(def tmp = [ 1 2 3 9 4 8])
I'm trying to create pairs of 2, then for each pair, subtract the second number from the first. desired result: (1 6 4)
Here is what I was trying:
(map #(apply - %2 %1) (partition 2 tmp))
how can I do this?
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From the example it looks like you subtract the first number from the second, not the other way around, right?Rafał Dowgird– Rafał Dowgird2012-10-09 20:30:31 +00:00Commented Oct 9, 2012 at 20:30
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3 Answers
Partition produces a sequence of sequences so the function you map over them needs to expect a sequence of two items. There are several ways to express this:
(def tmp [ 1 2 3 9 4 8])
user> (map #(- (second %) (first %)) (partition-all 2 tmp ))
(1 6 4)
user> (map #(apply - (reverse %)) (partition-all 2 tmp ))
(1 6 4)
user> (map (fn [[small large]] (- large small)) (partition-all 2 tmp ))
(1 6 4)
The version using apply is different because it will still "work" on odd length lists:
user> (map #(apply - (reverse %)) (partition-all 2 [1 2 3 4 5 6 7] ))
(1 1 1 -7)
The others will crash on invalid input, which you may prefer.
Comments
Here's a solution using reduce
(reduce #(conj %1 (apply - (reverse %2))) [] (partition-all 2 [1 2 3 9 4 8]))
=> [1 6 4]
1 Comment
Arthur Ulfeldt
(reduce (conj ...)) ia a cool way to express a non-lazy version of map. Check out the new reducers library for a bunch more on this topic: clojure.com/blog/2012/05/08/… if you are interested
I wonder why this solution was overlooked...
Since switching the order of subtraction is simply the negative of the original subtraction, (a-b=-(b-a)), the solution becomes more efficient (only in this particular case!!)
(map #(- (apply - %)) (partition-all 2 [1 2 3 9 4 8]))
Pedagogically, Arthur's solution is correct. This is just a solution that is more suited the specfic question.
