Multiplying array


#1

HI,

Does anyone know of a function that multiplies the contents of an array.
For example:

one = [1,2,3]
two = [[2],[3],[4]]

output = [[2],[6],[12]]

I’ve written a simple function that does this, however I’m sure there is
a better way, instead of casting the item to a float.

def multiplyArray(arr1,arr2)
newArr = []
i=0
arr1.each do |x|
x = x.to_s.to_f
item = arr2[i].to_s.to_f
ele = x * item
newArr.push(ele)
i = i + 1
end
puts newArr
end

Thanks,


#2

On 23.01.2007 16:05, WKC CCC wrote:

Does anyone know of a function that multiplies the contents of an array.
For example:

one = [1,2,3]
two = [[2],[3],[4]]

output = [[2],[6],[12]]

I’ve written a simple function that does this, however I’m sure there is
a better way, instead of casting the item to a float.

Why do you cast to float? You can multiply integers directly.

First with a straightforward array “two”:

irb(main):001:0> require ‘enumerator’
=> true
irb(main):002:0> one = [1,2,3]
=> [1, 2, 3]
irb(main):003:0> two = [2,3,4]
=> [2, 3, 4]
irb(main):004:0> one.to_enum(:zip, two).map {|a,b| a*b}
=> [2, 6, 12]

Now with your array:

irb(main):005:0> two.map! {|i| [i]}
=> [[2], [3], [4]]
irb(main):006:0> one.to_enum(:zip, two).map {|a,b| a * b[0]}
=> [2, 6, 12]

Kind regards

robert


#3

On Wed, 24 Jan 2007, WKC CCC wrote:

I’ve written a simple function that does this, however I’m sure there is
i = i + 1
end
puts newArr
end

Thanks,

if you are doing lots of this then check out narray:

harp:~ > cat a.rb
require ‘narray’

a = NArray.to_na [1,2,3]
b = NArray.to_na [2,3,4]

p( a * b )

harp:~ > ruby a.rb
NArray.int(3):
[ 2, 6, 12 ]

regards.

-a


#4

WKC CCC wrote:

Does anyone know of a function that multiplies the contents of an array.

module Enumerable
def product
inject{ |piece, prod| prod*piece }
end
def sum
inject(0){ |piece, total| total+piece }
end
end

a = (1…10)
p a.sum
#=> 55
p a.product
#=> 3628800


#5

Rob B. wrote:

final result wasn’t affected because the operations are commutative.
Man, I swear that every time I use #inject I get it backwards, and
think that I’ve fixed the answer in my mind for the next time. Thanks
for the correction. (Anyone got a good _why-like mnemonic they use for
remembering the order?)

I also added the argument to the inject within product for symmetry.
You could equally remove the 0 argument from the sum method’s use of
inject.

I thought about that, but in my mind, the sum of an empty array is
zero, but the product of an empty array is nil (or 0?), not 1. Of
course, totally up the the OP as to how s/he wants to handle this edge
case.


#6

On Jan 23, 2007, at 11:25 AM, Phrogz wrote:

end
end

a = (1…10)
p a.sum
#=> 55
p a.product
#=> 3628800

module Enumerable
def product
inject(1) { |prod, piece| prod*piece }
end
def sum
inject(0) { |total, piece| total+piece }
end
end

True, except the block parameters were reversed. In these cases, the
final result wasn’t affected because the operations are commutative.

I also added the argument to the inject within product for symmetry.
You could equally remove the 0 argument from the sum method’s use of
inject.

-Rob

Rob B. http://agileconsultingllc.com
removed_email_address@domain.invalid


#7

How can the inject function work on 2 arrays?


#8

WKC CCC wrote:

How can the inject function work on 2 arrays?

It can’t; I only read the first line of your post (which described a
single array) and rushed off my answer. As penance, I give you another
solution:

a = [ 1, 2, [3], [4], 5]
b = [ 6, [7], [8], 9, 10 ]

p a.flatten.zip(b.flatten).map{ |a,b| a*b }
#=> [6, 14, 24, 36, 50]


#9

Hi,

Am Mittwoch, 24. Jan 2007, 01:57:36 +0900 schrieb Patrick G.:

[inject confusion]

(Anyone got a good _why-like mnemonic they use for
remembering the order?)

not good, but it helps me:

“memo” comes first (alphabetically), so it is

inject(memo, obj)

I’m trying to help myself with the order invariance in

a == a.inject([]) { |r,e| r+[e] }
a == a.inject([]) { |r,e| r << e }

Bertram


#10

WKC CCC wrote:

I’ve written a simple function that does this, however I’m sure there is
i = i + 1
end
puts newArr
end

Thanks,


Posted via http://www.ruby-forum.com/.

[1,2,3].zip([2,3,4]).map{|a,b| a*b}
==>[2, 6, 12]


#11

[inject confusion]

(Anyone got a good _why-like mnemonic they use for

remembering the order?)

not good, but it helps me:

“memo” comes first (alphabetically), so it is

inject(memo, obj)

P.