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27 articles found matching your search request ‘argmin’…

  • …evaluating min and max values in the places where argmin and argmax values are evaluated for the sake of … evaluate minValue(a,b,c) and argMin(a,b,c) … argMin As Integer = 0 … 0 <> argMin And 0 <> argMax Then … 1 <> argMin And 1 <> argMax Then …

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  • …minValue = a;     var argMin = 0;     if (minValue > b) …         argMin = 1;     } …         argMin = 2;     } … midleValue = c;     if (0 != argMin && 0 != argMax) … = a;     if (1 != argMin && 1 != argMax) …

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  • …$minValue = $a;     my $argMin = 0;     if ($minValue > …         $argMin = 1;     } …         $argMin = 2;     } … = $c;     if (0 != $argMin && 0 != $argMax) …     if (1 != $argMin && 1 != $argMax) …

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  • …evaluate minValue(a,b,c) and argMin(a,b,c)     minValue := a …     argMin := 0     if minValue > b { … b         argMin = 1     } … c         argMin = 2     } … := c     if 0 != argMin && 0 != argMax { …

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  • …been interesting to see how they compare to the argmin/argmax functions of 5 arguments, but it will have … evaluate min(x,y,z)     double min = x[0];    int argmin = 0;    if (min > x[1])    {        min = … x[1];        argmin = 1;    }    if (min > x[2])    {        min = … x[2];        argmin = 2;    }    // evaluate max(x,y,z)     double …     double midleValue = x[2];    if (0 != argmin && 0 != argmax)        midleValue = x[0];  …

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  • …midleValue: real;     argMin, argMax: integer; begin     { … := a;     argMin := 0;     if (minValue > …         argMin := 1;     end; …         argMin := 2;     end; … := c;     if ((0 <> argMin) and (0 <> argMax)) then …

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  • …        argMin: integer := 0; … ((0 /= argMin) and then (0 /= argMax)) then … ((1 /= argMin) and then (1 /= argMax)) then …         argMin: integer := 0; … ((0 /= argMin) and then (0 /= argMax)) then …

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  • …with 3-4 discrete outputs that matched the argmin/argmax functions in performance), and both of … evaluate min(x,y,z)     double min = x[0];    int argmin = 0;    if (min > x[1])    {        min = … x[1];        argmin = 1;    }    if (min > x[2])    {        min = … x[2];        argmin = 2;    }    // evaluate max(x,y,z)     double …     double midleValue = x[2];    if (0 != argmin && 0 != argmax)        midleValue = x[0];  …

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  • …minValue As Double     Dim argMin As Long     minValue = a …     argMin = 0     If minValue > b … b         argMin = 1     End If … c         argMin = 2     End If … = c     If 0 <> argMin And 0 <> argMax Then …

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  • …minValue As Double     Dim argMin As Long     minValue = a …     argMin = 0     If minValue > b … b         argMin = 1     End If … c         argMin = 2     End If … = c     If 0 <> argMin And 0 <> argMax Then …

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