Question 262792 in Javascript Question posted in
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Javascript – How would I check if the significant digits of a "floored" value and another value are the same?

Suppose a value is "floored" (floored in some arbitrary digit) when its displayed (7.785->7.7, 110.6->110). Let this floored value be called flooredTarget. Suppose I have another number variable named toCheck. How would I check if the significant digits of flooredTarget and toCheck are equal?

NOTE: the first parameter is flooredTarget.

checkSignificant(7.7, 7.785); // true as 7.7 is common
checkSignificant(7.785, 7.8); // false as toCheck doesn't contain 7.785

Should I just check if toCheck.toString().beginsWith(flooredTarget.toString())? Or is there some other better way (Assume any language)

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2

Answers


  1. 0

    You can use regexp

    function checkSignificant(flooredTarget, toCheck) {
  const [digits] = flooredTarget.toString().match(/(?<=.)d+/)
  return new RegExp(`(?<=\.)${digits}`).test(toCheck.toString())
}

console.log(checkSignificant(7.7, 7.785));
console.log(checkSignificant(7.785, 7.8));
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  2. 0

    Converting to text is a performance killer there are other ways for example I see it like this in C++:

    //---------------------------------------------------------------------------
double align_to_exp(double x,int e) // align number x so its most significant digit is at 10^e weight
    {
    bool sig=false;
    if (x==0.0) return 0.0;         // special case
    if (x<0.0){ sig=true; x=-x; }   // ignore sign
    x*=pow(10,e-floor(log10(x)));   // shift digit positions to 10^e
    if (sig) x=-x;                  // restore sign
    return x;
    }
//---------------------------------------------------------------------------
int count_equal_digits(double a,double b) // return number of common significant digits between a,b or -1 if both are zero
    {
    int i;
    double aa,bb;
    a=align_to_exp(fabs(a),0);      // align numbers to 10^0
    b=align_to_exp(fabs(b),0);
    if ((a==0.0)&&(b==0.0)) return -1;  // special case
    for (i=0;i<17;i++)              // limit to 17 digits for double
        {
        aa=floor(a);                // aa is compared digit
        bb=floor(b);                // bb is compared digit
        if ((aa!=bb)||(a==0.0)||(b==0.0)) break;    // stop on difference
        a-=aa; a*=10.0;             // move to next digit
        b-=bb; b*=10.0;             // move to next digit
        }
    return i;
    }
//---------------------------------------------------------------------------

    there is still a lot of room for improvement like making align_to_exp brunchless but In this form the code is more understandable.

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