I'm trying to randomize an array, and this is what I have:
arr_set_unordered = randomizer(arr_set)
def randomizer(arr)
result = arr.collect { arr.slice!(rand arr.length) }
end
It does randomize it, but it only returns 3 values instead of the 5
values that I'm expecting. Any ideas why?
Thank you!
jwcooper wrote:
I'm trying to randomize an array...
Try this.
class Array
def randomize
duplicated_original, new_array = self.dup, self.class.new
new_array <<
duplicated_original.slice!(rand(duplicated_original.size)) until
new_array.size.eql?(self.size)
new_array
end
def randomize!
self.replace(randomize)
end
end
Enjoy!
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jwcooper wrote:
I'm trying to randomize an array, and this is what I have:
arr_set_unordered = randomizer(arr_set)
def randomizer(arr)
result = arr.collect { arr.slice!(rand arr.length) }
endIt does randomize it, but it only returns 3 values instead of the 5
values that I'm expecting. Any ideas why?Thank you!
ilans-Mac:~ ilan$ irb
irb(main):001:0> (1..10).to_a.sort {rand}
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
irb(main):002:0>
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Sorry, I guess a use-case would be helpful.
Call #randomize or #randomize! on any array to mix up its elements.
[ 5, 10, 15, 20, 25 ].randomize # [25, 15, 5, 20, 10]
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Ilan Berci wrote:
jwcooper wrote:
I'm trying to randomize an array, and this is what I have:
[...]
ilans-Mac:~ ilan$ irb
irb(main):001:0> (1..10).to_a.sort {rand}
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
irb(main):002:0>
Don't do this, better use (1..10).sort_by { rand }. Your version is equivalent to (1..10).sort { 1 } and *always* creates the same permutation for this array.
--
Florian Frank
I found a library called 'rand.rb':
http://raa.ruby-lang.org/project/rand/\. It allows you to select an
element at random from an array and 'scramble' (randomize the order)
of them as well.
Description:
Helper methods for Enumerable, Array, Hash, and String
that let you pick a random item or shuffle the order of items.
Examples with Array:
[1,2,3,4].pick
# => 2 (pick random element)
[1,2,3,4].shuffle
# => [2, 4, 1, 3] (return random-sorted version)
a = [1,2,3,4]
a.pick!
# => 3
a
# => [1,2,4] (remove the picked element)
On 29/12/06, Daniel Waite <rabbitblue@gmail.com> wrote:
Sorry, I guess a use-case would be helpful.
Call #randomize or #randomize! on any array to mix up its elements.
[ 5, 10, 15, 20, 25 ].randomize # [25, 15, 5, 20, 10]
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Florian Frank wrote:
irb(main):001:0> (1..10).to_a.sort {rand}
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
Wow. That is squeaky clean, and a lot faster than my method. Thanks for
sharing. 
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Florian Frank wrote:
Don't do this, better use (1..10).sort_by { rand }. Your version is
equivalent to (1..10).sort { 1 } and *always* creates the same
permutation for this array.
It's random for me. Both work.
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Daniel Waite wrote:
Florian Frank wrote:
Don't do this, better use (1..10).sort_by { rand }. Your version is
equivalent to (1..10).sort { 1 } and *always* creates the same
permutation for this array.It's random for me. Both work.
Actually Frank is correct, my mistake, it should definetely be sort_by
and not sort.. I should have reviewed it more thoroughly before
responding..
ilan
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Daniel Waite wrote:
Florian Frank wrote:
Don't do this, better use (1..10).sort_by { rand }. Your version is
equivalent to (1..10).sort { 1 } and *always* creates the same
permutation for this array.It's random for me. Both work.
No, really, it isn't. It may "look random", but try sorting (1..10) many times and ponder the coincidence.
--
Florian Frank
Florian Frank wrote:
Daniel Waite wrote:
Florian Frank wrote:
Don't do this, better use (1..10).sort_by { rand }. Your version is
equivalent to (1..10).sort { 1 } and *always* creates the same
permutation for this array.It's random for me. Both work.
No, really, it isn't. It may "look random", but try sorting (1..10) many
times and ponder the coincidence.
That's because sort{ foo } expects foo to be in -1,0,1.
rand() is always [0, 1).
rand(2)-1 is (-1,1) however and thus will work just fine.
One could do sort{rand <=> rand}, but that's ~9x slower by my testing.
irb(main):163:0> (1..10).to_a.sort{rand }
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
irb(main):164:0> (1..10).to_a.sort{rand }
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
irb(main):165:0> (1..11).to_a.sort{rand }
=> [11, 6, 1, 7, 3, 8, 5, 9, 4, 10, 2]
irb(main):166:0> (1..12).to_a.sort{rand }
=> [12, 7, 1, 8, 3, 9, 5, 10, 2, 11, 6, 4]
vs
irb(main):186:0> (1..12).to_a.sort{rand(2) -1}
=> [1, 2, 6, 11, 9, 4, 8, 7, 5, 10, 3, 12]
irb(main):187:0> (1..12).to_a.sort{rand(2) -1}
=> [1, 9, 8, 3, 5, 6, 11, 4, 2, 10, 7, 12]
- Sai http://saizai.com, who is too lazy to remember his actual account
pass right now
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Florian Frank wrote:
Daniel Waite wrote:
Florian Frank wrote:
Don't do this, better use (1..10).sort_by { rand }. Your version is
equivalent to (1..10).sort { 1 } and *always* creates the same
permutation for this array.It's random for me. Both work.
No, really, it isn't. It may "look random", but try sorting (1..10) many
times and ponder the coincidence.That's because sort{ foo } expects foo to be in -1,0,1.
I don't think so - the sign is important:
irb(main):001:0> (1..10).sort { -1 }
=> [1, 2, 3, 7, 4, 9, 5, 8, 6, 10]
irb(main):002:0> (1..10).sort { 1 }
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
irb(main):003:0> (1..10).sort { 10 }
=> [10, 6, 1, 7, 3, 8, 5, 9, 4, 2]
irb(main):004:0> (1..10).sort { -10 }
=> [1, 2, 3, 7, 4, 9, 5, 8, 6, 10]
rand() is always [0, 1).
Did you mean [0,1[?
rand(2)-1 is (-1,1) however and thus will work just fine.
rand(2)-1 is (-1,0). I would rather do rand(3)-1 or rand(2) - 0.5 -
But all this is worse than sort_by { rand } IMHO.
Kind regards
robert
2008/5/15 Joao Silva <rubyforum@thisisnotmyrealemail.com>:
--
use.inject do |as, often| as.you_can - without end