How to make a javascript comparison more concise?

Question

Consider the following two javascript objects:

Object A:

{
  specialPropertyA: string,
  partTime: boolean,
  role: { key: string, label: string }
}

Object B:

{
  specialPropertyB: string,
  partTime: boolean,
  role: { key: number, label: string }
}

Now, consider an arrayA: ObjectA[] and an arrayB: ObjectB[].

I am looking for a concise way to determine:

If any of the ObjectA's have partTime === true AND role.key === 1, are there any ObjectBs, which fulfill the same requirements? I want to check the same for role.key === 2.

I know I could do sth like:

if(
    arrayA.filter(objectA => objectA.partTime === true && objectA.role.key === 1) 
    && arrayB.filter(objectB => objectB.partTime === true && objectB.role.key === 1) 
    || arrayA.filter(objectA => objectA.partTime === true && objectA.role.key === 2)
    && arrayB.filter(objectB => objectB.partTime === true && objectB.role.key === 2)
  )

But I don't like this solution at all, especially since I am repeating the same code for key 2. Any suggestions as on how to make this more concise?

CertainPerformance · Accepted Answer · 2022-06-03T20:36:35.050

2

One way would be to make a higher-order function that you can pass the key number you want to find.

Also, because partTime is a boolean, obj.partTime === true simplifies to obj.partTime.

const makeCb = keyNum => obj => obj.partTime && obj.role.key === keyNum;

if (
  (arrayA.some(makeCb(1)) && arrayB.some(makeCb(1))) ||
  (arrayA.some(makeCb(2)) && arrayB.some(makeCb(2)))
) {

You'll want .some (or .find, if you want to put the result into a variable). .filter won't work in an if because even if there are no results, empty arrays are truthy.

edited Jun 03 '22 at 20:36

answered Jun 03 '22 at 20:35

CertainPerformance

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this looks very interesting! I am not familiar with this nested notation though. The way I understand it is: `makeCb()` takes in a parameter `keyNum` and returns a function, which takes in the parameter `obj` and returns `obj.partTime && obj.role.key === keyNum`. Is that correct? – Luk Jun 03 '22 at 21:03
1

It's [just currying](https://stackoverflow.com/questions/36314/what-is-currying) - a higher-order function, a function that returns a function, yes – CertainPerformance Jun 03 '22 at 21:19

score 0 · Answer 2 · answered Jun 03 '22 at 20:44

If your arrays have many possible key values, which could match, then maybe first collect the found key values in the first array before scanning the second:

const keysA = new Set(arrayA.map(({partTime, key}) => partTime && key));
if (arrayB.some(({partTime, key}) => partTime && keysA.has(key))) {
  /* ... */
}

This assumes that false is not a possible value of key.

How to make a javascript comparison more concise?

2 Answers2