Benchmark: Ways of returning a derived object 2

Tests:

...rest
const orig = { value1: "Hello", value2: 5, value3: 6 }; const fn = x => x + 1; const { value2, value3, ...rest} = orig; const out = { ...rest, value2: fn(value2), value3: fn(value3)};
Spread original
const orig = { value1: "Hello", value2: 5, value3: 6 }; const fn = x => x + 1; const { value2, value3 } = orig; const out = { ...orig, value2: fn(value2), value3: fn(value3)};
Don't destructure
const orig = { value1: "Hello", value2: 5, value3: 6 }; const fn = x => x + 1; const out = { ...orig, value2: fn(orig.value2), value3: fn(orig.value3)};
Object.assign
const orig = { value1: "Hello", value2: 5, value3: 6 }; const fn = x => x + 1; const out = Object.assign({}, orig, { value2: fn(orig.value2), value3: fn(orig.value3)});
Assign to spread individually
const orig = { value1: "Hello", value2: 5, value3: 6 }; const fn = x => x + 1; const { value2, value3, ...rest} = orig; rest.value2 = fn(value2); rest.value3 = fn(value3);

Rendered benchmark preparation results:

Previous results

Test case name	Result
...rest
Spread original
Don't destructure
Object.assign
Assign to spread individually

Ways of returning a derived object 2 (version: 1)