# Re: processing a sequence

André Thieme wrote:

joswig@xxxxxxxxxxxxxxxxxxxxxxx schrieb:

(defun collect-sublists (list)
(flet ((collect-sublist ()
(loop for element = (pop list)
while element collect element
while (and list
(<= element (first list))))))
(loop for sublist = (collect-sublist)
while sublist collect sublist)))

"We don't need no stinkin' loops!"

Ruby:

list = [0,1,2,3] * 3

def collect_sublists list, prev=nil, accum=[[]]
return accum if not element = list.first
if prev and element < prev
accum << [element]
else
accum[-1] << element
end
collect_sublists( list[1..-1], element, accum )
end

p collect_sublists( list )

# This could be slower if getting the last element in a list
# is expensive.

def collect_sublists_2 list, accum=[[]]
return accum if not element = list.first
if accum != [[]] and accum[-1][-1] > element
accum << [element]
else
accum[-1] << element
end
collect_sublists_2( list[1..-1], accum )
end

Nearly 15 minutes passed since you posted your solution, and William
James still didn?t show his Ruby one-liner that outperforms the Lisp
solution by a factor of 20 (speedwise). ;-)

André

Ruby is about the slowest "scripting language".
JavaScript is faster.

SpiderMonkey and jslibs:

list = [0,1,2,3,0,1,2,3,1,2,3]

function collect_groups( list, prev, accum )
{ accum = accum || [[]]
var element = list[0]
if ( typeof element == "undefined" )
return accum
if (prev && element < prev)
accum.push( [element] )
else
accum.slice(-1)[0].push( element )
return collect_groups( list.slice(1), element, accum )
}

Print( collect_groups(list).toSource(), '\n')

--- output ---
[[0, 1, 2, 3], [0, 1, 2, 3], [1, 2, 3]]
.

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