Find last index of element in list with duplicate elements python

Question

I have a list like this : [1,12,3,4,4,5,12,15,13,11]

I want to find index of 12 i.e 6. I have tried linear approach but it is not efficient.

item = 12
for i in range(len(mylist)):
    if mylist[i] == item:
        index = i
return index

Any efficient way to get this ?

What is wrong with a linear approach? You could approach from the right, maybe. But the solution is fundamentally linear. So what do you mean by "efficient"? — juanpa.arrivillaga, Jun 25 '17 at 06:35
What i mean by efficient is can we apply binary search for this ? — Kishan Mehta, Jun 25 '17 at 06:54

score 8 · Accepted Answer · answered Jun 25 '17 at 06:32

8

Go from right to left:

mylist = [1,12,3,4,4,5,12,15,13,11]
item = 12
for i in range(len(mylist)-1,-1,-1):
    if mylist[i] == item:
        index = i
        print(index)
        break

answered Jun 25 '17 at 06:32

Dmitriy Fialkovskiy

3,065
8
32
47

vZ10 · Answer 2 · 2017-06-25T06:45:10.180

3

len(mylist) - list(reversed(mylist)).index(item) -  1

will be enough

edited Jun 25 '17 at 06:45

answered Jun 25 '17 at 06:29

vZ10

2,468
2
23
33

1

The last occurrence, not the first. – zhenguoli Jun 25 '17 at 06:31
OK, reverse it. See my changes – vZ10 Jun 25 '17 at 06:34
2

you need to convert the index back to the regular indexing, not the "reversed" indexing. – Shai Jun 25 '17 at 06:36
You must be kidding me. It's really insolvable problem... reverse the index??? how??? – vZ10 Jun 25 '17 at 06:39
3

this will gie `AttributeError: 'listreverseiterator' object has no attribute 'index'` – Dmitriy Fialkovskiy Jun 25 '17 at 06:42
3

fix `len(mylist) - list(reversed(mylist)).index(item) - 1` – Dmitriy Fialkovskiy Jun 25 '17 at 06:43

gaurav · Answer 3 · 2017-06-25T07:12:11.697

2

The last occurrence of an element in a list is same as the first occurrence of that element in the reversed list.

So the index of last occurrence of an element in original list is equal to (length_of_list - index_in_reversed_list - 1).

You can use the list.reverse() method to reverse the list and then find the index of first occurrence of the required element in reversed list by list.index() method.

For example:

>>> mylist = [1,12,3,4,4,5,12,15,13,11]
>>> temp = mylist[:]
>>> temp.reverse()
>>> index = len(temp) - temp.index(12) - 1
>>> print(index)
6

edited Jun 25 '17 at 07:12

answered Jun 25 '17 at 06:40

gaurav

136
1
6

How is this more "efficient" than the OPs solution? – juanpa.arrivillaga Jun 25 '17 at 06:44
@juanpa.arrivillaga thanks for the help. I did not notice that. My mistake. But I corrected it now. – gaurav Jun 25 '17 at 06:50
@juanpa.arrivillaga It may not be as much more efficient than the OPs solution, but it provides an elegant way to solve the problem using inbuilt python functions that are anyday faster than your own implementation. – gaurav Jun 25 '17 at 06:52
@juanpa.arrivillaga Noted. Thanks again – gaurav Jun 25 '17 at 07:12
With most optimize way with respect to time and space complexity: Step-1: Start travelling the original array from last, Step-2: Once you find the given element. Step-3: Return the index = l - i -1; where l = total length of the array, i = index of the element in step-2. – ArunDhwaj IIITH Sep 04 '18 at 17:37

Find last index of element in list with duplicate elements python

3 Answers3