different size 2d array multiply in scala

Question

I have two 2d-array and I want to multiply such as

val x = Array(Array(2.0, 5.0))
val y = Array(Array(1.0, -1.0), Array(-1.0, 1.0), Array(2.0, -2.0))

I want to get 1d-Array like

Array(2.0 * 1.0 + 5.0 * -1.0, 2.0 * -1.0 + 5.0 * 1.0, 2.0 * 2.0 + 5.0 * -2.0)

I used x.zip(y) map (_.zipped map (_ * _)) map (_.sum)

But I only get Array(-3.0)

What am I supposed to do?

Thanks for your time.

Sorry, my mean is that x array's size will always be 1 item like Array(Array(2.0, 5.0))

@ccheneson If I assign x.zip(y) map (.zipped map ( * )) map (.sum) to z val z = x.zip(y) map (.zipped map ( * )) map (.sum) I will get z = Array[Double](-3.0) — Joey Kim
– Joey Kim, Commented Apr 25, 2016 at 18:24
What I mean is if x could be for example Array(Array(2.0, 5.0), Array(6.0, 2.0)) or there will always be 1 item like Array(Array(2.0, 5.0)) ? — ccheneson
– ccheneson, Commented Apr 25, 2016 at 18:45
Please elaborate on your question. What would happen if x size is different than 1? Are nested arrays always of the same size? — Łukasz
– Łukasz, Commented Apr 25, 2016 at 19:23

ryan · Accepted Answer · 2016-04-25 18:49:36Z

2

For the x and y defined above, the following should work. However, if x is an Array of Arrays, then the answer would be different.

scala> y map {x.flatten zip _ map {case(a,b) => a*b} reduce (_+_)}
res5: Array[Double] = Array(-3.0, 3.0, -6.0)

answered Apr 25, 2016 at 18:49

ryan

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3 Comments

Kestemont Max Over a year ago

The generic answer, regardless of the size of the sub arrays would be something like val z = y map { y1 => x map{ _ zip y1 map { case(a,b) => a*b } reduce (_ + _ ) } } that you could then flatten if needed.

0__ Over a year ago

reduce(_+_) can be replaced by sum

Joey Kim Over a year ago

Thanks for helping me

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