How do I recursively use Array.prototype.find() while returning a single object?

Question

The bigger problem I am trying to solve is, given this data:

var data = [
    { id: 1 },
    { id: 2 },
    { id: 3 },
    { id: 4, children: [
        { id: 6 },
        { id: 7, children: [
            {id: 8 },
            {id: 9 }
        ]}
    ]},
    { id: 5 }
]

I want to make a function findById(data, id) that returns { id: id }. For example, findById(data, 8) should return { id: 8 }, and findById(data, 4) should return { id: 4, children: [...] }.

To implement this, I used Array.prototype.find recursively, but ran into trouble when the return keeps mashing the objects together. My implementation returns the path to the specific object.

For example, when I used findById(data, 8), it returns the path to { id: 8 }:

 { id: 4, children: [ { id: 6 }, { id: 7, children: [ { id: 8}, { id: 9] } ] }

Instead I would like it to simply return

{ id: 8 }

---

Implementation (Node.js v4.0.0)

jsfiddle

var data = [
    { id: 1 },
    { id: 2 },
    { id: 3 },
    { id: 4, children: [
        { id: 6 },
        { id: 7, children: [
            {id: 8 },
            {id: 9 }
        ]}
    ]},
    { id: 5 }
]

function findById(arr, id) {
    return arr.find(a => {
        if (a.children && a.children.length > 0) {
            return a.id === id ? true : findById(a.children, id)
        } else {
            return a.id === id
        }
    })
    return a
}

console.log(findById(data, 8)) // Should return { id: 8 }

// Instead it returns the "path" block: (to reach 8, you go 4->7->8)
//
// { id: 4,
//   children: [ { id: 6 }, { id: 7, children: [ {id: 8}, {id: 9] } ] }

Answer 1

I would just use a regular loop and recursive style search:

function findById(data, id) {
    for(var i = 0; i < data.length; i++) {
        if (data[i].id === id) {
            return data[i];
        } else if (data[i].children && data[i].children.length && typeof data[i].children === "object") {
            findById(data[i].children, id);
        }
    }
}

//findById(data, 4) => Object {id: 4, children: Array[2]}
//findById(data, 8) => Object {id: 8}

Answer 2

The problem what you have, is the bubbling of the find. If the id is found inside the nested structure, the callback tries to returns the element, which is interpreted as true, the value for the find.

The find method executes the callback function once for each element present in the array until it finds one where callback returns a true value. [MDN]

Instead of find, I would suggest to use a recursive style for the search with a short circuit if found.

var data = [{ id: 1 }, { id: 2 }, { id: 3 }, { id: 4, children: [{ id: 6 }, { id: 7, children: [{ id: 8 }, { id: 9 }] }] }, { id: 5 }];

function findById(data, id) {
    function iter(a) {
        if (a.id === id) {
            result = a;
            return true;
        }
        return Array.isArray(a.children) && a.children.some(iter);
    }

    var result;
    data.some(iter);
    return result
}

console.log(findById(data, 8));

Answer 3

Let's consider the implementation based on recursive calls:

function findById(tree, nodeId) {
  for (let node of tree) {
    if (node.id === nodeId) return node

    if (node.children) {
      let desiredNode = findById(node.children, nodeId)
      if (desiredNode) return desiredNode
    }
  }
  return false
}

Usage

var data = [
  { id: 1 }, { id: 2 }, { id: 3 },
  { id: 4, children: [
      { id: 6 },
      { id: 7,
        children: [
          { id: 8 }, 
          { id: 9 }
        ]}]},
  { id: 5 }
]

findById(data,  7 ) // {id: 7, children: [{id: 8}, {id: 9}]}
findById(data,  5 ) // {id: 5}
findById(data,  9 ) // {id: 9}
findById(data,  11) // false

To simplify the picture ?, imagine that:

• you are the monkey sitting on the top of a palm tree ?;
• and searching for a ripe banana, going down the tree
• you are in the end and searches aren't satisfied you;
• come back to the top of the tree and start again from the next branch;
• if you tried all bananas on the tree and no one is satisfied you, you just assert that ripe bananas don't grow on this this palm;
• but if the banana was found you come back to the top and get pleasure of eating it.

Now let's try apply it to our recursive algorithm ?:

1. Start iteration from the top nodes (from the top of the tree);
2. Return the node if it was found in the iteration (if a banana is ripe);
3. Go deep until item is found or there will be nothing to deep. Hold the result of searches to the variable (hold the result of searches whether it is banana or just nothing and come back to the top);
4. Return the searches result variable if it contains the desired node (eat the banana if it is your find, otherwise just remember not to come back down by this branch);
5. Keep iteration if node wasn't found (if banana wasn't found keep testing other branches);
6. Return false if after all iterations the desired node wasn't found (assert that ripe bananas doesn't grow on this tree).

Keep learning recursion it seems not easy at the first time ? but this technique allows you to solve daily issues in elegant way ?.

Answer 4

I know this is an old question, but as another answer recently revived it, I'll another version into the mix.

I would separate out the tree traversal and testing from the actual predicate that we want to test with. I believe that this makes for much cleaner code.

A reduce-based solution could look like this:

const nestedFind = (pred) => (xs) =>
  xs .reduce (
    (res, x) => res ? res : pred(x) ? x : nestedFind (pred) (x.children || []), 
    undefined
  )
 
const findById = (testId) => 
  nestedFind (({id}) => id == testId)


const data = [{id: 1}, {id: 2}, {id: 3}, {id: 4, children: [{id: 6}, {id: 7, children: [{id: 8}, {id: 9}]}]}, {id: 5}]

console .log (findById (8) (data))
console .log (findById (4) (data))
console .log (findById (42) (data))

.as-console-wrapper {min-height: 100% !important; top: 0}

There are ways we could replace that reduce with an iteration on our main list. Something like this would do the same:

const nestedFind = (pred) => ([x = undefined, ...xs]) =>
  x == undefined
    ? undefined
  : pred (x)
    ? x
  : nestedFind (pred) (x.children || []) || nestedFind (pred) (xs)

And we could make that tail-recursive without much effort.

While we could fold the two functions into one in either of these, and achieve shorter code, I think the flexibility offered by nestedFind will make other similar problems easier. However, if you're interested, the first one might look like this:

const findById = (id) => (xs) =>
  xs .reduce (
    (res, x) => res ? res : x.id === id ? x : findById (id) (x.children || []), 
    undefined
  )

Answer 5

const data = [
  { id: 1 },
  { id: 2 },
  { id: 3 },
  {
    id: 4,
    children: [{ id: 6 }, { id: 7, children: [{ id: 8 }, { id: 9 }] }]
  },
  { id: 5 }
];

// use Array.flatMap() and Optional chaining to find children
// then Filter undefined results

const findById = (id) => (arr) => {
  if (!arr.length) return null;

  return (
    arr.find((obj) => obj.id === id) ||
    findById(id)(arr.flatMap((el) => el?.children).filter(Boolean))
  );
};

const findId = (id) => findById(id)(data);

console.log(findId(12)); /* null */
console.log(findId(8)); /* { id: 8 } */

Answer 6

Based on Purkhalo Alex solution,

I have made a modification to his function to be able to find the ID recursively based on a given dynamic property and returning whether the value you want to find or an array of indexes to recursively reach to the object or property afterwards.

This is like find and findIndex together through arrays of objects with nested arrays of objects in a given property.

findByIdRecursive(tree, nodeId, prop = '', byIndex = false, arr = []) {
    for (let [index, node] of tree.entries()) {
        if (node.id === nodeId) return byIndex ? [...arr, index] : node;

        if (prop.length && node[prop].length) {
            let found = this.findByIdRecursive(node[prop], nodeId, prop, byIndex, [
                ...arr,
                index
            ]);
            if (found) return found;
        }
    }
    return false;
}

Now you can control the property and the type of finding and get the proper result.

Answer 7

This can be solved with reduce.

const foundItem = data.reduce(findById(8), null)
function findById (id) {
  const searchFunc = (found, item) => {
    const children = item.children || []
    return found || (item.id === id ? item : children.reduce(searchFunc, null))
  }
  return searchFunc
}

Answer 8

You can recursively use Array.prototype.find() in combination with Array.prototype.flatMap()

const findById = (a, id, p = "children", u) => 
  a.length ? a.find(o => o.id === id) || findById(a.flatMap(o => o[p] || []), id) : u; 


const tree = [{id:1}, {id:2}, {id:3}, {id:4, children:[{id: 6}, {id:7, children:[{id:8}, {id:9}]}]}, {id:5}];
console.log(findById(tree, 9));  // {id:9}
console.log(findById(tree, 10)); // undefined

Answer 9

If one wanted to use Array.prototype.find this is the option I chose:

    findById( my_big_array, id ) {
        var result;

        function recursiveFind( haystack_array, needle_id ) {
            return haystack_array.find( element => {
                if ( !Array.isArray( element ) ) {
                    if( element.id === needle_id ) {
                        result = element;
                        return true;
                    }
                } else {
                    return recursiveFind( element, needle_id );
                }
            } );
        }
        recursiveFind( my_big_array, id );

        return result;
    }

You need the result variable, because without it, the function would return the top level element in the array that contains the result, instead of a reference to the deeply nested object containing the matching id, meaning you would need to then filter it out further.

Upon looking through the other answers, my approach seems very similar to Nina Scholz's but instead uses find() instead of some().

Answer 10

Here is a solution that is not the shortest, but divides the problem into recursive iteration and finding an item in an iterable (not necessarily an array).

You could define two generic functions:

• deepIterator: a generator that traverses a forest in pre-order fashion
• iFind: a finder, like Array#find, but that works on an iterable

function * deepIterator(iterable, children="children") {
    if (!iterable?.[Symbol.iterator]) return;
    for (let item of iterable) {
        yield item;
        yield * deepIterator(item?.[children], children);
    }
}

function iFind(iterator, callback, thisArg) {
    for (let item of iterator) if (callback.call(thisArg, item)) return item;
}
  
// Demo
var data = [{ id: 1 }, { id: 2 }, { id: 3 }, { id: 4, children: [{ id: 6 }, { id: 7, children: [{ id: 8 }, { id: 9 }] }] }, { id: 5 }];

console.log(iFind(deepIterator(data), ({id}) => id === 8));

Answer 11

In my opinion, if you want to search recursively by id, it is better to use an algorithm like this one:

function findById(data, id, prop = 'children', defaultValue = null) {
  for (const item of data) {
    if (item.id === id) {
      return item;
    }

    if (Array.isArray(item[prop]) && item[prop].length) {
      const element = this.findById(item[prop], id, prop, defaultValue);

      if (element) {
        return element;
      }
    }
  }

  return defaultValue;
}

findById(data, 2);

But I strongly suggest using a more flexible function, which can search by any key-value pair/pairs:

function findRecursive(data, keyvalues, prop = 'children', defaultValue = null, _keys = null) {
  const keys = _keys || Object.keys(keyvalues);

  for (const item of data) {
    if (keys.every(key => item[key] === keyvalues[key])) {
      return item;
    }

    if (Array.isArray(item[prop]) && item[prop].length) {
      const element = this.findRecursive(item[prop], keyvalues, prop, defaultValue, keys);

      if (element) {
        return element;
      }
    }
  }

  return defaultValue;
}

findRecursive(data, {id: 2});

Answer 12

you can use this function:

If it finds the item so the item returns. But if it doesn't find the item, tries to find the item in sublist.

list: the main/root list

keyName: the key that you need to find the result up to it for example 'id'

keyValue: the value that must be searched

subListName: the name of 'child' array

callback: your callback function which you want to execute when item is found

function recursiveSearch(
      list,
      keyName = 'id',
      keyValue,
      subListName = 'children',
      callback
    ) {
      for (let i = 0; i < list.length; i++) {
        const x = list[i]
        if (x[keyName] === keyValue) {
          if (callback) {
            callback(list, keyName, keyValue, subListName, i)
          }
          return x
        }
        if (x[subListName] && x[subListName].length > 0) {
          const item = this.recursiveSearch(
            x[subListName],
            keyName,
            keyValue,
            subListName,
            callback
          )
          if (!item) continue
          return item
        }
      }
    },

Answer 13

Roko C. Buljan's solution, but more readable one:

function findById(data, id, prop = 'children', defaultValue = null) {
  if (!data.length) {
    return defaultValue;
  }

  return (
    data.find(el => el.id === id) ||
    findById(
      data.flatMap(el => el[prop] || []),
      id
    )
  );
}