Fivenum: Difference between revisions
Fivenum in FreeBASIC
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=={{header|BASIC}}==
==={{header|FreeBASIC}}===
{{trans|Phix}}
<syntaxhighlight lang="vb">#define floor(x) ((x*2.0-0.5) Shr 1)
Sub rapidSort (array()As Single, l As Integer, r As Integer)
Dim As Integer n, wert, nptr, rep
Dim As Single arr, LoVal = array(l), HiVal = array(r)
For n = l To r
If LoVal > array(n) Then LoVal = array(n)
If HiVal < array(n) Then HiVal = array(n)
Next n
Redim SortArray(LoVal To HiVal) As Single
For n = l To r
wert = array(n)
SortArray(wert) += 1
Next n
nptr = l-1
For arr = LoVal To HiVal
rep = SortArray(arr)
For n = 1 To rep
nptr += 1
array(nptr) = arr
Next n
Next arr
Erase SortArray
End Sub
Function median(tbl() As Single, lo As Integer, hi As Integer) As Single
Dim As Integer l = hi-lo+1
Dim As Integer m = lo+floor(l/2)
If l Mod 2 = 1 Then Return tbl(m)
Return (tbl(m-1)+tbl(m))/2
End Function
Sub fivenum(tbl() As Single)
rapidSort(tbl(), Lbound(tbl), Ubound(tbl))
Dim As Integer l = Ubound(tbl)
Dim As Single m = floor(l/2) + (l Mod 2)
Dim As Single r1,r2,r3,r4,r5
r1 = tbl(1)
r2 = median(tbl(),1,m)
r3 = median(tbl(),1,l)
r4 = median(tbl(),m+1,l)
r5 = tbl(l)
Print "[" & r1; ","; r2; ","; r3; ","; r4; ", "; r5 & "]"
End Sub
Dim As Single x1(1 To ...) = {15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43}
Dim As Single x2(1 To ...) = {36, 40, 7, 39, 41, 15}
Dim As Single x3(1 To ...) = {_
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, _
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772, _
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469, _
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578}
fivenum(x1())
fivenum(x2())
fivenum(x3())
Sleep</syntaxhighlight>
{{out}}
<pre>[6, 25.5, 40, 43, 49]
[7, 15, 37.5, 40, 41]
[-1.950596,-0.950596, 0.04940403, 1.049404, 0.3256658]</pre>
==={{header|VBA}}===
Uses [[Sorting_algorithms/Quicksort#VBA|Quicksort]].
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