Collapse duplicates in python list

Question

sorry if this is a duplicate thread but I couldn't find one.

I'm currently trying to make an inventory for a game that is in the works. The problem I run into is that, when you have a lot of a particular item, it kind of wastes a lot of space.

Here's my code: (inv is the inventory)

inv = ["item1","item2","item3","item2","item2","item1"]

print(', '.join(inventory))

Output:

item1, item2, item3, item2, item2, item1

I'd like to, instead of doing that, do this:

item1 x2, item2 x3, item3

I don't really care if you need item3 to be x1, either way would be fine, thanks.

Where did the x2, x3 come from? There is no mention of it in your code? What is the idea behind what it is you are exactly trying to generate? What is the *problem* you are trying to solve here? — idjaw, Jun 29 '17 at 23:18
Well, that's why I want it to do that... Please read my question over again :) — Ryan, Jun 29 '17 at 23:19
https://stackoverflow.com/questions/10806866/counting-each-letters-frequency-in-a-string — idjaw, Jun 29 '17 at 23:21

score 2 · Answer 1 · answered Jun 29 '17 at 23:20

2

There are many options actually. The main goal is to count your items, and print them out in order.

One option is to use collections.OrderedDict. Others are collections.Counter/collections.defaultdict, but I'll give you an example with the former:

from collections import OrderedDict

counter = OrderedDict()

inv = ["item1", "item2", "item3", "item2", "item2", "item1"]
for i in inv:
    if i not in counter:
        counter[i] = 0

    counter[i] += 1

for c in counter:
    if counter[c] > 1:
        print(c + ' x{}'.format(counter[c]))
    else:
        print(c)

This prints:

item1 x2
item2 x3
item3

You're free to modify this example as needed if you want to print them out in one line. This code was written in python3.x.

answered Jun 29 '17 at 23:20

cs95

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@RyanTartaglia Quite so. If you're on python3, you can use `print(..., end=', ')`. If you're on python 2, first import print using `from __future__ import print_function` and do that. – cs95 Jun 29 '17 at 23:29
@RyanTartaglia You'll have to do that in both print statements :) – cs95 Jun 29 '17 at 23:32
@RyanTartaglia Noticed you revoked acceptance. What's wrong? Anything I can help with? – cs95 Jun 30 '17 at 12:54
@RyanTartaglia Hey, no worries. As an asker you're not obligated to accept. But thank you for doing so. :) – cs95 Jul 05 '17 at 00:58
Also, my bad... did mean to upvote your question before but it slipped my mind. Did it now. – cs95 Jul 05 '17 at 00:59
@RyanTartaglia Yes... I have an idea you might be unknowingly using the same referenced list in multiple places... can't say without your code. Open a new question and put your code there and I'd be happy to help. – cs95 Jul 31 '17 at 23:19
@RyanTartaglia Was it a problem with my answer? Or how you adapted it? If it was the former, I won't stop you from unmarking it... :) – cs95 Jul 31 '17 at 23:24

score 0 · Answer 2 · answered Jun 29 '17 at 23:21

0

Here is an example with Counter:

from collections import Counter

inv = ["item1", "item2", "item3", "item2", "item2", "item1"]

print (', '.join('%s x%d' % (item, count)
                 for item, count in sorted(Counter(inv).items())))

answered Jun 29 '17 at 23:21

Robᵩ

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Alter · Answer 3 · 2017-06-29T23:24:47.387

0

out = ['{} x{}'.format(item, inv.count(item)) if inv.count(item) > 1 else item 
   for item in set(inv)]

out:

['item1 x2', 'item2 x3', 'item3']

edited Jun 29 '17 at 23:24

answered Jun 29 '17 at 23:24

Alter

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Same as Prune's answer, but less readable... – cs95 Jun 29 '17 at 23:24
I'd tend to agree with @Coldspeed here. When you have to began splitting list comprehensions across several lines, you'd want to consider refactoring it into a vanilla `for` loop. – Christian Dean Jun 29 '17 at 23:27
I think that's a valid point, I don't like taking up a lot of space for simple operations though – Alter Jun 29 '17 at 23:31

Collapse duplicates in python list

3 Answers3