按公共属性过滤对象数组

Question

Is it possible in some way to filter let's say such an array of object arrays:

[[{id: 1}, {id: 2}, {id: 3}], [{id:6}, {id: 2}], [{id: 2}, {id: 1}, {id: 9}, {id: 3}]]

To get array of objects which all arrays have the same property (id), so in this case it output will be:

[{id: 2}] // becasue this id is the same in all three subarrays

I've only try intersectionBy from loadash but it seems to work in completely other way :/

Answer 1

I would take one array (it's enough to take one because if the property is not there its already not common), in this example I'm taking the first one but probably more efficient will be picking the shortest one.

iterate over the array and check for each object if its common to all other arrays.

 const arr = [[{id: 1}, {id: 2}, {id: 3}], [{id:6}, {id: 2}], [{id: 2}, {id: 1}, {id: 9}, {id: 3}]]; let firstArray = arr.shift(); const result = firstArray.reduce((common, item)=>{ if (arr.every(inner => inner.some(_item => _item.id === item.id))) { common.push(item); } return common; },[]) console.log(result);

Answer 2

Using Ramda :

const input = [[{id: 1}, {id: 2}, {id: 3}], [{id:6}, {id: 2}], [{id: 2}, {id: 1}, {id: 9}, {id: 3}]];
R.intersection(...input);

Answer 3

You can use array reduce , forEach , findIndex and sort to get the most common object. In first inside the reduce callback use forEach to iterate each of the child array and then use findIndex to find if in accumulator array , there exist an object with same id. If it does not exist create a new object with key id & occurrence . If it exist then increase the value of occurrence. This will give the most common id , even if an id is not present in few child array

 let data = [ [{id: 1}, {id: 2}, { id: 3}], [{id: 6}, {id: 2}], [{id: 2}, {id: 1}, {id: 9}, { id: 3}] ]; let obj = data.reduce((acc, curr) => { curr.forEach((item) => { let getInd = acc.findIndex((elem) => { return elem.id === item.id }); if (getInd === -1) { acc.push({ id: item.id, occurence: 1 }) } else { acc[getInd].occurence += 1; } }) return acc; }, []).sort((a, b) => { return b.occurence - a.occurence; }); console.log(obj[0])

Answer 4

 var arr = [
   [{id: 1}, {id: 2}, {id: 3}], 
   [{id:6}, {id: 2}], 
   [{id: 2}, {id: 1}, {id: 9}, {id: 3}]
 ]
 var obj = {};
 var arrLength = arr.length;

 arr.forEach((val,index) => {
  val.forEach((item) =>{
    if(index == 0){
        if(!obj.hasOwnProperty(item.id)){
            obj[item.id] = 1;
        }
    }else{
        if(obj.hasOwnProperty(item.id)){
            obj[item.id] = obj[item.id] + 1;
        }else{
           return;
        }
      }
   });
});

var output = [];

for (const property in obj) {
 if(obj[property] == arrLength){
   output.push({
      id: property
   })
 }
}

console.log(output);

Answer 5

My approach is similar to that of naortor, but with an attempt to be more generic.

 const intersection = (pred) => (as, bs) => as .filter (a => bs .some (b => pred (a, b))) const intersectionAll = (pred) => (xs) => xs.length ? xs .reduce (intersection (pred)) : [] const input = [[{id: 1}, {id: 2}, {id: 3}], [{id:6}, {id: 2}], [{id: 2}, {id: 1}, {id: 9}, {id: 3}]] const eqIds = (a, b) => a .id == b .id console .log ( intersectionAll (eqIds) (input) )

 .as-console-wrapper {min-height: 100% !important}

This version requires you to say how you identify two equal values. (We will check that they have the same id, but any binary predicate function is allowed.) This function is passed to intersection which returns a function that takes two arrays and finds all the element in common between those two. intersectionAll wraps this behavior up, folding intersection over an array of arrays.

This breakdown is useful, as intersection is a useful function on its own too. And abstracting out the id check into a function you need to supply means these functions are much more generic.