Averages/Median: Difference between revisions
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return list.get((list.size()-1)/2); |
return list.get((list.size()-1)/2); |
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}</lang> |
}</lang> |
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=={{header|Python}}== |
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<lang python>def median(aray): |
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srtd = sorted(aray) |
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alen = len(srtd) |
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return 0.5*( srtd[(alen-1)//2] + srtd[alen//2]) |
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a = (4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2) |
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print a, median(a) |
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a = (4.1, 7.2, 1.7, 9.3, 4.4, 3.2) |
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print a, median(a)</lang> |
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=={{header|Ruby}}== |
=={{header|Ruby}}== |
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Revision as of 15:59, 12 June 2009
You are encouraged to solve this task according to the task description, using any language you may know.
Write a program to find the median value of a vector of floating-point numbers. The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements.
See also: Mean
Java
<lang java5>public static double median(List<Double> list){
Collections.sort(list);
if(list.size() % 2 == 0){
return (list.get((list.size()) / 2) + list.get((list.size()) / 2 - 1)) / 2;
}
return list.get((list.size()-1)/2);
}</lang>
Python
<lang python>def median(aray):
srtd = sorted(aray) alen = len(srtd) return 0.5*( srtd[(alen-1)//2] + srtd[alen//2])
a = (4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2) print a, median(a) a = (4.1, 7.2, 1.7, 9.3, 4.4, 3.2) print a, median(a)</lang>
Ruby
<lang ruby>def median(ary)
return nil if ary.empty? mid, rem = ary.length.divmod(2) if rem == 0 ary.sort[mid-1,2].inject(:+) / 2.0 else ary.sort[mid] end
end
p median([]) # => nil p median([5,3,4]) # => 4 p median([5,4,2,3]) # => 3.5 p median([3,4,1,-8.4,7.2,4,1,1.2]) # => 2.1</lang>
Tcl
<lang tcl>proc median args {
set list [lsort -real $args]
set len [llength $list]
# Odd number of elements
if {$len & 1} {
return [lindex $list [expr {($len-1)/2}]]
}
# Even number of elements
set idx2 [expr {$len/2}]
set idx1 [expr {$idx2-1}]
return [expr {
([lindex $list $idx1] + [lindex $list $idx2])/2.0
}]
}
puts [median 3.0 4.0 1.0 -8.4 7.2 4.0 1.0 1.2]; # --> 2.1</lang>