Run results for: jAg's comparison of unzipping, possibly modifying, and rezipping objects with 250 fields

You've got an object with a bunch of fields. You want to iterate through each field and if it matches a certain set of values/state, you want to modify it, and then get the final updated object (ie not just the updated values, but all the unchanged values too. The only thing that doesn't matter is the original value).

Go to the benchmark

Run details:

User agent: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/122.0.0.0 Safari/537.36

Browser: Chrome 122

Operating system: Linux

Device Platform: Desktop

Date tested: 8 months ago

Test name	Executions per second
Object.entries -> map -> Object.fromEntries	1960.4 Ops/sec
Object.entries -> reduce (single object)	11012.0 Ops/sec
Object.entries -> map -> reduce (single object)	10919.3 Ops/sec
Object.entries -> reduce (temporary objects)	5931.0 Ops/sec
Object.entries -> map -> reduce (temporary objects)	5628.8 Ops/sec
new map -> loop over object.entries -> Object.fromEntries	1693.9 Ops/sec

Script Preparation code:

data = {};
const ITERATIONS = 250;
for (i = 0; i < ITERATIONS; ++i) {
  data[i] = Math.random();
}

function switcher(i) {
  return i < 0.5 ? i : i.toString();
}

Tests:

Object.entries -> map -> Object.fromEntries
Object.fromEntries(Object.entries(data).map((key, value) => [key, switcher(value)]));
Object.entries -> reduce (single object)
Object.entries(data).reduce((acc, [k, v]) => { acc[k] = switcher(v) return acc; }, {});
Object.entries -> map -> reduce (single object)
Object.entries(data).map(([k, v]) => [k, switcher(v)]).reduce((acc, [k, v]) => { acc[k] = v return acc; }, {});
Object.entries -> reduce (temporary objects)
Object.entries(data).reduce((acc, [k, v]) => ({ ...acc, [k]: switcher(v) }), {});
Object.entries -> map -> reduce (temporary objects)
Object.entries(data).map(([k, v]) => [k, switcher(v)]).reduce((acc, [k, v]) => ({ ...acc, [k]: v }), {});
new map -> loop over object.entries -> Object.fromEntries
m = new Map(Object.entries(data)) for([k, v] of Object.entries(data)) { const newV = switcher(v); if (newV !== v) { m.set(k, newV); } } Object.fromEntries(m)

Run results for: jAg's comparison of unzipping, possibly modifying, and rezipping objects with 250 fields

Object.entries -> map -> Object.fromEntries

Object.entries -> reduce (single object)

Object.entries -> map -> reduce (single object)

Object.entries -> reduce (temporary objects)

Object.entries -> map -> reduce (temporary objects)

new map -> loop over object.entries -> Object.fromEntries