CODEWITHSUNDEEP
pythonlist
user308827
user308827

Reputation: 21961

Find last occurrence of any item of one list in another list in python

I have the following 2 lists in python:

ll = [500,500,500,501,500,502,500]
mm = [499,501,502]

I want to find out the position of last occurence of any item in mm, in the list ll. I can do this for a single element like this:

len(ll) - 1 - ll[::-1].index(502)
>> 5

Here ll[::-1].index(502) provides position of last occurence of 502 and len(ll) - 1 gives the total length.

How do I extend this to work for the entire list mm? I know I can write a function, but is there a more pythonic way

Upvotes: 0

Views: 424

Answers (3)

Jon Clements
Jon Clements

Reputation: 142106

If you want all the last indices of each item in ll present in mm, then:

ll = [500,500,500,501,500,502,500]
mm = [499,501,502]

d = {v:k for k,v in enumerate(ll) if v in mm}
# {501: 3, 502: 5}

It's probably worth creating a set from mm first to make it an O(1) lookup, instead of O(N), but for three items, it's really not worth it.

Following @Apero's concerns about retaining missing indices as None and also using a hash lookup to make it an O(1) lookup...

# Build a key->None dict for all `mm`
d = dict.fromkeys(mm)
# Update `None` values with last index using a gen-exp instead of dict-comp
d.update((v,k) for k,v in enumerate(ll) if v in d)
# {499: None, 501: 3, 502: 5}

Upvotes: 4

DevLounge
DevLounge

Reputation: 8437

results = {}
reversed = ll[::-1]

for item in mm:
    try:
        index = ((len(ll) - 1) - reversed.index(item))
    except ValueError:
        index = None
    finally:
        results[item] = index

print results

Output:

{499: None, 501: 3, 502: 5}

Upvotes: 1

mdml
mdml

Reputation: 22872

You can do this with a list comprehension and the max function. In particular, for each element in ll, iterate through mm to create a list of indices where ll[i] = mm[i], and take the max of this list.

>>> indices = [ max([-1] + [i for i, m in enumerate(mm) if m == n])  for n in ll ]
>>> indices
[-1, -1, -1, 1, -1, 2, -1]

Note that you need to add [-1] in case there are no matching indices in mm.

All this being said, I don't think it's more Python-ic than writing a function and using a list comprehension (as you alluded to in your question).

Upvotes: 0

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