您可以使用此函數,它接受一個多維數組並返回其n個最小值的數組,其中n是一個參數。重要的是,返回數組中的元素是一個數據結構Type Point
,包含每個找到的點的座標和值。
您可以輕鬆地調整它找到N MAX值,僅僅通過在代碼改變兩個字符,如在評論中指出(初始化和比較)
Option Explicit
Type Point
X As Long
Y As Long
Z As Long
value As Double
End Type
Function minVals(ar() As Double, nVals As Long) As Point()
Dim i As Long, j As Long, k As Long, m As Long, n As Long, pt As Point
'Initialize returned array with max values.
pt.value = 9999999# ' <-------- change to -9999999# for finding max
ReDim ret(1 To nVals) As Point
For i = LBound(ret) To UBound(ret): ret(i) = pt: Next
For i = LBound(ar, 1) To UBound(ar, 1)
For j = LBound(ar, 2) To UBound(ar, 2)
For k = LBound(ar, 3) To UBound(ar, 3)
' Find first element greater than this value in the return array
For m = LBound(ret) To UBound(ret)
If ar(i, j, k) < ret(m).value Then ' <------- change to > for finding max
' shift the elements on this position and insert the current value
For n = UBound(ret) To m + 1 Step -1: ret(n) = ret(n - 1): Next n
pt.X = i: pt.Y = j: pt.Z = k: pt.value = ar(i, j, k)
ret(m) = pt
Exit For
End If
Next m
Next k
Next j
Next i
minVals = ret
End Function
Sub Test()
Dim i As Long, j As Long, k As Long, pt As Point
Const n As Long = 11
ReDim CC(1 To n, 1 To n, 1 To n) As Double
For i = 1 To n
For j = 1 To n
For k = 1 To n
CC(i, j, k) = Application.RandBetween(100, 100000)
Next k
Next j
Next i
' Testing the function: get the smalles 5 values and their coordinates
Dim mins() As Point: mins = minVals(CC, 5)
' Printing the results
For i = LBound(mins) To UBound(mins)
Debug.Print mins(i).value, mins(i).X, mins(i).Y, mins(i).Z
Next
End Sub
你需要跟蹤多少個數值 - 只是一個(相對)小的數字,還是你需要對它們進行排序? –
能夠輕鬆調整我想要的最小數量會很好,但10可能是最大值。 –
此外,您可以將此代碼添加到您之前的問題中,而不是發佈一個新的問題... –