Python how to count groups of duplicates?

Question

I'm new to python and the problem I'm facing is how can I count groups of duplicates. For example given a list like this:

['down', 'down', 'down', 'up', 'right', 'right', 'down', 'down']

I have to calculate the following:

[('down', 3), ('up', 1), ('right', 2), ('down', 2)]

Or rather how can I achieve this in a pythonic way coming from languages like java/c#?

EDIT: Since it seems that I didn't clarify my my problem enough, I don't want to count all the occurrences of for example down in the list, but only those that are neighbors (if that is the correct phrasing), so Counter(my_list) doesn't give the desired output.

What is the data structure you expect to get? because what you asked is not a valid one. — Maor Refaeli
– Maor Refaeli, Commented Oct 21, 2018 at 9:27
It should have been {'down' : 3, 'up' : 1, 'right' : 2, 'down' : 2} right? Like a Hashmap in java — Vishnudev Krishnadas
– Vishnudev Krishnadas, Commented Oct 21, 2018 at 9:28
@MaorRefaeli sorry, I hope that it's good now with using a tuple. — мајдејсаремех
– мајдејсаремех, Commented Oct 21, 2018 at 9:32
List of tuples is a better data structure for your implementation — Vishnudev Krishnadas
– Vishnudev Krishnadas, Commented Oct 21, 2018 at 9:34

Aran-Fey · Accepted Answer · 2018-10-21 09:29:42Z

3

Use itertools.groupby:

>>> import itertools
>>> lst = ['down', 'down', 'down', 'up', 'right', 'right', 'down', 'down']
>>> [(value, sum(1 for _ in group)) for value, group in itertools.groupby(lst)]
[('down', 3), ('up', 1), ('right', 2), ('down', 2)]

answered Oct 21, 2018 at 9:29

Aran-Fey

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Comments

Rohit-Pandey · Accepted Answer · 2018-10-21 09:56:51Z

1

Try This:

from itertools import groupby
[(c,len(list(cgen))) for c,cgen in groupby(a)]

answered Oct 21, 2018 at 9:56

Rohit-Pandey

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2 Comments

мајдејсаремех Over a year ago

could you please elaborate as to have that solution works?

Rohit-Pandey Over a year ago

groupby gives group of same element like:

[('down', <itertools._grouper at 0x7f4438b1b320>),  ('up', <itertools._grouper at 0x7f4438b1b6d8>),  ('right', <itertools._grouper at 0x7f4438b1b5f8>),  ('down', <itertools._grouper at 0x7f4438b1b278>)]

So for each print the element and length of group.

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