How to avoid repeating "a.condition_A==b.condition_A" in if statements about "compare condition A first, if equals, then B, then C..."?

For example, there is a piece of code to find the best student according to their exam results, with maths result first, if having the same maths score, then use en result and then es result:

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const students=[
  {"name":"Ada","maths":58,"en":70,"es":60},
  {"name":"Bob","maths":70,"en":54,"es":42},
  {"name":"Tom","maths":60,"en":50,"es":50}
];
let bestIndex=0;
let bestMaths=0;
let bestEn=0;
let bestEs=0;
let index=0;
for(const s of students){
  if(s.maths>bestMaths || 
    (s.maths==bestMaths && s.en>bestEn) ||
    (s.maths==bestMaths && s.en==bestEn && s.es>bestEs)){
    bestIndex=index;
    bestMaths=s.maths;
    bestEn=s.en;
    bestEs=s.es
  }
  index++
}
document.write(bestIndex);

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While the code works fines currently, the coding style is not good looking because it is repeating

s.maths==bestMaths

. Also when adding a new requirement : "select student with highest score in all language subjects before comparing en result", new "a==b" statements is required to repeat (repeating "s.en+s.es==bestTotalLang" at the following):

 s.maths>bestMaths || 
(s.maths==bestMaths && s.en+s.es>bestTotalLang) ||
(s.maths==bestMaths && s.en+s.es==bestTotalLang && s.en>bestEn) ||
(s.maths==bestMaths && s.en+s.es==bestTotalLang && s.en==bestEn && s.es>bestEs)

Also it is required to rewrite the "pyramid" of code above when the priority of condition changed, for example : select student with best en first, then maths and then es.

My question is, is there any coding style to rewrite the code above, which can avoid repeating s.maths==bestMaths (ie: allow "s.maths==bestMaths" appears once only), and also for other "a==b" statements?

For the general case of a repetitive comparison, as long as impure getters aren't involved, you can store the compare result in a variable and then use that variable. So

s.maths>bestMaths || 
(s.maths==bestMaths && ...
(s.maths==bestMaths && ...
(s.maths==bestMaths && ...

could become

const thisBest = s.maths==bestMaths;
s.maths>bestMaths || 
(thisBest && ...
(thisBest && ...
(thisBest && ...

But for this particular logic you're going with, it would be easier to perform the s.maths==bestMaths check first, once, in a single branch, then list out into the other possibilities in a big && that follows.

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const students=[{"name":"Ada","maths":58,"en":70,"es":60},{"name":"Bob","maths":70,"en":54,"es":42},{"name":"Tom","maths":60,"en":50,"es":50}];
let bestIndex=0;
let bestMaths=0;
let bestEn=0;
let bestEs=0;
let index=0;
for (const s of students) {
  if (s.maths > bestMaths || (s.maths == bestMaths && (
    s.en > bestEn
    || (s.en == bestEn && s.es > bestEs)
    // add other conditions here
  ))) {
    bestIndex = index;
    bestMaths = s.maths;
    bestEn = s.en;
    bestEs = s.es
  }
  index++
}
console.log(bestIndex);

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A better way to approach this would be to .reduce while keeping the best student so far in the accumulator, then log its index at the end.

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const students=[{"name":"Ada","maths":58,"en":70,"es":60},{"name":"Bob","maths":70,"en":54,"es":42},{"name":"Tom","maths":60,"en":50,"es":50}];
const bestStudent = students.reduce((a, s) => (
  (
    s.maths > a.maths ||
    (s.maths === a.maths && s.en > a.en) ||
    (s.maths === a.maths && s.en === a.en && s.es > a.es)
  ) ? s : a
));
const bestIndex = students.indexOf(bestStudent);
console.log(bestIndex);

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It may be a bit repetitive, but repetitiveness is not always the enemy in itself - it's OK to be somewhat repetitive if the alternative is sacrificing readability.

A readable solution can be achieved without repetitions. Here is my take on it:

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const students=[{"name":"Ada","maths":58,"en":70,"es":60},{"name":"Bob","maths":70,"en":54,"es":42},{"name":"Tom","maths":60,"en":50,"es":50},{"name":"Mary","maths":70,"en":53,"es":65}];

const best=students.reduce((a,c)=>
  c.maths>a.maths
   ?c
   :c.maths==a.maths
    ?c.en>a.en
     ?c
     :c.en==a.en&&c.es>a.es
      ?c
      :a
    :a );

console.log(best);
console.log(students.indexOf(best))

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