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…min and max values in the places where argmin and argmax values are evaluated for the sake of uniformity …
evaluate maxValue(a,b,c) and argMax(a,b,c) …
argMax As Integer = 0 …
0 <> argMin And 0 <> argMax Then …
1 <> argMin And 1 <> argMax Then …
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…maxValue = a; var argMax = 0; if (maxValue < b) …
argMax = 1; } …
argMax = 2; } …
(0 != argMin && 0 != argMax) …
(1 != argMin && 1 != argMax) …
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…$maxValue = $a; my $argMax = 0; if ($maxValue < …
$argMax = 1; } …
$argMax = 2; } …
(0 != $argMin && 0 != $argMax) { …
(1 != $argMin && 1 != $argMax) { …
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…with 3-4 discrete outputs that matched the argmin/argmax functions in performance), and both of them …
perform better than the argmax function of 4 arguments (98% hits for B, 99% for …
C, and 96% for the argmax). I've already talked about the Map4B function in …
but nonetheless performs slightly better than the argmax function of 4 arguments, even though it's only a …
function of 2 arguments (97% hits vs 96% for the argmax)! Like I said, the Map4A function (and all mapper …
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…98% for Map5B, 99% for Map5C against 96% for the argmax), behaving very similarly to their counterparts …
that show a performance comparable to the argmax function of 4 arguments (it would have been …
interesting to see how they compare to the argmin/argmax functions of 5 arguments, but it will have to …
evaluate max(x,y,z) double max = x[0]; int argmax = 0; if (max < x[1]) { max = …
x[1]; argmax = 1; } if (max < x[2]) { max = …
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…evaluate maxValue(a,b,c) and argMax(a,b,c) maxValue := a …
argMax := 0 if maxValue < b { …
b argMax = 1 } …
c argMax = 2 } …
0 != argMin && 0 != argMax { …
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…midleValue: real; argMin, argMax: integer; begin { …
:= a; argMax := 0; if (maxValue < …
argMax := 1; end; …
argMax := 2; end; …
((0 <> argMin) and (0 <> argMax)) then …
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…one go with these functions (and also the argmin/argmax functions of 4 arguments) using the current …
for Map6B, and 99% for Map6C against 96% for the argmax. // Map6A(x0,x1): 6-Output Mapper Function …
evaluate max(x,y,z) double max = x[0]; int argmax = 0; if (max < x[1]) { max = …
x[1]; argmax = 1; } if (max < x[2]) { max = …
x[2]; argmax = 2; } // evaluate midleValue(x,y,z) …
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… argMax: integer := 0; …
((0 /= argMin) and then (0 /= argMax)) then …
((1 /= argMin) and then (1 /= argMax)) then …
argMax: integer := 0; …
((0 /= argMin) and then (0 /= argMax)) then …
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…maxValue As Double Dim argMax As Long maxValue = a …
argMax = 0 If maxValue < b …
b argMax = 1 End If …
c argMax = 2 End If …
0 <> argMin And 0 <> argMax Then …