Check if array of objects containg sub-array all have the same values

Question

  for (let material of submission.materials) {
      const { docs } = await getDocs(
        query(collection(db, "materials"), where("name", "==", material.name))
      );
      const materialKpis = docs[0].data().kpis.map((kpi) => {
        return { kpis: kpi.id, value: kpi.value };
      });
      allKpis = [
        ...allKpis,
        {
          material: material.name,
          kpis: materialKpis,
        },
      ];
    }



[{"kpis": [{id: "A4yYU8r9hFxCgEDV77ps", value: 2}], "material": "Test Materiaal"}, 
     {"kpis": [{id: "A4yYU8r9hFxCgEDV77ps", value: 2}, {id: "HhoomonFUYGVv4H4RTj3", value: 3}], 
      "material": "Test Materiaal 2"}
    ]

I have this array of objects called allKpis, containing a nested array of KPI objects with an ID and a value. I want to create a function that would return

   [{"kpis": [{id: "A4yYU8r9hFxCgEDV77ps", value: 2}], "material": "Test Materiaal"}, 
     {"kpis": [{id: "A4yYU8r9hFxCgEDV77ps", value: 2}], 
      "material": "Test Materiaal 2"}
    ]

Since the other KpiId does not exist on all objects. How would i go about doing this?

can you share how can you convert from the original array to the new array? I can't figure out the logic that you want to implement here. Also, What have you tried so far? — Tree Nguyen
– Tree Nguyen, Commented Dec 19, 2022 at 10:40
why is there an inner kpis field {"kpis": [{"kpis": [{id: "A4yYU8r9hFxCgEDV77ps", value: 2}] ` — cmgchess
– cmgchess, Commented Dec 19, 2022 at 10:48

trincot · Accepted Answer · 2022-12-19 11:07:19Z

1

You can:

Iterate allKpis and count how many occurrences you have of each ID
Filter that list of IDs to only keep those that have a count that is equal to the length of allKpis.
Turn this list into a Set.
Map the objects in allKpis and apply a filter to their kpis arrays, requiring that the id is in that set.

Code to execute after your loop:

const ids = new Set(
    Object.entries(
        allKpis.flatMap(({kpis}) => kpis.map(({id}) => id))
               .reduce((acc, id) => (acc[id] = (acc[id] ?? 0) + 1, acc), {})
    ).filter(([id, count]) => count == allKpis.length)
     .map(([id]) => id)
);
 
allKpis = allKpis.map(o => ({
    ...o,
    kpis: o.kpis.filter(({id}) => ids.has(id))
}));

answered Dec 19, 2022 at 11:07

trincot

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1 Comment

Bowis Over a year ago

That did the trick! Now i have to decipher what it does..

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