Grouping the same recurring items that occur in a row from list

Question

For instance, we have a list like this:

L = ["item1", "item2", "item3", "item3", "item3", "item1", "item2", "item4", "item4", "item4"]

I want to pack them into list of tuples of the form:

[("item1", 1), ("item2", 1), ("item3", 3),... ("item1", 1)]

I've already developed an algorithm which does something similar, to get:

{item1: 2, item2: 2, ...}

(it finds all the occurrences and counts them, even if they aren't neighbours...)

However, I want it to groups only those items which have the same and are neighbours (i.e. occur in a row together), how could I accomplish this?

It's not that I don't know how to do it but I tend to write code that is long and I want an elegant and uncomplicated solution in this case.

`item1: 1, item2:1, item3:3....item1: 1` would not be a dictionary... (it has more than one of the same key) — Andy Hayden, Nov 26 '12 at 12:59
Sorry about that. That is true. Ignore the fact that I've written that must be a dictionary. It is not a requirement. It can be in the form of tuples. So the order is important ofcourse. — RecursionSnake, Nov 26 '12 at 13:03

score 5 · Answer 1 · answered Nov 26 '12 at 13:02

5

This is also using itertools.groupby (a generator version):

from itertools import groupby
counts = ((k, sum(1 for _ in g)) for k, g in groupby(l))
>>> list(counts)
[('item1', 1),
 ('item2', 1),
 ('item3', 3),
 ('item1', 1),
 ('item2', 1),
 ('item4', 3)]

answered Nov 26 '12 at 13:02

Lev Levitsky

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`len(list(g))` is shorter than `sum(1 for _ in g)`, +1 anyway. – Ashwini Chaudhary Nov 26 '12 at 13:13
@AshwiniChaudhary it's shorter, but I figured it could be faster; I tend to think it's a good idea to avoid creating a list just to count its elements. Thanks for the upvote :) – Lev Levitsky Nov 26 '12 at 13:15
2

Good point, jut timed them, `sum(1 for _ in g)`<`len(tuple(g))`<`len(list(g))`, learned something new today. :) – Ashwini Chaudhary Nov 26 '12 at 13:22

Ashwini Chaudhary · Accepted Answer · 2012-11-26T13:15:20.253

4

using itertools.groupby(), items are repeated so you might not be able to store all values in a dictionary, as item1 & item2 are repeated:

In [21]: l = ["item1", "item2", "item3", "item3", "item3", "item1", "item2", "item4", "item4", "item4"]

In [22]: for k,g in groupby(l):
    print "{0}:{1}".format(k,len(list(g)))
   ....:     
item1:1
item2:1
item3:3
item1:1
item2:1
item4:3

edited Nov 26 '12 at 13:15

answered Nov 26 '12 at 13:00

Ashwini Chaudhary

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No need for a key function in this case. – Steven Rumbalski Nov 26 '12 at 13:05
Great.Thanks.I was looking for the groupby function indeed. And yes it is no use for key lambda function. It works absolutely flawlessly:) – RecursionSnake Nov 26 '12 at 13:11
You can also use [ilen](http://funcy.readthedocs.org/en/latest/seqs.html#ilen) from [funcy](https://github.com/Suor/funcy) library instead of `len(list(...))` for speed. – Suor Jun 04 '14 at 19:46

score 0 · Answer 3 · edited Nov 26 '12 at 14:01

0

python 3.2
from itertools import groupby

>>> [(i,(list(v)).count(i)) for i,v in groupby(L)]

edited Nov 26 '12 at 14:01

bluish

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answered Nov 26 '12 at 13:44

raton

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Grouping the same recurring items that occur in a row from list

3 Answers3

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