Javascript Variable with Highest Number

Question

How do I select the variable associated with the highest number? If I have p1 =1 , p2=4 and p3=3, how can I make the program choose the highest one(in this case p2?)

You asked how the program can choose a given variable (which for me makes little sense), but I reckon you simply want the value instead: consr foo = Math.max(p1, p2, p3). — Gerardo Furtado
– Gerardo Furtado, Commented Dec 6, 2019 at 3:00

Mr. Polywhirl · Accepted Answer · 2019-12-06 03:50:07Z

1

You should apply a scope to the parameters, so that you reduce the number of variables you need to scan for.

let myScope = {};

myScope.p1 = 1;
myScope.p2 = 4;
myScope.p3 = 3;

console.log(getHighest(myScope)); // 4

function getHighest(scope) {
  return Math.max.apply(Math, Object.values(scope));
}

Note: A better alternative would be to have a variable called p to store the values e.g.

var p = [ 1, 4, 3 ];
var max = Math.max.apply(Max, p);

NOT ADVISED: You could also access them from the window, but this is tricker...

var p1 = 1;
var p2 = 4;
var p3 = 3;

console.log(getHighest(window, /^p\d$/)); // 4

function getHighest(scope, pattern) {
  return Math.max.apply(
    Math,
    Object.keys(scope)
      .filter(k => k.match(pattern))
      .map(k => scope[k])
  );
}

edited Dec 6, 2019 at 3:50

answered Dec 6, 2019 at 3:00

Mr. Polywhirl

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6 Comments

Gerardo Furtado Over a year ago

This is not necessary.

yqlim Over a year ago

In case anyone reading this asks "why not just Math.max(p1, p2, p3)", this approach does not require you to manually key in your keys everytime you add values to the scope. This can also be done using Array instead of object literal.

Gerardo Furtado Over a year ago

@YongQuan indeed, but look at the word need: "You need to apply a scope to the parameters...". And also a comment in the question, by the same answerer: "You need to scope them". It's not necessary.

yqlim Over a year ago

@GerardoFurtado I agree. You don't need to do this if your context doesn't require dynamically adding values.

Mr. Polywhirl Over a year ago

@MarkMeyer I went ahead and updated/annotated the response. Thanks.

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