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How can I find the index of the duplicate element in the array?

mylist = [1,2,3,4,5,6,7,8,9,10,1,2]

How can I find the index number of 1 and 2 repeated array elements here? So, not the first duplicate element, I want to find and delete the trailing duplicate element from the array. 1 2 values at the end. Thanks for your responses.

Apo1
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  • If you want to get unique values you can use ```set()```. ```myList = list(set(myList))``` will return a list with unique values. – Arthur Aug 08 '22 at 16:03
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    `list(dict.fromkeys(mylist))`, if you need to ensure their order in the initial list. – Mechanic Pig Aug 08 '22 at 16:04
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    Does this answer your question? [How do I remove duplicates from a list, while preserving order?](https://stackoverflow.com/questions/480214/how-do-i-remove-duplicates-from-a-list-while-preserving-order) – ouroboros1 Aug 08 '22 at 16:20

5 Answers5

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Single pass remove duplicates:

mylist = [1,2,3,4,5,6,7,8,9,10,1,2]

def remove_duplicates(l):
    seen = {}
    res = []
    for item in l:
        if item not in seen:
            seen[item] = 1
            res.append(item)
    return res

print(remove_duplicates(mylist))

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Which also preserves order:

mylist = [1,10,3,4,5,6,7,8,9,2,1,2]
print(remove_duplicates(mylist))
[1, 10, 3, 4, 5, 6, 7, 8, 9, 2]
Tom McLean
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    IMO this is the right approach, but `seen` is a dict where values don't matter. Make it a set instead: `seen = set()` and `seen.add(item)`. – VPfB Aug 08 '22 at 16:28
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The simplest way to do this (unless you want to log where the duplicates are) is to convert the list to a set.

A set can contain only one instance of each value.

You can do this very easily as shown below.

Note: A set is respresented with curly braces { like a dictionary, but does not have key,value pairs.

mylist = [1,2,3,4,5,6,7,8,9,10,1,2]
myset = set(mylist)
// myset = {1,2,3,4,5,6,7,8,9,10}

EDIT: If the order is important, this method will not work as a set does not store the order.

HPringles
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    Don't do this, if you want to preserve the order. `set()` does *not* preserve order, even if in this example it **looks** like it does. E.g. with `mylist = [1,2,3,4,5,11,7,8,9,10,1,2]`, myset will become: `{1, 2, 3, 4, 5, 7, 8, 9, 10, 11}`. – ouroboros1 Aug 08 '22 at 16:06
  • Ah yes, that does make sense. I'll update my answer to make that clear – HPringles Aug 08 '22 at 16:07
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    @HSPringles I made an answer that preserves order, its the classic leetcode remove duplicates problem – Tom McLean Aug 08 '22 at 16:20
  • You could check out the latest post below. – Daniel Hao Aug 08 '22 at 17:27
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You can find duplicates in a cycle, memorizing previously seen values in a set.

mylist = [1,2,3,4,5,6,7,8,9,10,1,2]

myset = set()
indices_of_duplicates = []

for ind, val in enumerate(mylist):
    if val in myset:
        indices_of_duplicates.append(ind)
    else:
        myset.add(val)

Then you can delete duplicate elements with

for ind in reversed(indices_of_duplicates):
    del mylist[ind]

Note that we are deleting elements starting from the end of the list because otherwise we would shift elements and we would have to update the indices we have found.

Василий
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eh, I did this:

a=[1,2,3,4,5,5,6,7,7,8]
b=list(set(a))

tr=0
al=len(a)
bl=len(b)
iz=[]

if al != bl:
    for ai in range (0,al):
        ai0=a[ai]
        bi0=b[ai]
        if ai0 != bi0:
            iz.append(ai)
            b.insert(ai,ai0)

then the list iz is all the i values where they don't match, so you just plug those back in. just wanted to share!

Marlon D. Alessandra
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If you have to remove duplicates you can use Set()

mylist = [x for x in set([1,2,3,4,5,6,7,8,9,10,1,2])]
Talon
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  • Same problem as with other answer, on top of being verbose. This is just: `list(set([1,2,3,4,5,11,7,8,9,10,1,2]))`, which becomes: `[1, 2, 3, 4, 5, 7, 8, 9, 10, 11]`. So, order is again not preserved. – ouroboros1 Aug 08 '22 at 16:09