array to find all positions where count of unique elements exceeds 1, and find their indices.

import numpy as np
arr = np.array([[10,10],[3,6],[2,4],[10,10],[0,0],[2,4]])
vals, inverse, count = np.unique(arr,
                                 return_inverse=True,
                                 return_counts=True,
                                 axis=0)
out = np.where(count[inverse] > 1)[0] #find all indices where counts > 1
print(out) #array([0, 2, 3, 5], dtype=int64)

Using .index() with a list that has repeated elements

Use enumerate.

>>> top = [1, 3, 7, 8, 3, -3, 3, 0]
>>> hits = (i for i,value in enumerate(top) if value == 3)

This is a generator that will yield all indices i where top[i] == 3.

>>> for i in hits:
...     print(i)
... 
1
4
6

how to index duplicates in a list?

You can use a dictionary where the key is an item from your id list and the value is the count. Then when you find a new id, append the id to the output list. However, if the id is already in the dictionary, then increment the count and append the id along with that count to the output list. Finally, return the list. Here is an example code in python:

ids = ["a", "b", "c", "a", "a", "d", "c"]
output_ids = []
id_count = {}

# iterate through ids
for item in ids:
    # if it's a new id
    if item not in id_count:
        # add it to dictionary
        id_count[item] = 0
        # add it to the output as is
        output_ids.append(item)
    # if it's not new
    else:
        # increment its count
        id_count[item] += 1
        # add it to the output along with its count
        output_ids.append(f"{item}_{id_count[item]}")

print(output_ids)  # ['a', 'b', 'c', 'a_1', 'a_2', 'd', 'c_1']

---

Related Topics