How to sort a multi-elements array

I know how to sort an array with 2 elements… that’s the only example
I could fin on Array#sort…
but what if … I need to sort an array like this one :

anArray = [ [1, “joe”, 4], [2, “arthur”, 2], [5, “william”, 3], [16,
“bob”, 1], [20, “ernesto”, 17], [27, “julia”, 1], [28, “barbara”, 2] ]

sorting on third-element (highest ranking) , and second-element
(nickname)

aSortedArray = [ [20, “ernesto”, 17], [1, “joe”, 4], 5, “william”, 3],
[2, “arthur”, 2], [28, “barbara”, 2], [16, “bob”, 1] ]

any doc link where this is explained ?

thanks a lot for your tip

On Saturday 31 May 2008, Erwin wrote:

aSortedArray = [ [20, “ernesto”, 17], [1, “joe”, 4], 5, “william”, 3],
[2, “arthur”, 2], [28, “barbara”, 2], [16, “bob”, 1] ]

any doc link where this is explained ?

thanks a lot for your tip

You can do this:

aSortedArray = anArray.sort do |a, b|
if a[2] == b[2] then a[1] <=> b[1]
else a[2] <=> b[2]
end
end

If the third element of the two subarrays are equal, then the block
returns
the result of the comparison of the second elements (using <=>),
otherwise it
returns the result of the comparison of the third elements, again using
<=>.

I hope this helps

Stefano

On 31.05.2008 19:31, Erwin wrote:

aSortedArray = [ [20, “ernesto”, 17], [1, “joe”, 4], 5, “william”, 3],
[2, “arthur”, 2], [28, “barbara”, 2], [16, “bob”, 1] ]

any doc link where this is explained ?

As always, use #sort with a block or #sort_by - the number of elements
in those arrays is irrelevant, you just need to decide which elements
you want to pick as sort key.

an_array.sort_by {|a,b,c| [c,b]}
an_array.sort {|(a,b,c),(d,e,f)| [c,b] <=> [f,e]}
an_array.sort do |(a,b,c),(d,e,f)|
x = c <=> f
x == 0 ? b <=> e : x
end

Btw, your data looks like it could benefit from using Struct and also
benefit from the automatically implemented <=>, i.e.

Person = Struct.new :age, :name, :position

an_array = [
Person.new(4, “joe”, 1),
Person.new(2, “arthur”, 2),
Person.new(3, “william”, 5),
]

Kind regards

robert

On 31 mai, 19:40, Robert K. [email protected] wrote:

(nickname)
an_array.sort_by {|a,b,c| [c,b]}

an_array = [
Person.new(4, “joe”, 1),
Person.new(2, “arthur”, 2),
Person.new(3, “william”, 5),
]

Kind regards

    robert

thanks Robert , I’ll write that in my NoteTaker Ruby book !!

Hi,

Was going through these older posts and have a similar question. I
have a nested array that contains a date and a filename. I’m trying to
sort by file date.

The array is in the form:

SortedFiles = [ [200904031109, “filea.txt”], [200904031337,
“fileb.txt”], [200904031110, “filec.xt”] ]

What I was trying to do was:

SortedFiles.sort { |[[a,b],[x,y]] a < x | }

That, unfortunately, did not work. I’m using Ruby 1.8.5 btw. Can anyone
help out?

Thanks,
Alex

Stefano C. wrote:

On Saturday 31 May 2008, Erwin wrote:

aSortedArray = [ [20, “ernesto”, 17], [1, “joe”, 4], 5, “william”, 3],
[2, “arthur”, 2], [28, “barbara”, 2], [16, “bob”, 1] ]

any doc link where this is explained ?

thanks a lot for your tip

You can do this:

aSortedArray = anArray.sort do |a, b|
if a[2] == b[2] then a[1] <=> b[1]
else a[2] <=> b[2]
end
end

If the third element of the two subarrays are equal, then the block
returns
the result of the comparison of the second elements (using <=>),
otherwise it
returns the result of the comparison of the third elements, again using
<=>.

I hope this helps

Stefano

On 4/6/09, Alexey Z. [email protected] wrote:

What I was trying to do was:

SortedFiles.sort { |[[a,b],[x,y]] a < x | }

That, unfortunately, did not work. I’m using Ruby 1.8.5 btw. Can anyone
help out?

SortedFiles.sort_by {|date, fname| date}
or:
SortedFiles.sort {|record1,record2| record1[0]<=>record2[0]}

On Tue, Apr 7, 2009 at 12:31 AM, Christopher D.
[email protected]wrote:

“fileb.txt”], [200904031110, “filec.xt”] ]
SortedFiles.sort {|record1,record2| record1[0]<=>record2[0]}

Since you are sorting on the first element of each array you can just
use
SortedFiles.sort

Array comparison with another array works by using the comparison of the
first pair of elements which aren’t equal, so [1,1] < [2,0] < [2,1]


Rick DeNatale

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