Averages/Root mean square: Difference between revisions
Underscore (talk | contribs) (Added Haskell.) |
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>>> qmean(numbers) |
>>> qmean(numbers) |
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6.2048368229954285</lang> |
6.2048368229954285</lang> |
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=={{header|Tcl}}== |
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{{works with|Tcl|8.5}} |
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<lang tcl>proc qmean list { |
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set sum 0.0 |
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foreach value $list { set sum [expr {$sum + $value**2}] } |
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return [expr { sqrt($sum / [llength $list]) }] |
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} |
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puts "RMS(1..10) = [qmean {1 2 3 4 5 6 7 8 9 10}]"</lang> |
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Output: |
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<pre> |
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RMS(1..10) = 6.2048368229954285 |
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</pre> |
Revision as of 17:05, 21 February 2010
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You are encouraged to solve this task according to the task description, using any language you may know.
Compute the Root mean square of the numbers 1..10.
The root mean square is also known by its initial RMS (or rms), and as the quadratic mean.
The RMS is calculated as the mean of the squares of the numbers, square-rooted:
C.f. Averages/Pythagorean means
Haskell
Given the mean
function defiend in Averages/Pythagorean means:
<lang haskell>main = print $ mean 2 [1 .. 10]</lang>
Lua
<lang lua>function sumsq(a, ...) return a and a^2 + sumsq(...) or 0 end function rms(t) return (sumsq(unpack(t)) / #t)^.5 end
print(rms{1, 2, 3, 4, 5, 6, 7, 8, 9, 10})</lang>
PL/I
<lang PL/I> declare A(10) fixed decimal static initial (1,2,3,4,5,6,7,8,9,10); n = hbound(A,1); RMS = sqrt(sum(A**2)/n); </lang>
Python
<lang Python>>>> from __future__ import division >>> from math import sqrt >>> def qmean(num): return sqrt(sum(n*n for n in num)/len(num))
>>> numbers = range(1,11) # 1..10 >>> qmean(numbers) 6.2048368229954285</lang>
Tcl
<lang tcl>proc qmean list {
set sum 0.0 foreach value $list { set sum [expr {$sum + $value**2}] } return [expr { sqrt($sum / [llength $list]) }]
}
puts "RMS(1..10) = [qmean {1 2 3 4 5 6 7 8 9 10}]"</lang> Output:
RMS(1..10) = 6.2048368229954285