由多个属性组成的无点组

Question

I'm struggling a bit with implementing a variant groupBy that would allow grouping by multiple properties in a point-free style. (I'm using typescript & ramda).

I want to group some elements say of type A by properties returned from function getProperties :: A a -> [b] . In imperative paradigm the implementation could look like that:

const getProperties = (item: Item): Array<keyof Item> => [item.x, item.y.z];

const groupByMany = (items: Item[]): Dictionary<Item[]> => {
  let groupped: Dictionary<Item[]> = {};
  for (let item of items) {
    for (let key of getProperties(item)) {
      if (groupped[key]) {
        groupped[key].push(item);
      } else {
        groupped[key] = [item];
      }
    }
  }
}

Example:

const items = [
  { i: 1, x: 'A', y: { z: 'B' } },
  { i: 2, x: 'A' },
  { i: 3, x: 'B', y: { z: 'B' } },
];
const expectedOutput = {
  A: [ { i: 1, ... }, { i: 2, ... }],
  B: [ { i: 1, ... }, { i: 3, ... }],
};

Answer 1

I'll get you started -

 const reduce = (f, init, xs) => xs .reduce (f, init) const upsert = (m, k, v) => m .has (k) ? m .get (k) .push (v) : m .set (k, [ v ]) const groupByMany = (f, xs) => reduce ( (m, x) => ( f (x) .forEach (k => k && upsert (m, k, x)) , m ) , new Map , xs ) const items = [ { i: 1, x: 'A', y: { z: 'B' } } , { i: 2, x: 'A' } , { i: 3, x: 'B', y: { z: 'B' } } ] const result = groupByMany ( item => [ item.x, item.y && item.yz ] , items ) console.log(Object.fromEntries(result.entries())) 

Notice how the last item has a B for .x and .yz so it get's inserted into the B group twice . We change upsert so it will not insert a duplicate value -

const upsert = (m, k, v) =>
  m .has (k)
    ? m .get (k) .includes (v)
      ? m
      : m .get (k) .push (v)
    : m .set (k, [ v ])

Expand the snippet below to see the final result in your own browser -

 const reduce = (f, init, xs) => xs .reduce (f, init) const upsert = (m, k, v) => m .has (k) ? m .get (k) .includes (v) ? m : m .get (k) .push (v) : m .set (k, [ v ]) const groupByMany = (f, xs) => reduce ( (m, x) => ( f (x) .forEach (k => k && upsert (m, k, x)) , m ) , new Map , xs ) const items = [ { i: 1, x: 'A', y: { z: 'B' } } , { i: 2, x: 'A' } , { i: 3, x: 'B', y: { z: 'B' } } ] const result = groupByMany ( item => [ item.x, item.y && item.yz ] , items ) console.log(Object.fromEntries(result.entries())) 

---

A note on SO's peculiar output: SO will not display the same object twice, instead it will give an object a reference, and print that reference where the duplicate object would appear. For example /**id:3**/ in the program's output -

{
  "A": [
    {
      /**id:3**/
      "i": 1,
      "x": "A",
      "y": {
        "z": "B"
      }
    },
    {
      "i": 2,
      "x": "A"
    }
  ],
  "B": [
    /**ref:3**/,
    {
      "i": 3,
      "x": "B",
      "y": {
        "z": "B"
      }
    }
  ]
}

Which matches your expected output -

const expectedOutput = {
  A: [ { i: 1, ... }, { i: 2, ... }],
  B: [ { i: 1, ... }, { i: 3, ... }],
};

It's not point-free like you asked for, but I only said I'd get you started ...

Answer 2

I couldn't tell from the question whether you wanted something that made it easy for you to code point-free or if for some reason you were looking for an actual point-free implementation. If it's the latter, then I'm afraid this will be no help. But it's a fairly simple

 const groupByMany = (fn) => (xs) => xs .reduce ( (a, x) => [...new Set ( fn(x) )] . reduce ( (a, k) => k ? {...a, [k]: [... (a [k] || []), x ] } : a , a ) , {} ) // const getProperties = (item) => [path(['x'], item), path(['y', 'z'], item)] const getProperties = juxt ( [path (['x']), path (['y', 'z']) ] ) const items = [{ i: 1, x: 'A', y: { z: 'B' } }, { i: 2, x: 'A'}, { i: 3, x: 'B', y: { z: 'B' } }] console .log ( groupByMany (getProperties) (items) ) 

 <script src="https://bundle.run/ramda@0.26.1"></script></script> <script>const { juxt, path } = ramda </script> 

Running the keys through [... new Set ( fn(x) ) ] is just a quick way to eliminate duplicates from the array returned by fn (x) . The rest of the function should be pretty clear.