Benchmark: Group by single pass vs multiple

Script Preparation code:

var data = [
  { name: 'john', savings: 900, firm: 'chase', advisor: 'holly' },
  {
    name: 'carol',
    savings: 500,
    checking: 600,
    firm: 'chase',
    advisor: 'mark',
  },
  {
    name: 'sarah',
    savings: 500,
    checking: 300,
    firm: 'chase',
    advisor: 'holly',
  },
  { name: 'toby', savings: 500, checking: 300, firm: 'chase', advisor: 'mark' },
  {
    name: 'casey',
    savings: 500,
    checking: 200,
    firm: 'citiBank',
    advisor: 'cindy',
  },
  { name: 'bruce', checking: 200, firm: 'chase', advisor: 'holly' },
  { name: 'jim', checking: 200, firm: 'chase', advisor: 'holly' },
  { name: 'nich', firm: 'chase', advisor: 'mark' },
];

Tests:

n
var keys = [ ['firm', 'advisors'], ['advisor', 'clients'], ]; var result = data.reduce( (r, o) => { keys .reduce((q, [key, children]) => { let name; ({ [key]: name, ...o } = o); if (!q[name]) { q[name] = { _: [] }; q._.push({ name, [children]: q[name]._ }); } return q[name]; }, r) ._.push(o); return r; }, { _: [] } )._;
b
function group(arr, accumulate) { let accumulator = {}; for (const obj of arr) { accumulator = accumulate(accumulator, obj); } return Object.values(accumulator); } function by(parentProp, childProp) { return function (acc, obj) { const { [parentProp]: name, ...child } = obj; (acc[name] ??= { name, [childProp]: [] })[childProp].push(child); return acc; }; } var result = group(data, by('firm', 'advisors')).map( ({ advisors, ...firm }) => ({ ...firm, advisors: group(advisors, by('advisor', 'clients')), }) );

Rendered benchmark preparation results:

Suite status: <idle, ready to run>

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Related benchmarks:

Group by single pass vs multiple (version: 0)

Comparing performance of: n vs b

Created: 3 years ago by: Guest

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n

b

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