Got a range from an array, but looking for other ways of doing it

ruby-talk
1

a is an array. I'm trying to chop the first and last third off. or,
choose only the middle third. can be a rough third.

a.length # 4808
range = a.length * 0.3 # 1442.4
range = range.to_i # 1442
rangetop = a.length / 2 + range # 3846
rangebottom = a.length / 2 - range # 962

a[rangebottom..rangetop].length # 2885

2

irb(main):002:0> a = Array(1..100)
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36,
37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53,
54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87,
88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100]
irb(main):003:0> a[a.length/3,a.length/3]
=> [34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49,
50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66]

···

On 30/09/2007, Simon Schuster <significants@gmail.com> wrote:

a is an array. I'm trying to chop the first and last third off. or,
choose only the middle third. can be a rough third.

a.length # 4808
range = a.length * 0.3 # 1442.4
range = range.to_i # 1442
rangetop = a.length / 2 + range # 3846
rangebottom = a.length / 2 - range # 962

a[rangebottom..rangetop].length # 2885

3

That's a very rough third :slight_smile:

arr = (1..100).to_a
bot = arr.length/3
p arr[bot..bot*2]

[34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67]

Harry

···

On 10/1/07, Simon Schuster <significants@gmail.com> wrote:

a is an array. I'm trying to chop the first and last third off. or,
choose only the middle third. can be a rough third.

a.length # 4808
range = a.length * 0.3 # 1442.4
range = range.to_i # 1442
rangetop = a.length / 2 + range # 3846
rangebottom = a.length / 2 - range # 962

a[rangebottom..rangetop].length # 2885

--
A Look into Japanese Ruby List in English

4

Simon Schuster wrote:

a is an array. I'm trying to chop the first and last third off. or,
choose only the middle third. can be a rough third.

a.length # 4808
range = a.length * 0.3 # 1442.4
range = range.to_i # 1442
rangetop = a.length / 2 + range # 3846
rangebottom = a.length / 2 - range # 962

a[rangebottom..rangetop].length # 2885

if the size is very big you can also do it inline:

a = Array(1..100)
    ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 2
1, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
39, 40, 4
1, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58,
59, 60, 6
1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78,
79, 80, 8
1, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,
99, 100]

len = a.length
    ==>100

a[0..len / 3] = nil
    ==>nil

a
    ==>[35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52,
53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72,
73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90,
91, 92,
93, 94, 95, 96, 97, 98, 99, 100]

a[len / 3..-1] = nil
    ==>nil

a
    ==>[35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52,
53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67]

···

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

5

What exactly do you want to do with the slice?

Kind regards

robert

···

2007/9/30, Simon Schuster <significants@gmail.com>:

a is an array. I'm trying to chop the first and last third off. or,
choose only the middle third. can be a rough third.

a.length # 4808
range = a.length * 0.3 # 1442.4
range = range.to_i # 1442
rangetop = a.length / 2 + range # 3846
rangebottom = a.length / 2 - range # 962

a[rangebottom..rangetop].length # 2885

6

SpringFlowers AutumnMoon wrote:

Simon Schuster wrote:

a is an array. I'm trying to chop the first and last third off. or,
choose only the middle third. can be a rough third.

that's strange, if i do

a = Array(1..100)
    ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1
6, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 4
6, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6
1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 7
6, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
1, 92, 93, 94, 95, 96, 97, 98, 99, 100]

a[0..a.size/2] = a[a.size*2/3..-1] = nil
    ==>nil
a
    ==>[52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64,
65, 66]

a = Array(1..100)
    ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1
6, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 4
6, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6
1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 7
6, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
1, 92, 93, 94, 95, 96, 97, 98, 99, 100]

a[0..a.size/3] = a[a.size*2/3..-1] = nil
    ==>nil
a
    ==>[35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47,
48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62,
63, 64, 65, 66]

shouldn't it be a[0..a.size/2] = a[a.size*2/3..-1] = nil
really?

because it is equiv to

a[0..a.size/2] = (a[a.size*2/3..-1] = nil)
a[0..a.size/2] = (nil)

by the time the Right most got evaluated, the array is roughly 66
elements, so it should be half the size to be deleted. looks like the
interpreter precalculate a.size / 2 before the final assignment.

···

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

7

No it's not because the leftmost = in the first statement is'nt doing
what you think it is:

a = Array(1..100) # => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,
32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65,
66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82,
83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99,
100]

b = a[a.size*2/3..-1] = nil # => nil

a # => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52,
53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66]

qri Array#=
-------------------------------------------------------------- Array#=
     array[index] = obj -> obj
     array[start, length] = obj or an_array or nil -> obj or an_array
     or nil
     array[range] = obj or an_array or nil -> obj or an_array
     or nil

Note that Array#= returns its second argument, NOT the resulting array.

ALSO

The value of a ruby assignment expression is the right hand side, NOT
the left hand side.

class Foo
  def a=(b)
    10
  end
end

f = Foo.new

f.send(:a=, 3) # => 10
b = f.a = 3 # => 3

···

On 9/30/07, SpringFlowers AutumnMoon <summercoolness@gmail.com> wrote:

SpringFlowers AutumnMoon wrote:
> Simon Schuster wrote:
>> a is an array. I'm trying to chop the first and last third off. or,
>> choose only the middle third. can be a rough third.

that's strange, if i do

a = Array(1..100)
    ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1
6, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 4
6, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6
1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 7
6, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
1, 92, 93, 94, 95, 96, 97, 98, 99, 100]

a[0..a.size/2] = a[a.size*2/3..-1] = nil
    ==>nil
a
    ==>[52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64,
65, 66]

a = Array(1..100)
    ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1
6, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 4
6, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6
1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 7
6, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
1, 92, 93, 94, 95, 96, 97, 98, 99, 100]

a[0..a.size/3] = a[a.size*2/3..-1] = nil
    ==>nil
a
    ==>[35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47,
48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62,
63, 64, 65, 66]

shouldn't it be a[0..a.size/2] = a[a.size*2/3..-1] = nil
really?

because it is equiv to

a[0..a.size/2] = (a[a.size*2/3..-1] = nil)
a[0..a.size/2] = (nil)

--
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/