Chunk, ChunkC -- nice simple way(s) to write these in Python?

(Chunk '(a a b a a a b b))
==> ((a a) (b) (a a a) (b b))

(Chunk '(a a a a b c c a a d e e e e))
==> ((a a a a) (b) (c c) (a a) (d) (e e e e))

(Chunk '(2 2 foo bar bar j j j k baz baz))
==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))

_________________

(ChunkC '(a a b b b))
==> ((a 2) (b 3))

(ChunkC '(a a b a a a b b))
==> ((a 2) (b 1) (a 3) (b 2))

On 2024-06-09 22:20, HenHanna via Python-list wrote:
>
> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>
>
> (Chunk '(a a b a a a b b))
> ==> ((a a) (b) (a a a) (b b))
>
>
> (Chunk '(a a a a b c c a a d e e e e))
> ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>
>
> (Chunk '(2 2 foo bar bar j j j k baz baz))
> ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>
> _________________
>
> (ChunkC '(a a b b b))
> ==> ((a 2) (b 3))
>
> (ChunkC '(a a b a a a b b))
> ==> ((a 2) (b 1) (a 3) (b 2))

You can make use of itertools.groupby.

On 6/9/2024 3:50 PM, MRAB wrote:
> On 2024-06-09 22:20, HenHanna via Python-list wrote:
>>
>> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>>
>>
>> (Chunk  '(a a   b    a a a   b b))
>>       ==> ((a a) (b)  (a a a) (b b))
>>
>>
>> (Chunk  '(a a a a   b   c c   a a   d   e e e e))
>>       ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>>
>>
>> (Chunk  '(2 2   foo   bar bar   j j j   k   baz baz))
>>       ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>>
>> _________________
>>
>> (ChunkC  '(a a   b b b))
>>        ==> ((a 2)  (b 3))
>>
>> (ChunkC  '(a a   b      a a a   b b))
>>        ==> ((a 2)  (b 1)  (a 3)   (b 2))
>
> You can make use of itertools.groupby.
>

Thanks! i'll try it.

Scheme (Gauche)

(use srfi-1) ; span

(define (gp x)
(if (null? x) '()
(let-values (((F L) (span (cut equal? (car x) <>) x)))
(cons F (gp L)))))

(print (gp '(a b b a a a b b b b)))
(print (gp '(c c c a d d d d a e e e e e)))

HH,

Before bothering, it might be helpful if you spelled out a bit more in the
way of requirements that would satisfy you and perhaps show your attempts.

Your examples suggest it would be fairly simple to create an iterator, for
example, that would yield as it examined one item at a time until it ran out
or the next item was not the same. It could then either return a tuple it
had collected so far since the last yield, or for what you call chunkC,
return a tuple containing one copy of the item and a count of how many.

But if you want something else, such as a list all at once, that too would
be trivial, perhaps leveraging the above.

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On
Behalf Of HenHanna via Python-list
Sent: Sunday, June 9, 2024 5:20 PM
To: python-list@python.org
Subject: in Python? -- Chunk -- (ChunkC '(a a b b b)), ==> ((a 2) (b 3))

Chunk, ChunkC -- nice simple way(s) to write these in Python?

(Chunk '(a a b a a a b b))
==> ((a a) (b) (a a a) (b b))

(Chunk '(a a a a b c c a a d e e e e))
==> ((a a a a) (b) (c c) (a a) (d) (e e e e))

(Chunk '(2 2 foo bar bar j j j k baz baz))
==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))

_________________

(ChunkC '(a a b b b))
==> ((a 2) (b 3))

(ChunkC '(a a b a a a b b))
==> ((a 2) (b 1) (a 3) (b 2))
--
https://mail.python.org/mailman/listinfo/python-list

I remembered that HenHanna had been hard to deal with in the past and when
my reply to him/her/them bounced as a bad/fake address it came back to me
that I am better off not participating in this latest attempt to get us to
perform then probably shoot whatever we say down.

A considerate person would ask questions more clearly and perhaps explain
what language they are showing us code from and so on.

Life is too short to waste.

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On
Behalf Of HenHanna via Python-list
Sent: Sunday, June 9, 2024 5:20 PM
To: python-list@python.org
Subject: in Python? -- Chunk -- (ChunkC '(a a b b b)), ==> ((a 2) (b 3))

Chunk, ChunkC -- nice simple way(s) to write these in Python?

(Chunk '(a a b a a a b b))
==> ((a a) (b) (a a a) (b b))

(Chunk '(a a a a b c c a a d e e e e))
==> ((a a a a) (b) (c c) (a a) (d) (e e e e))

(Chunk '(2 2 foo bar bar j j j k baz baz))
==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))

_________________

(ChunkC '(a a b b b))
==> ((a 2) (b 3))

(ChunkC '(a a b a a a b b))
==> ((a 2) (b 1) (a 3) (b 2))
--
https://mail.python.org/mailman/listinfo/python-list

On 6/9/2024 7:05 PM, avi.e.gross@gmail.com wrote:
> I remembered that HenHanna had been hard to deal with in the past and when
> my reply to him/her/them bounced as a bad/fake address it came back to me
> that I am better off not participating in this latest attempt to get us to
> perform then probably shoot whatever we say down.
>
> A considerate person would ask questions more clearly and perhaps explain
> what language they are showing us code from and so on.
>
> Life is too short to waste.
>
> -----Original Message-----
> From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On
> Behalf Of HenHanna via Python-list
> Sent: Sunday, June 9, 2024 5:20 PM
> To: python-list@python.org
> Subject: in Python? -- Chunk -- (ChunkC '(a a b b b)), ==> ((a 2) (b 3))
>
> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>
>
> (Chunk '(a a b a a a b b))
> ==> ((a a) (b) (a a a) (b b))
>
>
> (Chunk '(a a a a b c c a a d e e e e))
> ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>
>
> (Chunk '(2 2 foo bar bar j j j k baz baz))
> ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>
> _________________
>
> (ChunkC '(a a b b b))
> ==> ((a 2) (b 3))
>
> (ChunkC '(a a b a a a b b))
> ==> ((a 2) (b 1) (a 3) (b 2))

i was just curiuos about simple, clever way to write it in Python

in Scheme (Gauche)

(use srfi-1) ;; span

(define (gp x)
(if (null? x) '()
(let-values (((F L) (span (cut equal? (car x) <>) x)))
(cons F (gp L)))))

(print (gp '(a b b a a a b b b b)))
(print (gp '(c c c a d d d d a e e e e e)))

(define (gpC x) (map (lambda (x) (list (car x) (length x))) (gp x)))

(print (gpC '(a b b a a a b b b b)))
(print (gpC '(c c c a d d d d a e e e e e)))

> i was just curiuos about simple, clever way to write it in Python

It depends on what you mean by "clever".

For some, it was like a suggestion on using something already available such
as itertools.groupby, perhaps even better if it is actually compiled in from
a language like C and perhaps more efficient.

For some who like to Scheme, only a recursive solution may be considered
clever!

For some who have absorbed ideas in languages like python, perhaps they
would implement it the way I described as an iterator that is efficient in
the sense that it delivers things just in time and quits if not iterated
further. An example might be if you asked to get groups of consecutive
digits in the calculated indefinite value of a number like pi or e or looked
at how many digits in a row were the same for prime numbers but wanted to
quit after the first billion.

Some may like some kind of functional programming method. Some want short
code and some want efficiency and some even like something written in a
subtle way so others have trouble understanding it, as in the obfuscated C
contests.

What you do need to consider is what exactly are your requirements. Your
examples are trivial so maybe we need to think about what it means to gather
together what is supplied in one container, into a container containing
other grouped containers.

First, what python containers should be handled? Yes, you probably mean a
list but if handed a tuple, should it return the same, or always return a
list for the top level and perhaps even deeper levels? There are many
containers in python including (ordered) dictionaries where the keys may be
unique, but you may want to see if the contents have some order, numpy
arrays, and so on.

And, are all the contents required to be atomic? What does it mean for items
in a row to be equal? If I have a sublist, should I unlist to make it flat
first, or should it be an error, or should each such sublist be compared for
full equality or even relative equality so that (a (b c) (c b) d) actually
accepts (b c) and (c b) as the same for the purpose? For that matter, is 1.0
matched to 1 or even "1" or perhaps an object that has that value in some
way such as when comparing arbitrary objects?

There can be many such questions and some elegant methods get less elegant
if they need to handle too much. You need to explain what to do when a bad
unintended case is found, such as perhaps an empty list or ...

To just do what your examples ask, again, seems easy enough. It is the usual
code if not done too elegantly. Write a function that accepts a list (or
tries to coerce what it gets into a copy as a list. It returns an empty list
if it has nothing, otherwise pops off the first item and then loops on the
rest. When items match the current first item, extend a sublist and when
they don't, yield what you have and start a new sublist. After the loop,
anything remaining is returned.

Simple enough?

Note to get your result properly, since it is an iterator, you need to
either be calling it iteratively, or do something like:

Result = list(chunk(something))

To force it to run to completion.

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On
Behalf Of HenHanna via Python-list
Sent: Sunday, June 9, 2024 10:37 PM
To: python-list@python.org
Subject: Re: in Python? -- Chunk -- (ChunkC '(a a b b b)), ==> ((a 2) (b 3))

On 6/9/2024 7:05 PM, avi.e.gross@gmail.com wrote:
> I remembered that HenHanna had been hard to deal with in the past and when
> my reply to him/her/them bounced as a bad/fake address it came back to me
> that I am better off not participating in this latest attempt to get us to
> perform then probably shoot whatever we say down.
>
> A considerate person would ask questions more clearly and perhaps explain
> what language they are showing us code from and so on.
>
> Life is too short to waste.
>
> -----Original Message-----
> From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org>
On
> Behalf Of HenHanna via Python-list
> Sent: Sunday, June 9, 2024 5:20 PM
> To: python-list@python.org
> Subject: in Python? -- Chunk -- (ChunkC '(a a b b b)), ==> ((a 2) (b 3))
>
> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>
>
> (Chunk '(a a b a a a b b))
> ==> ((a a) (b) (a a a) (b b))
>
>
> (Chunk '(a a a a b c c a a d e e e e))
> ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>
>
> (Chunk '(2 2 foo bar bar j j j k baz baz))
> ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>
> _________________
>
> (ChunkC '(a a b b b))
> ==> ((a 2) (b 3))
>
> (ChunkC '(a a b a a a b b))
> ==> ((a 2) (b 1) (a 3) (b 2))

i was just curiuos about simple, clever way to write it in Python

in Scheme (Gauche)

(use srfi-1) ;; span

(define (gp x)
(if (null? x) '()
(let-values (((F L) (span (cut equal? (car x) <>) x)))
(cons F (gp L)))))

(print (gp '(a b b a a a b b b b)))
(print (gp '(c c c a d d d d a e e e e e)))

(define (gpC x) (map (lambda (x) (list (car x) (length x))) (gp x)))

(print (gpC '(a b b a a a b b b b)))
(print (gpC '(c c c a d d d d a e e e e e)))
--
https://mail.python.org/mailman/listinfo/python-list

import itertools

def chunk1(seq):
    return [ ch * len(list(grp)) for (ch, grp) in itertools.groupby(s) ]

def chunk2(seq):
    return [ (ch, len(list(grp))) for (ch, grp) in itertools.groupby(s) ]

s='aaabbccccaa'
print(chunk1(s))
print(chunk2(s))
###################################
Program output:
['aaa', 'bb', 'cccc', 'aa']
[('a', 3), ('b', 2), ('c', 4), ('a', 2)]

Rob Cliffe

On 09/06/2024 22:20, HenHanna via Python-list wrote:
>
> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>
>
> (Chunk  '(a a   b    a a a   b b))
>     ==> ((a a) (b)  (a a a) (b b))
>
>
> (Chunk  '(a a a a   b   c c   a a   d   e e e e))
>     ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>
>
> (Chunk  '(2 2   foo   bar bar   j j j   k   baz baz))
>     ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>
> _________________
>
> (ChunkC  '(a a   b b b))
>      ==> ((a 2)  (b 3))
>
> (ChunkC  '(a a   b      a a a   b b))
>      ==> ((a 2)  (b 1)  (a 3)   (b 2))

Rob,

That is a fairly straightforward and elegant solution if using an added mode called itertools.

I went a different way by creating my own focused iterator and calling it, when needed, to produce one or the other of the results requested in the functions named chunk() and chunkc(). But note my version operates to produce lists no matter what the input was. It produces lists nestled within lists. Presumably it could be adjusted to check if the input was a tuple and then return a grouping of tuples within a tuple.

The code, just for fun, is below and probably could be more elegant! LOL!

### CODE START

# Iterator version to be called as needed or all at once as with ChunkC
def ChunkIterator(seq):
# Purpose is to evaluate a list and return a list
# of lists with each one containing the longest run
# of each item.

# Handle an empty list
if len(seq) == 0: return container

# Initialize the first item in a copy.
# The first item is now in a initialized list.
gathered = [last := seq[0]]

# Loop over the remaining items and yield as needed.
for item in seq[1:]:
if len(gathered) == 0: gathered = [item]
elif item == last: gathered.append(item)
else: # not equal means yield result and start anew for next group
yield(gathered)
gathered = [last := item]

# After loop completes, any remaining data is returned as the function terminates.
if len(gathered) > 0: yield(gathered)

# Accesory function that runs the iterator to completion into a list.
def Chunk(seq): return list(ChunkIterator(seq))

# Function that simplifies the Chunks as an item and a number of instances.
def ChunkC(seq):
return [[group[0], len(group)] for group in ChunkIterator(seq)]
### CODE END

### EXAMPLES
>>> tuple(Chunk([1, 2, 2, 'c', 'c', 'c', 'singleton']))
([1], [2, 2], ['c', 'c', 'c'], ['singleton'])
>>> chunk([1, 2, 2, 'c', 'c', 'c', 'singleton'])
[[1], [2, 2], ['c', 'c', 'c'], ['singleton']]
>>> chunkC([1, 2, 2, 'c', 'c', 'c', 'singleton'])
[[1, 1], [2, 2], ['c', 3], ['singleton', 1]]

# COMMENTS
The current version has flaws I have not bothered correcting. Just for a demo.

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On Behalf Of Rob Cliffe via Python-list
Sent: Monday, June 10, 2024 9:29 AM
To: python-list@python.org
Subject: Re: in Python? -- Chunk -- (ChunkC '(a a b b b)), ==> ((a 2) (b 3))

import itertools

def chunk1(seq):
return [ ch * len(list(grp)) for (ch, grp) in itertools.groupby(s) ]

def chunk2(seq):
return [ (ch, len(list(grp))) for (ch, grp) in itertools.groupby(s) ]

s='aaabbccccaa'
print(chunk1(s))
print(chunk2(s))
###################################
Program output:
['aaa', 'bb', 'cccc', 'aa']
[('a', 3), ('b', 2), ('c', 4), ('a', 2)]

Rob Cliffe

On 09/06/2024 22:20, HenHanna via Python-list wrote:
>
> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>
>
> (Chunk '(a a b a a a b b))
> ==> ((a a) (b) (a a a) (b b))
>
>
> (Chunk '(a a a a b c c a a d e e e e))
> ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>
>
> (Chunk '(2 2 foo bar bar j j j k baz baz))
> ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>
> _________________
>
> (ChunkC '(a a b b b))
> ==> ((a 2) (b 3))
>
> (ChunkC '(a a b a a a b b))
> ==> ((a 2) (b 1) (a 3) (b 2))

--
https://mail.python.org/mailman/listinfo/python-list

On 6/10/2024 6:29 AM, Rob Cliffe wrote:
> import itertools
>
> def chunk1(seq):
>     return [ ch * len(list(grp)) for (ch, grp) in itertools.groupby(s) ]
>
> def chunk2(seq):
>     return [ (ch, len(list(grp))) for (ch, grp) in itertools.groupby(s) ]
>
> s='aaabbccccaa'
> print(chunk1(s))
> print(chunk2(s))
> ###################################
> Program output:
> ['aaa', 'bb', 'cccc', 'aa']
> [('a', 3), ('b', 2), ('c', 4), ('a', 2)]
>
> Rob Cliffe
>

thank you... OMG... For 10 minutes... i was SO mystified by
the question...
How can this code work??? , when it's
> def chunk1(seq):
and it's [s] within the def-body ?

it seemed as if the Compiler was doing a DWIM (Do what i mean) trick.

> On 09/06/2024 22:20, HenHanna via Python-list wrote:
>>
>> Chunk, ChunkC -- nice simple way(s) to write these in Python?
>>
>>
>> (Chunk  '(a a   b    a a a   b b))
>>     ==> ((a a) (b)  (a a a) (b b))
>>
>>
>> (Chunk  '(a a a a   b   c c   a a   d   e e e e))
>>     ==> ((a a a a) (b) (c c) (a a) (d) (e e e e))
>>
>>
>> (Chunk  '(2 2   foo   bar bar   j j j   k   baz baz))
>>     ==> ((2 2) (foo) (bar bar) (j j j) (k) (baz baz))
>>
>> _________________
>>
>> (ChunkC  '(a a   b b b))
>>      ==> ((a 2)  (b 3))
>>
>> (ChunkC  '(a a   b      a a a   b b))
>>      ==> ((a 2)  (b 1)  (a 3)   (b 2))
>

On 6/9/2024, HenHanna wrote:

> > (Chunk '(a a b a a a b b))
> > ==> ((a a) (b) (a a a) (b b))
> >

Gauche Scheme:

(use gauche.sequence)

(group-sequence '(2 4 4 3 0 5 8 6 6 6))
===>
((2) (4 4) (3) (0) (5) (8) (6 6 6))