Develop Reference

Transform dot notation to a tree data form

I have object oriented data in the form:
var alist = [
'foo',
'foo.lol1',
'foo.lol2',
'bar.lol1',
'bar.barbar.kk',
...
]
which I would like to transform into a tree structure, to be able to serve them with a tree component (https://github.com/vinz3872/vuejs-tree in particular). The require form is the following:
var ok = [
{
text: "foo",
state: { expanded: false },
nodes: [
{
id: 1,
path: "foo.lol1",
text: "lol1",
checkable: true,
state: { checked: false },
},
{
id: 2,
path: "foo.lol2",
text: "lol2",
checkable: true,
state: { checked: false },
},
]
},
{
text: "bar",
state: { expanded: false },
nodes: [
{
id: 3,
path: "bar.lol1",
text: "lol1",
checkable: true,
state: { checked: false },
},
]
},
{
text: "bar",
state: { expanded: false },
nodes: [
{
id: 3,
path: "bar.lol1",
text: "lol1",
checkable: true,
state: { checked: false },
},
{
text: "barbar",
state: { expanded: false },
nodes: [
{
id: 4,
path: "bar.barbar.kk",
text: "kk",
checkable: true,
state: { checked: false },
},
]
},
]
}
]
I am aware that I should use recursion and I have tried all relevan posts in stackoverflow, i.e. How to build a JSON tree structure using object dot notation.
My main problem is that I have to somehow preserve the information of the full path to the leaves of the tree. As a newbie in js I lost myself in counters and callback for days without any luck.
I would appreciate your help.
Thank you in advance

Basically you could use forEach then split each string into array and then use reduce on that. Then you build nested object where the keys are current paths and also ad to result array.
var alist = [
'foo',
'foo.lol1',
'foo.lol2',
'bar.lol1',
'bar.barbar.kk',
]
const result = []
const levels = {
result
}
let prev = ''
let id = 1
alist.forEach(str => {
str.split('.').reduce((r, text, i, arr) => {
const path = prev += (prev.length ? '.' : '') + text
if (!r[path]) {
r[path] = {result: []}
const obj = {
id: id++,
text,
}
if (i === 0) {
obj.state = {expanded: false}
} else {
obj.state = {checked: false}
obj.checkable = true
obj.path = path
}
obj.nodes = r[path].result
r.result.push(obj)
}
if (i === arr.length - 1) {
prev = ''
}
return r[path]
}, levels)
})
console.log(result)

I found that it was easiest to do this transformation in two steps. The first converts your input into this format:
{
foo: {
lol1: {},
lol2: {}
},
bar: {
barbar: {
kk: {}
},
lol1: {}
},
}
The second uses just this format to create your desired structure. This has two advantages. First, I have tools lying around that make it easy to create this structure from your input. Second, this structure embeds enough information to create your output, with only one branching construct: whether the value at a path is an empty object or has properties. This makes the generation code relatively simple:
const setPath = ([p, ...ps]) => (v) => (o) =>
p == undefined ? v : Object .assign (
Array .isArray (o) || Number .isInteger (p) ? [] : {},
{...o, [p]: setPath (ps) (v) ((o || {}) [p])}
)
const reformat = (o, path = [], nextId = ((id) => () => String (++ id)) (0)) =>
Object .entries (o) .map (([k, v]) => Object .entries (v) .length > 0
? {text: k, state: {exapanded: false}, nodes: reformat (v, [...path, k], nextId)}
: {id: nextId (), path: [...path, k] .join('.'), text: k, checkable: false, state: {checked: false}}
)
const transform = (pathTokens) =>
reformat (pathTokens
.map (s => s .split ('.'))
.reduce ((a, path) => setPath (path) ({}) (a), {})
)
const alist = ['foo', 'foo.lol1', 'foo.lol2', 'bar.lol1', 'bar.barbar.kk']
console .log (transform (alist))
.as-console-wrapper {max-height: 100% !important; top: 0}
We start with setPath, which takes a path, in a format such as ['bar', 'barbar', 'kk'], the value to set at that path, and an object to shallow clone with this new property along that path. Thus setPath (['foo', 'bar', 'baz']) (42) ({foo: {qux: 7}, corge: 6}) yields {foo: {qux: 7, bar: {baz: 42}}, corge: 6}. (There's a little more in this reusable function to also handle array indices instead of string object paths, but we can't reach that from this input format.)
Then we have reformat, which does the format conversion. It simply builds a different input object based upon whether the input value is an empty object.
Finally, transform maps a splitting function over your input array to get the path structure needed for setPath, folds the results into an initially empty object by setting every path value to an empty object, yielding our intermediate format, which we then pas to reformat.
There is one thing I really don't like here, and that is the nextId function, which is a stateful function. We could just have easily used a generator function, but whatever we do here, we're using state to build this output and that bothers me. If someone has a cleaner suggestion for this, I'd love to hear it.

Joining two arrays using a common property javascript

What I want to do is basically join the information from two arrays via userId. Until then, this solution that I made works only when there is little data, if they are very large arrays, this huge amount of filter is very impractical. Can anyone think of a more efficient solution?
PS: I'm using > 0 ? because sometimes one of the properties is empty.
const data01 = [
{ userId: 0, motorcycles: 'motorcycle01', cars: 'car01' },
{ userId: 1, motorcycles: 'motorcycle02', cars: 'car02' },
{ userId: 2, cars: 'car03' },
{ userId: 3, motorcycles: 'motorcycle04' },
]
items.forEach(
a =>
(
a.motorcylces = data01.filter(b => b.userId === a.userId).length > 0 ? data01.filter(b => b.userId === a.userId)[0].motorcylces : null,
a.cars = data01.filter(b => b.userId === a.userId).length > 0 ? data01.filter(b => b.userId === a.userId)[0].cars : null
)
);
console.log(items)
Expected Output:
[
{
...
motorcycles: 'motorcycle01',
cars: 'cars01'
},
{
...
motorcycles: 'motorcycle01',
cars: 'cars01'
}
]

You can speed up the process by creating a Map of data01, keyed by userId.
And with Object.assign you can copy the properties from a match. This will not create the property if it doesn't exist in the source, so there will be no null assignments (unless the source has an explicit null):
let map = new Map(data01.map(o => [o.userId, o]));
items.forEach(a => Object.assign(a, map.get(a.userId)));
If you are only interested in a selection of properties, then create objects that only have those properties:
let map = new Map(data01.map(o =>
[o.userId, { cars: o.cars, motorcycles: o.motorcycles }]
));
items.forEach(a => Object.assign(a, map.get(a.userId)));
This second solution will always create the specific properties, also when they didn't exist yet. In that case their values will be undefined.

If your arrays are arrays of objects, and you need to consolidate 2+ arrays based on some property in the object, seems like the best thing to do would be to make an intermediate map that has its keys as the userIDs, and then just code something that will non-destructively update the map as you iterate through the arrays.
const data01 = [
{ userId: 0, motocycles: 'motocycle01', cars: 'car01' },
{ userId: 1, motocycles: 'motocycle02', cars: 'car02' },
{ userId: 2, cars: 'car03' },
{ userId: 3, motocycles: 'motocycle04' },
]
const data02 = [
{ userId: 0, dogs: 'doggy', cats: 'car01' },
{ userId: 1, dogs: 'doggo', cats: 'car02' },
{ userId: 2, dogs: 'sheperd' },
{ userId: 3, cats: 'kitty' },
]
function combineArrFromUserId(arr1,arr2){
const idMap= new Map()
data01.forEach(item=>checkAndAdd(item,idMap))
data02.forEach(item=>checkAndAdd(item,idMap))
return idMap.values()
}
function checkAndAdd(item,map){
const current = map.get(item.userId)
if(current){
map.set(item.userId,Object.assign(current,item))
} else {
map.set(item.userId, item)
}
}
console.log(combineArrFromUserId(data01,data02))

Creating nested array from flat array - Data Structure

I need to create a nested array using the path as reference for the children.
E.g: 4.1 is a child of 4, 4.1.1 is a child of 4.1, 4.2 is a child of 4...
I have this flat array, with all the data and paths. How would be the best approach to create a nested array where the children are nested to its parent based on its path.
Input:
const list = [
{
location: 1,
path: '4'
},
{
location: 2,
path: '4.1'
},
{
location: 3,
path: '4.1.1'
},
{
location: 4,
path: '4.1.2'
},
{
location: 5,
path: '4.2'
},
{
location: 6,
path: '4.2.1'
},
{
location: 7,
path: '4.3'
},
{
location: 8,
path: '4.3.1'
}
];
Output:
const list = [
{
location: 1,
path: '4',
children: [
{
location: 2,
path: '4.1',
children: [
{
location: 3,
path: '4.1.1'
},
{
location: 4,
path: '4.1.2'
},
]
},
{
location: 5,
path: '4.2',
children: [
{
location: 6,
path: '4.2.1'
},
]
},
{
location: 7,
path: '4.3',
children: [
{
location: 8,
path: '4.3.1'
}
]
},
]
},
];
The best approach would be something recursive.
Any suggestions for this algorithm?

I was curious if the linked answer from Scott would be able to solve this problem without modification. It does!
import { tree } from './Tree'
import { bind } from './Func'
const parent = (path = "") =>
bind
( (pos = path.lastIndexOf(".")) =>
pos === -1
? null
: path.substr(0, pos)
)
const myTree =
tree // <- make tree
( list // <- array of nodes
, node => parent(node.path) // <- foreign key
, (node, children) => // <- node reconstructor
({ ...node, children: children(node.path) }) // <- primary key
)
console.log(JSON.stringify(myTree, null, 2))
[
{
"location": 1,
"path": "4",
"children": [
{
"location": 2,
"path": "4.1",
"children": [
{
"location": 3,
"path": "4.1.1",
"children": []
},
{
"location": 4,
"path": "4.1.2",
"children": []
}
]
},
{
"location": 5,
"path": "4.2",
"children": [
{
"location": 6,
"path": "4.2.1",
"children": []
}
]
},
{
"location": 7,
"path": "4.3",
"children": [
{
"location": 8,
"path": "4.3.1",
"children": []
}
]
}
]
}
]
The Tree module is shared in this post and here's a peek at the Func module that supplies bind -
// Func.js
const identity = x => x
const bind = (f, ...args) =>
f(...args)
const raise = (msg = "") => // functional throw
{ throw Error(msg) }
// ...
export { identity, bind, raise, ... }
Expand the snippet below to verify the results in your browser -
// Func.js
const bind = (f, ...args) =>
f(...args)
// Index.js
const empty = _ =>
new Map
const update = (r, k, t) =>
r.set(k, t(r.get(k)))
const append = (r, k, v) =>
update(r, k, (all = []) => [...all, v])
const index = (all = [], indexer) =>
all.reduce
( (r, v) => append(r, indexer(v), v)
, empty()
)
// Tree.js
// import { index } from './Index'
function tree (all, indexer, maker, root = null)
{ const cache =
index(all, indexer)
const many = (all = []) =>
all.map(x => one(x))
const one = (single) =>
maker(single, next => many(cache.get(next)))
return many(cache.get(root))
}
// Main.js
// import { tree } from './Tree'
// import { bind } from './Func'
const parent = (path = "") =>
bind
( (pos = path.lastIndexOf(".")) =>
pos === -1
? null
: path.substr(0, pos)
)
const list =
[{location:1,path:'4'},{location:2,path:'4.1'},{location:3,path:'4.1.1'},{location:4,path:'4.1.2'},{location:5,path:'4.2'},{location:6,path:'4.2.1'},{location:7,path:'4.3'},{location:8,path:'4.3.1'}]
const myTree =
tree
( list // <- array of nodes
, node => parent(node.path) // <- foreign key
, (node, children) => // <- node reconstructor
({ ...node, children: children(node.path) }) // <- primary key
)
console.log(JSON.stringify(myTree, null, 2))

One way to do this is to use an intermediate index mapping paths to objects, then folding your list into a structure by looking up each node and its parent in the index. If there is no parent, then we add it to the root object. In the end, we return the children of our root object. Here's some code for that:
const restructure = (list) => {
const index = list .reduce(
(a, {path, ...rest}) => ({...a, [path]: {path, ...rest}}),
{}
)
return list .reduce((root, {path}) => {
const node = index [path]
const parent = index [path .split('.') .slice(0, -1) .join('.')] || root
parent.children = [...(parent.children || []), node]
return root
}, {children: []}) .children
}
const list = [{location: 1, path: '4'}, {location: 2, path: '4.1' }, {location: 3, path: '4.1.1'}, {location: 4, path: '4.1.2'}, {location: 5, path: '4.2'}, {location: 6, path: '4.2.1'}, {location: 7, path: '4.3'}, {location: 8, path: '4.3.1'}]
console.log (restructure (list))
.as-console-wrapper {min-height: 100% !important; top: 0}
Using the index means that we don't have to sort anything; the input can be in any order.
Finding the parent involves replacing, for instance, "4.3.1" with "4.3" and looking that up in the index. And when we try "4", it looks up the empty string, doesn't find it and uses the root node.
If you prefer regex, you could use this slightly shorter line instead:
const parent = index [path.replace (/(^|\.)[^.]+$/, '')] || root
But, you might also want to look at a more elegant technique in a recent answer on a similar question. My answer here, gets the job done (with a bit of ugly mutation) but that answer will teach you a lot about effective software development.

You can first sort the array of objects by path so that the parent will always be before it's children in the sorted array.
eg: '4' will be before '4.1'
Now, you can create an object where the keys are the paths. Let's assume '4' is already inserted in our object.
obj = {
'4': {
"location": 1,
"path": "4",
}
}
When we process '4.1', we first check if '4' is present in our object. If yes, we now go into its children (if the key 'children' isn't present, we create a new empty object) and check if '4.1' is present. If not, we insert '4.1'
obj = {
'4': {
"location": 1,
"path": "4",
"children": {
"4.1": {
"location": 2,
"path": "4.1"
}
}
}
}
We repeat this process for each element in list. Finally, we just have to recursively convert this object into an array of objects.
Final code:
list.sort(function(a, b) {
return a.path - b.path;
})
let obj = {}
list.forEach(x => {
let cur = obj;
for (let i = 0; i < x.path.length; i += 2) {
console.log(x.path.substring(0, i + 1))
if (x.path.substring(0, i + 1) in cur) {
cur = cur[x.path.substring(0, i + 1)]
if (!('children' in cur)) {
cur['children'] = {}
}
cur = cur['children']
} else {
break;
}
}
cur[x.path] = x;
})
function recurse (obj) {
let res = [];
Object.keys(obj).forEach((key) => {
if (obj[key]['children'] !== null && typeof obj[key]['children'] === 'object') {
obj[key]['children'] = recurse(obj[key]['children'])
}
res.push(obj[key])
})
return res;
}
console.log(recurse(obj));

was thinking along the same terms as Aadith but came up from an iterative approach. I think the most performant way to do it is to use a linked list structure and then flatten it though.
const list = [
{
location: 1,
path: '4'
},
{
location: 2,
path: '4.1'
},
{
location: 3,
path: '4.1.1'
},
{
location: 4,
path: '4.1.2'
},
{
location: 5,
path: '4.2'
},
{
location: 6,
path: '4.2.1'
},
{
location: 7,
path: '4.3'
},
{
location: 8,
path: '4.3.1'
}
];
let newList = [];
list.forEach((location) =>
{
console.log('Handling location ',location);
if(location.path.split('.').length==1)
{
location.children = [];
newList.push(location);
}
else
{
newList.forEach(loc => {
console.log('checking out: ',loc);
let found = false;
while(!found)
{
console.log(loc.path,'==',location.path.substring(0, location.path.lastIndexOf('.')));
found = loc.path == location.path.substring(0, location.path.lastIndexOf('.'));
if(!found)
{
for(let i=0;i<loc.children.length;i++)
{
let aloc = loc.children[i];
found = aloc.path == location.path.substring(0, location.path.lastIndexOf('.'));
if(found)
{
console.log('found it...', loc);
location.children = [];
aloc.children.push(location);
break;
}
}
}
else
{
console.log('found it...', loc);
location.children = [];
loc.children.push(location);
}
}
} );
}
});
console.log(newList);
This was my quick and dirty way of how to go about it

Thank you for all the suggestions!
I could definitely see really sophisticated solutions to my problem.
By the end of the day, I've ended up creating my own "dirty" solution.
It is definitely a slower approach, but for my application this list won't be long to the point where i should be too worry about optimization.
I had simplified the flatted list for the purpose of my question. Although, in reality the list was a little more complex then that. I could also pass the path I want to find its children.
This is the solution that worked for me.
function handleNested(list, location) {
return list.map((item) => {
if (item.children && item.children.length > 0) {
item.children = handleNested(item.children, location);
}
if (
location.path.startsWith(item.path) &&
location.path.split(".").length === item.path.split(".").length + 1
) {
return {
...item,
children: item.children ? [...item.children, location] : [location],
};
} else {
return item;
}
});
}
function locationList(path) {
// Filtering the list to remove items with different parent path, and sorting by path depthness.
const filteredList = list
.filter((location) => location.path.startsWith(path))
.sort((a, b) => a.path.length - b.path.length);
let nestedList = [];
// Looping through the filtered list and placing each item under its parent.
for (let i = 0; i < filteredList.length; i++) {
// Using a recursive function to find its parent.
let res = handleNested(nestedList, filteredList[i]);
nestedList = res;
}
return nestedList;
}
locationList("4");

TS/JS - Get "value" from an Array of Objects if a Property of an Object from the Array matches another Property from a separate Object

THE PROBLEM:
I have an array of Objects. And a currentObject, that I currently am viewing. I want to get the value from a Property that comes from the Array of Objects, if 2 other properties match.
Here is the Array, simplified:
ARRAY = [{
id: 1,
metadata: {author: "Company1"}
},
{
id: 2,
metadata: {author: "Company2"}
}
Here is the Object, simplified:
OBJECT = {
name: "Something
templateId: 2
}
So, basically, I want to return, the metdata.author information, if the ARRAY.id, matches the OBJECT.templateId..
Here is the code I wrote..
const getAuthorInfo = (connTemplates: ARRAY[], currentConn: ITEM_OBJECT) => {
connTemplates.find((connTemplate: ARRAY_ITEM_OBJECT) => connTemplate.id === currentConn.templateId);
};
console.log('Author Info:', connection); // This though returns the OBJECT, not the ARRAY_ITEM
Any ideas, on how to make this work? I tried to filter as well, with the same condition, but that returned undefined, when I called it in my ReactComponent.

is this what you need?
const arr = [{
id: 1,
metadata: { author: "Company1" }
},
{
id: 2,
metadata: { author: "Company2" }
}]
const obj = {
name: "Something",
templateId: 2
}
function getAuthorInfo(arr, obj) {
const arrItem = arr.find(item => item.id === obj.templateId)
return arrItem.metadata.author
}
console.log(getAuthorInfo(arr, obj))

You are on the right path:
const result = arr.find(f => f.id == obj.templateId).metadata.author;
const arr = [{
id: 1,
metadata: {author: "Company1"}
},
{
id: 2,
metadata: {author: "Company2"}
}]
const obj = {
name: "Something",
templateId: 2
}
const result = arr.find(f => f.id == obj.templateId);
console.log(result);

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

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] } ] }

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));

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:
Start iteration from the top nodes (from the top of the tree);
Return the node if it was found in the iteration (if a banana is ripe);
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);
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);
Keep iteration if node wasn't found (if banana wasn't found keep testing other branches);
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.

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}

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
)

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 } */

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.

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
}

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

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().

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));

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});

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
}
}
},

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
)
);
}