How to get the last occurrence of list as output?

Question

Can anyone please explain how to output the rightmost index from several most-same-values indexes in the list?

my function:

def last_index(xs,key):
i = 0
    for i in range(len(xs)):
        if xs[i] == key:
            if i != len(xs):
                return i            
            else: 
                return 'None'

for example,

xs = [3,4,5,6,4,4,5]                
key = 4

the rightmost index output should be a single 5. But I got all three all them which are index 1,4,5. Thanks for the help, and sorry I'm totally new.

what if the input as strings like:

 xs=[True,True,True,False]
 key = True

I believe the output is 2?

I think question is about counting occurrence of key in list — AlokThakur, Feb 27 '16 at 16:54
Duplicate of http://stackoverflow.com/questions/6890170/how-to-find-the-last-occurrence-of-an-item-in-a-python-list — SPKoder, Feb 27 '16 at 16:56
@AlokThakur Yes. I read the question too quickly. My mistake. I retracted it. — idjaw, Feb 27 '16 at 16:59

score 3 · Accepted Answer · edited May 23 '17 at 12:23

3

This simple solution should do:

def last_index(xs, key):
    index = None
    for i in xrange(len(xs)):
        if xs[i] == key:
            index = i  # override index, result in the rightmost index of key
    return index  # returns None if key is not on the list

A more efficient way to do this is by iterating from the end to start and returning the index when key is found, in worst case - key is not found and we will iterate over the entire list.

Check out the more efficient version:

def last_index(xs, key):
    index = None
    for i in xrange(len(xs)-1, 0, -1):  # iterate from last item to first
        if xs[i] == key:
            index = i
            break  # found the rightmost index, exit the loop
    return index

Notice you should prefer using xrange over range (unless in python 3 where range equals to xrange), also to avoid an edge case when items involve different types see Andriy's answer.

edited May 23 '17 at 12:23

Community

1
1

answered Feb 27 '16 at 16:57

Forge

6,538
6
44
64

Thanks a lot! Didn't think about the override! – markzzzz Feb 27 '16 at 17:08
@markzzzz you would loop thought list more than needed, traverse in reverse order – Andriy Ivaneyko Feb 27 '16 at 17:10
Nice, just want to point you to NOTE#2 within my answer, if you wanted you can point to that moment as your answer is accepted and future views would mainly be concentrated on your answer )) Have a good day ! – Andriy Ivaneyko Feb 27 '16 at 17:27

score 1 · Answer 2 · answered Feb 27 '16 at 16:55

You can try a function like this

def last_index(xs,key):
    index = -1
    for i in range(len(xs)):
        if xs[i] == key:
            index=i           
    if index!=-1:
        return index
    else:
        return "none"

This will get the last index that matches your key. If there is none will return "none".

score 1 · Answer 3 · answered Feb 27 '16 at 16:57

1

This should do the trick:

def last_index(xs,key):
    index = -1
    for i in range(len(xs)):
        if xs[i] != key:
            continue
        else:
            index = i
    return index if index != -1 else 'None'

answered Feb 27 '16 at 16:57

Ian

30,182
19
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score 1 · Answer 4 · edited May 23 '17 at 10:28

Traverse xs in reverse order and return first matched value, with reversed function:

def last_index(xs,key):
    for i in reversed(range(len(xs))):
        if xs[i] == key:
            return i


xs = [3,4,5,6,4,4,5]
key = 4
print last_index(xs, key) # output: 5

xs=[True,True,True,False]
key = True
print last_index(xs, key) # output: 2
print last_index(xs, 2) # output: None

NOTE#1

You can use xrange instead of range it would give you better performace and won't be deprecated since python3, see Should you always favor xrange() over range()? for more info.

Your comparison may be improved by replacing

if xs[i] == key

to

if xs[i] == key and type(a) == type(b)

NOTE#2

To avoid bug when your 1 == True would return you index of True however you wanna index of 1 whicn not exist, compare result for both if conditions when xs and key have values below

xs=[True,True,True,False]
key = 1

See Strict comparison for more information about that behaviour.

Just to note, if he is using python 3 then he has to use range (xrange does not exists anymore in python 3). — Hadrián, Feb 27 '16 at 17:06

score 0 · Answer 5 · answered Feb 27 '16 at 16:54

0

You can reverse the list and then use .index():

index = xs[len(xs) - list(reversed(xs)).index(key)]

By the way, in your second list, True and False are booleans, not strings.

answered Feb 27 '16 at 16:54

zondo

19,901
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Spyke · Answer 6 · 2016-03-01T20:43:42.010

0

Iterate from behind like this:

def last_index(xs,key):
    i= len(xs)-1
    while i >=0:
        if xs[i] == key:
            return i
         i -= 1

This way if the key does not exist the function will return the none value

edited Mar 01 '16 at 20:43

answered Feb 27 '16 at 17:02

Spyke

114
4

How to get the last occurrence of list as output?

6 Answers6