Looping idiom

[Brian Will]

Very basic question. What's the idiom for producing a new seq wherein
each val is based on pairs from an existing seq, e.g.:

; add together n0 n1, n1 n2, n2 n3, etc.
[1 2 3 4 5 6]
; giving...
[3 5 7 9 11]

The verbose way would be something like:

(loop [s origSeq n [ ])
(let [a (first origSeq)
b (second origSeq)]
(if (nil? b)
n
(recur (rest origSeq) (conj n (+ a b))))))

[Brian Will]

should read:

(loop [s origSeq n [ ])

(let [a (first s
b (second s)]
(if (nil? b)
n
(recur (rest s) (conj n (+ a b))))))

[Sean Devlin]

I'd use map, reduce, partial, and partition like so

user=>(map (partial reduce +) (partition 2 1 [1 2 3 4 5 6]))
(3 5 7 9 11)

[Timothy Pratley]

Yet another way :)

user=> (map + (rest a) a)

[songoku]

wow! i like your solution!

[Sean Devlin]

Very very slick :)

[Laurent PETIT]

2009/9/8 Timothy Pratley <timothy...@gmail.com>

Yet another way :)

user=> (map + (rest a) a)

(3 5 7 9 11)

Haskell background ? :-)
 

[atreyu]

uau cutting the Gordian knot