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Display the unique items in a list
Display the unique items in a list, e.g. given ages =
perl
perl
perl
perl
perl
java
clojure 1.0.0
fantom
csharp .NET 3.5
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There are 16 other solutions in additional languages (cpp, erlang, fsharp, groovy ...)
[18, 16, 17, 18, 16, 19, 14, 17, 19, 18], display the unique elements, i.e. with duplicates removed.
@ages = (18, 16, 17, 18, 16, 19, 14, 17, 19, 18);
@seen{@ages} = ();
@unique = keys %seen;
print join(', ', @unique);
@seen{@ages} = ();
@unique = keys %seen;
print join(', ', @unique);
@ages = (18, 16, 17, 18, 16, 19, 14, 17, 19, 18);
@unique = grep(!$seen{$_}++, @ages);
print join(', ', @unique);
@unique = grep(!$seen{$_}++, @ages);
print join(', ', @unique);
@ages = (18, 16, 17, 18, 16, 19, 14, 17, 19, 18);
print join(', ', grep(!$seen{$_}++, @ages));
print join(', ', grep(!$seen{$_}++, @ages));
@ages = (18, 16, 17, 18, 16, 19, 14, 17, 19, 18);
for (@ages) {
push(@unique, $_) unless $seen{$_}++;
}
print join(', ', @unique);
for (@ages) {
push(@unique, $_) unless $seen{$_}++;
}
print join(', ', @unique);
use List::MoreUtils qw(uniq);
@ages = (18, 16, 17, 18, 16, 19, 14, 17, 19, 18);
print join(', ', uniq(@ages));
@ages = (18, 16, 17, 18, 16, 19, 14, 17, 19, 18);
print join(', ', uniq(@ages));
Set<Integer> ages = new TreeSet<Integer>(Arrays.asList(new Integer[]{18, 16, 17, 18, 16, 19, 14, 17, 19, 18}));
System.out.println(ages);
System.out.println(ages);
;; returns a set
(set [18, 16, 17, 18, 16, 19, 14, 17, 19, 18])
;;#{14 16 17 18 19}
;; returns a lazy sequence of the unique elements
(distinct [18, 16, 17, 18, 16, 19, 14, 17, 19, 18])
;;(18 16 17 19 14)
(set [18, 16, 17, 18, 16, 19, 14, 17, 19, 18])
;;#{14 16 17 18 19}
;; returns a lazy sequence of the unique elements
(distinct [18, 16, 17, 18, 16, 19, 14, 17, 19, 18])
;;(18 16 17 19 14)
uniqueAges := [18, 16, 17, 18, 16, 19, 14, 17, 19, 18].unique
echo(uniqueAges)
echo(uniqueAges)
using System.Collections.Generic;
using System.Linq;
public class UniqueElements {
public static void Main() {
var list = new List<int>() { 18, 16, 17, 18, 16, 19, 14, 17, 19, 18 };
var uniques = list.Distinct();
}
}
using System.Linq;
public class UniqueElements {
public static void Main() {
var list = new List<int>() { 18, 16, 17, 18, 16, 19, 14, 17, 19, 18 };
var uniques = list.Distinct();
}
}
Submit a new solution for perl, java, clojure, fantom ...
There are 16 other solutions in additional languages (cpp, erlang, fsharp, groovy ...)
