Find every element's first index in a list

Question

Here is a list a=[1,1,1,2,4,2,4,32,1,4,35,23,24,23]

I do this in python:

unique_number=list(set(a))
ans=map(lambda x:a.index(x),unique_number)

output:

<map at 0x2b98b307828>

I want to know what's wrong with my code and find an more efficient way to achieve this.

What exactly is your desired output? Also you don't need lambda to get `list(map(a.index,list(set(a))))` — Chris_Rands, Aug 29 '17 at 08:31
Possible duplicate of https://stackoverflow.com/q/1303347/2301450 — vaultah, Aug 29 '17 at 08:32
If you are trying to get the indexes of the unique values, then @khelwood is right, hence the output. If you are trying to get the Unique value itself, then remove `ans=map(lambda x:a.index(x),unique_number)` and directly print `unique_number` — SajidSalim, Aug 29 '17 at 08:33
Possible duplicate of [Handling map function in python2 & python3](https://stackoverflow.com/questions/41090310/handling-map-function-in-python2-python3) — Netwave, Aug 29 '17 at 08:36

score 2 · Answer 1 · answered Aug 29 '17 at 08:30

2

This code would work as you expected in Python 2. In Python 3, map returns an iterator. You could, e.g., convert it to a list:

>>> ans=map(lambda x:a.index(x),unique_number)
>>> list(ans)
[7, 0, 3, 10, 4, 11, 12]

answered Aug 29 '17 at 08:30

Mureinik

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No need for lambda here – Chris_Rands Aug 29 '17 at 08:37
@Chris_Rands this is copied from the OP. I was just explaining why he got the output he got. – Mureinik Aug 29 '17 at 08:42
1

Might be worth noting for future reference that the whole lambda can go when using map... Just `list(map(a.index, set(a))` will do it... – Jon Clements Aug 29 '17 at 08:51

score 1 · Answer 2 · answered Aug 29 '17 at 08:29

1

Try this:

for value in map(lambda x:a.index(x),unique_number):
     print(value)

or append this:

for var in ans:
     print(var)

answered Aug 29 '17 at 08:29

matsbauer

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score 1 · Accepted Answer · answered Aug 29 '17 at 08:38

1

You can avoid keep re-indexing and building a set first - simply build a dict iterating over a backwards as the dictionary will only keep the last value for a key (in this case - the earliest appearing index), eg:

a=[1,1,1,2,4,2,4,32,1,4,35,23,24,23]

first_index = {v:len(a) - k for k,v in enumerate(reversed(a), 1)}
# {1: 0, 2: 3, 4: 4, 23: 11, 24: 12, 32: 7, 35: 10}

This way you're only scanning the sequence once.

answered Aug 29 '17 at 08:38

Jon Clements

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maybe take `len(a)` outside the dict comp – Chris_Rands Aug 29 '17 at 08:43
@Chris_Rands yup... or zip `a` with `range(len(a), -1, -1)` instead of `enumerate`... the idea can be tweaked however :) – Jon Clements Aug 29 '17 at 08:46

Find every element's first index in a list

3 Answers3