How to sort array ascending, except zero?


#1

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


#2

On Mon, Jun 8, 2009 at 10:40 AM, Paganoniremoved_email_address@domain.invalid 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


#3

2009/6/8 Andrew T. removed_email_address@domain.invalid:

On Mon, Jun 8, 2009 at 10:40 AM, Paganoniremoved_email_address@domain.invalid 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


#4

On Mon, Jun 8, 2009 at 2:10 PM, Paganoniremoved_email_address@domain.invalid 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


#5

2009/6/8 Martin DeMello removed_email_address@domain.invalid:

On Mon, Jun 8, 2009 at 2:10 PM, Paganoniremoved_email_address@domain.invalid 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: :slight_smile:

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


#6

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? !


#7

On Mon, Jun 8, 2009 at 5:40 PM, Paganoniremoved_email_address@domain.invalid 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


#8

On Mon, Jun 8, 2009 at 2:57 PM, Robert
Klemmeremoved_email_address@domain.invalid 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: :slight_smile:

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 :slight_smile:

martin


#9

On Mon, Jun 8, 2009 at 7:59 PM, Harry K.removed_email_address@domain.invalid
wrote:

Oops!

tmp = arr.partition{|x| x != 0 }
p (tmp[0].sort + tmp[1])

The second line should be like this, of course.

Harry


#10

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.


#11

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


#12

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


#13

On Mon, Jun 8, 2009 at 6:40 PM, Paganoniremoved_email_address@domain.invalid 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


#14

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


#15

Robert K. removed_email_address@domain.invalid 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]

http://myretrocomputing.altervista.org


#16

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


#17

On 08.06.2009 20:55, removed_email_address@domain.invalid 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


#18

Am Montag 08 Juni 2009 20:55:17 schrieb removed_email_address@domain.invalid:

Assuming a stable sorting algorithm

Neither sort nor sort_by use a stable sorting algorithm though.


#19

On 6/8/09, Robert K. removed_email_address@domain.invalid 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]


#20

On 6/8/09, Robert K. removed_email_address@domain.invalid wrote:

On 08.06.2009 20:55, removed_email_address@domain.invalid 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 :slight_smile:

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

Linux 2.6.28-11-generic #42-Ubuntu SMP Fri Apr 17 01:57:59 UTC 2009
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