Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]
A simple ascending sort but with the zero values to the end
I really don’t see how to do
Thanks for your help
On Mon, Jun 8, 2009 at 10:40 AM, Paganoni[email protected] wrote:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]A simple ascending sort but with the zero values to the end
I really don’t see how to do
Thanks for your help
Maybe not the most elegant solution but it gets the job done:
arr = [1,4,2,0,8,9]
arr = (arr - [0]).sort << 0
Otherwise you want to override <=> on Fixnum
Andrew T.
http://ramblingsonrails.com
http://MyMvelope.com - The SIMPLE way to manage your savings
2009/6/8 Andrew T. [email protected]:
On Mon, Jun 8, 2009 at 10:40 AM, Paganoni[email protected] wrote:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]A simple ascending sort but with the zero values to the end
Otherwise you want to override <=> on Fixnum
Definitively not! Changing the default Fixnum ordering is dangerous
because it will likely break a lot of other code and it is
superfluous, too. There are better tools for that, i.e. defining the
sort order in the place where it is needed (see Martin’s solution).
Kind regards
robert
On Mon, Jun 8, 2009 at 2:10 PM, Paganoni[email protected] wrote:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]A simple ascending sort but with the zero values to the end
max = a.max + 1
or max = 2 ** 31, say, if you don’t want the extra pass
but sorting is O(n log n) and max is O(n) so it doesn’t really matter
a.sort_by {|x| x.zero? ? max : x}
martin
2009/6/8 Martin DeMello [email protected]:
On Mon, Jun 8, 2009 at 2:10 PM, Paganoni[email protected] wrote:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]A simple ascending sort but with the zero values to the end
max = a.max + 1
or max = 2 ** 31, say, if you don’t want the extra pass
but sorting is O(n log n) and max is O(n) so it doesn’t really matter
Still traversing the array twice just to get the max beforehand does
not /feel/ right. I’d rather use your “large constant” - maybe even
with a really large number: 
irb(main):020:0> a = [1,4,2,0,8,9]
=> [1, 4, 2, 0, 8, 9]
irb(main):021:0> INF = 1.0 / 0.0
=> Infinity
irb(main):022:0> a.sort_by {|x| x == 0 ? INF : x}
=> [1, 2, 4, 8, 9, 0]
Another good alternative is to use the block form of #sort:
irb(main):023:0> a.sort do |x,y|
irb(main):024:1* case
irb(main):025:2* when x == 0 then 1
irb(main):026:2> when y == 0 then -1
irb(main):027:2> else x <=> y
irb(main):028:2> end
irb(main):029:1> end
=> [1, 2, 4, 8, 9, 0]
Kind regards
robert
Hi,
Am Montag, 08. Jun 2009, 17:40:06 +0900 schrieb Paganoni:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]
A simple ascending sort but with the zero values to the end
I don’t know what you want to do further with the sorted values
but maybe this is an approach to consider:
s = a.map { |x| x.nonzero? }
s.compact!
s.sort!
or
s, z = a.map { |x| x.nonzero? }.partition { |x| x }
s.sort!
puts z.length
Bertram
P.S.: Still, my opinion is that there should be an equivalent
String#notempty? to Numeric#nonzero? !
On Mon, Jun 8, 2009 at 5:40 PM, Paganoni[email protected] wrote:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]A simple ascending sort but with the zero values to the end
I really don’t see how to do
Thanks for your help
tmp = arr.partition{|x| x != 0 }
(tmp[0].sort + tmp[1])
Harry
On Mon, Jun 8, 2009 at 2:57 PM, Robert
Klemme[email protected] wrote:
Still traversing the array twice just to get the max beforehand does
not /feel/ right. I’d rather use your “large constant” - maybe even
with a really large number:irb(main):020:0> a = [1,4,2,0,8,9]
=> [1, 4, 2, 0, 8, 9]
irb(main):021:0> INF = 1.0 / 0.0
=> Infinity
Ah yes, forgot you could compare floats with fixnums
martin
On Mon, Jun 8, 2009 at 7:59 PM, Harry K.[email protected]
wrote:
Oops!
tmp = arr.partition{|x| x != 0 }
p (tmp[0].sort + tmp[1])
The second line should be like this, of course.
Harry
le 08/06/2009 10:38, Paganoni nous a dit:
Hello, I need to sort
[1,4,2,0,8,9] to [1,2,4,8,9,0]A simple ascending sort but with the zero values to the end
I really don’t see how to do
Thanks for your help
I forgot to mention that the array can contain several 0 values not only
one.
Giampiero Z. wrote:
ciao
if you are sure not to have negative numbers in your input, then you can
map each number to its negative, sort descending and then map again to
positive
sorry; it cannot work, of course
ciao
if you are sure not to have negative numbers in your input, then you can
map each number to its negative, sort descending and then map again to
positive
On Mon, Jun 8, 2009 at 6:40 PM, Paganoni[email protected] wrote:
le 08/06/2009 10:38, Paganoni nous a dit:
I forgot to mention that the array can contain several 0 values not only
one.
My solution took that into account. Give it a try.
martin
Paganoni schrieb:
I forgot to mention that the array can contain several 0 values not only
one.
Infinity = 1.0/0.0
[1,4,2,0,8,9].sort_by {|x| x == 0 ? Infinity : x }
Or:
a = [1,4,2,0,8,9]
max = a.max + 1
a.sort_by {|x| x == 0 ? max : x }
Regards,
Michael
Robert K. [email protected] wrote:
Another good alternative is to use the block form of #sort:
irb(main):023:0> a.sort do |x,y|
irb(main):024:1* case
irb(main):025:2* when x == 0 then 1
irb(main):026:2> when y == 0 then -1
irb(main):027:2> else x <=> y
irb(main):028:2> end
irb(main):029:1> end
=> [1, 2, 4, 8, 9, 0]
isn’t this simplified version fine too?
irb(main):003:0> a=[0,3,2,-5,0,4]
=> [0, 3, 2, -5, 0, 4]
irb(main):004:0> a.sort{|x,y| x.zero? ? 1 : x<=>y}
=> [-5, 2, 3, 4, 0, 0]
–
On 08.06.2009 17:07, andrea wrote:
irb(main):029:1> end
=> [1, 2, 4, 8, 9, 0]isn’t this simplified version fine too?
irb(main):003:0> a=[0,3,2,-5,0,4]
=> [0, 3, 2, -5, 0, 4]
irb(main):004:0> a.sort{|x,y| x.zero? ? 1 : x<=>y}
=> [-5, 2, 3, 4, 0, 0]
irb(main):003:0> a=[1,0]
=> [1, 0]
irb(main):004:0> a.sort{|x,y| x.zero? ? 1 : x<=>y}
=> [0, 1]
irb(main):005:0>
Not what the OP wanted.
Cheers
robert
On 08.06.2009 20:55, [email protected] wrote:
irb(main):005:0>
Assuming a stable sorting algorithm, sort twice is an option:
irb(main):001:0> [1, 0].sort.sort_by{|n| n.zero? ? 1 : 0}
=> [1, 0]
Sorting twice only to make the block to sort simpler does not sound like
a good operation.
Kind regards
robert
Am Montag 08 Juni 2009 20:55:17 schrieb [email protected]:
Assuming a stable sorting algorithm
Neither sort nor sort_by use a stable sorting algorithm though.
On 6/8/09, Robert K. [email protected] wrote:
irb(main):005:0>
Assuming a stable sorting algorithm, sort twice is an option:
irb(main):001:0> [1, 0].sort.sort_by{|n| n.zero? ? 1 : 0}
=> [1, 0]
irb(main):002:0> [0,3,2,-5,0,4].sort.sort_by{|n| n.zero? ? 1 : 0}
=> [-5, 4, 3, 2, 0, 0]
irb(main):003:0> a = (1…10).map{|n| rand(5) - 2}
=> [-1, 2, -1, 0, 0, 2, 0, 2, 2, -2]
irb(main):004:0> a.sort.sort_by{|n| n.zero? ? 1 : 0}
=> [-2, -1, -1, 2, 2, 2, 2, 0, 0, 0]
irb(main):033:0> a.sort.sort_by{|n| n.zero?.to_s}
=> [-2, -1, -1, 2, 2, 2, 2, 0, 0, 0]
On 6/8/09, Robert K. [email protected] wrote:
On 08.06.2009 20:55, [email protected] wrote:
Assuming a stable sorting algorithm, sort twice is an option:
irb(main):001:0> [1, 0].sort.sort_by{|n| n.zero? ? 1 : 0}
=> [1, 0]Sorting twice only to make the block to sort simpler does not sound like
a good operation.
Especially the to_s version
Error: Your application used more memory than the safety cap of 500m.
Specify -J-Xmx####m to increase it (#### = cap size in MB).
Specify -w for full OutOfMemoryError stack trace
require ‘benchmark’
Infinity = 1.0/0.0
Benchmark.bm(25) do |b|
a = (1…500000).map{|n| rand(10) - 20}
b.report(“A”) {
a.sort.sort_by{|n| n.zero? ? 1 : 0}
}
b.report(“B”) {
a.sort_by {|x| x == 0 ? Infinity : x }
}
b.report(“C”) {
max = a.max + 1
a.sort_by {|x| x == 0 ? max : x }
}
b.report(“D”) {
tmp = a.partition{|x| x != 0 }
tmp[0].sort + tmp[1]
}
b.report(“E”) {
a.sort do |x,y|
case
when x == 0 then 1
when y == 0 then -1
else x <=> y
end
end
}
end
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