Counting occurrence of all characters in string but only once if character is repeated

Question

Note: Please make sure this is not duplicate, i am trying to find one liner way.

I am trying to turn string filled with ascii letters to string filled with occurrence of ascii letters.

For example: string1 = 'abc'.

To turn this into string filled with occurrence of ascii letters, i would simply use one line command:

string2 = [string1.count(l) for l in string1]

And this would output:

[1, 1, 1]

But if i had multiple repeated characters in string1, it would look different.

For example: string1 = 'aaabbbccc'

And transforming it:

string2 = [string1.count(l) for l in string1]

And output would be:

[3, 3, 3, 3, 3, 3, 3, 3, 3]

So as you see, character occurrences are repeated, and i don't know how can i make a generator that prints single occurrence of single letter only once.

So for example:

Instead of:

[3, 3, 3, 3, 3, 3, 3, 3, 3]

Is there a way to output this with single line generator?

[3, 3, 3]

what is the expected output of `aabaab`, where there are repeating characters not next to each other? — depperm, Feb 06 '17 at 16:50
Use `collections.Counter` or a `dict` . I don't see any generators in your question. — juanpa.arrivillaga, Feb 06 '17 at 16:51
`Counter('aaabbbccc')` -> `Counter({'b': 3, 'a': 3, 'c': 3})` — Cristian Ciupitu, Feb 06 '17 at 17:04

wim · Answer 1 · 2017-02-06T16:53:55.223

4

You can do it in one line with an OrderedCounter.

>>> from collections import Counter, OrderedDict
>>> class OrderedCounter(Counter, OrderedDict): 
...     pass
...
>>> OrderedCounter('aaabbbccc').values()
[3, 3, 3]

edited Feb 06 '17 at 16:53

answered Feb 06 '17 at 16:51

wim

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Uriel · Accepted Answer · 2017-02-06T17:09:00.290

4

Use collections.Counter:

>>> from collections import Counter
>>> Counter('aaabbbccc')
Counter({'a': 3, 'b': 3, 'c': 3})

You can get the counts as a sorted list easily by iterating the counter with string.ascii_lowercase:

>>> import string
>>> c = Counter('aaabbbccc')
>>> [c[l] for l in string.ascii_lowercase]
[3, 3, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

edited Feb 06 '17 at 17:09

answered Feb 06 '17 at 16:51

Uriel

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Thank you but i am trying to turn it more easily to string type, is there a simple way for that? – ShellRox Feb 06 '17 at 16:55
@ShellRox you are trying to turn what to a string type? I didn't got that. Do you want the result as `'333'` in this case? – Moinuddin Quadri Feb 06 '17 at 16:56
@MoinuddinQuadri I am trying to output numbers without array form, sorry i haven't mentioned in question. – ShellRox Feb 06 '17 at 16:58
@ShellRox check [Joining a list that has Integer values with Python](http://stackoverflow.com/questions/3590165/joining-a-list-that-has-integer-values-with-python) – Moinuddin Quadri Feb 06 '17 at 17:00
@Uriel Eli You are true mind reader – ShellRox Feb 06 '17 at 17:00
It is not necessary to `c.get(l, 0)` with a counter. Simply `c[l]` will default to `0`. – wim Feb 06 '17 at 17:07
@wim I actually needed those 0's for specific solution, but i shrinked it to nessecary amount: `k = [c.get(l, 0) for l in string.ascii_lowercase]; k[:len('aaabbbccc')]` – ShellRox Feb 06 '17 at 17:09
@wim thanks, fixed – Uriel Feb 06 '17 at 17:09

hansaplast · Answer 3 · 2017-02-06T19:42:01.793

0

Use set:

>>> [string1.count(l) for l in set(string1)]
[3, 3, 3]

Set turns a string into a set of distinct characters: e.g. aabc and abbc but also acb will turn into {'a', 'b', 'c'}

This is the output you looked for, but you would not know which 3 goes to which character, so in the end you'd be better off using the Counter methods suggested

edited Feb 06 '17 at 19:42

answered Feb 06 '17 at 16:50

hansaplast

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Hello, do you know how could this output also be turned into a string easily? without brackets, thanks! – ShellRox Feb 06 '17 at 16:58

Counting occurrence of all characters in string but only once if character is repeated

3 Answers3