Develop Reference

Sort by Capacity and Name in ReactJS using array sort()

I have one question. How can I sort array items first by their capacity and then by the first letter of their name? The main idea is first to be sorted by capacity from low to high or high to low and then if two items have the same capacity, those two items to be sorted by the first letter of their name.
Here is a basic code sample from me
myArray.sort((eventA, eventB) => {
if (this.state.sort.highToLow.enabled) {
return eventA.capacity > eventB.capacity ? -1 : 1;
} else if (this.state.sort.lowToHigh.enabled) {
return eventA.capacity > eventB.capacity ? 1 : -1;
} else { return 0; }
})

you can use lodash sortBy method
_.sortBy(myArray, ['capacity', 'name']);

try some thing like this. sort method, check for name when capacity is same.
const data = [
{
name: "z",
capacity: 2,
},
{
name: "b",
capacity: 1,
},
{
name: "a",
capacity: 1,
},
];
data.sort((a, b) => {
if (a.capacity === b.capacity) {
return a.name > b.name ? 1 : -1;
}
return a.capacity - b.capacity;
});
console.log(data);

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
)
);
}

Array of objects

I have an array of objects, an example of a few of them being
[
{
"lang_code": "eng",
"site_language": "1",
"name": "English"
},
{
"lang_code": "afr",
"site_language": "1",
"name": "Afrikaans"
},
{
"lang_code": "ale",
"site_language": "0",
"name": "Aleut"
},
]
I want to be able to search the whole array for a specific lang_code, let's say I use eng. I want to search the whole array for eng. If it's there, I want it to return English, if not, I want it to return invalid. Any ideas on this?

A generic and more flexible version of the findById function above:
// array = [{key:value},{key:value}]
function objectFindByKey(array, key, value) {
for (var i = 0; i < array.length; i++) {
if (array[i][key] === value) {
return array[i];
}
}
return null;
}
var array = your array;
var result_obj = objectFindByKey(array, 'lang_code', 'eng');
Click here for demo

You could use filter:
function findLang(arr, code) {
var filtered = arr.filter(function (el) {
return el.lang_code === code;
});
// filter returns an array of objects that match the criteria
// if the array is not empty return the language,
// otherwise return 'invalid'
return filtered.length > 0 ? filtered[0].name : 'invalid';
}
findLang(arr, 'eng'); // English
DEMO
If you wanted to add map into the mix instead of using that ternary operation (but which would most likely be slower and doesn't really provide any additional benefit):
function findLang(arr, code) {
return arr.filter(function (el) {
return el.lang_code === code;
}).map(function (el) {
return el.name;
})[0] || 'invalid';
}
DEMO

How about a for loop. Something like:
function find_lang(input_arr, lang_code) {
for(var i = 0; i < input_arr.length; i++) {
var o = input_arr[i];
if( o.lang_code === lang_code ) {
return o.name;
}
}
return "invalid";
}

Underscore.js has some useful helpers for this kind of thing: http://underscorejs.org/
E.g Find:
Looks through each value in the list, returning the first one that
passes a truth test (predicate), or undefined if no value passes the
test. The function returns as soon as it finds an acceptable element,
and doesn't traverse the entire list.
var even = _.find([1, 2, 3, 4, 5, 6], function(num){ return num % 2 == 0; });
=> 2
Where:
Looks through each value in the list, returning an array of all the
values that contain all of the key-value pairs listed in properties.
_.where(listOfPlays, {author: "Shakespeare", year: 1611});
=> [{title: "Cymbeline", author: "Shakespeare", year: 1611},
{title: "The Tempest", author: "Shakespeare", year: 1611}]

You could do it this way:
// configure your parameters
function checkArr(){
for (var i = 0; i < x.length; i++){
if (x[i].lang_code == "eng")
return "English";
}
return "invalid";
}
var x = [
{
"lang_code": "eng",
"site_language": "1",
"name": "English"
},
{
"lang_code": "afr",
"site_language": "1",
"name": "Afrikaans"
},
{
"lang_code": "ale",
"site_language": "0",
"name": "Aleut"
}
];

Merge two arrays with objects

I plan to merge two objects:
var c = {
name: "doo",
arr: [
{
id: 1,
ver: 1
},
{
id: 3,
ver: 3
}
]
};
var b = {
name: "moo",
arr: [
{
id: 1,
ver: 0
},
{
id: 2,
ver: 0
}
]
};
When using Object.assign({},b,c) what happens is, that the b.arr is simply being replaced with c.arr.
My question is, how do I preserve objects inside the b.arr that are not in c.arr but still merge objects from that array when they match b.arr[0].id === c.arr[0].id. The desired outcome would look like:
{
name: "doo",
arr: [
{
id: 1,
ver: 1
},
{
id: 2,
ver: 0
},
{
id: 3,
ver: 3
}
]
}
Thanks.

You could have a look at ArrayUtils.addAll() from the apache commons

As soon as you use lodash - you may use a combination of lodash's functions. It may look a bit complex but it's not:
_.assign({}, b, c, function(objectValue, sourceValue, key, object, source) {
//merging array - custom logic
if (_.isArray(sourceValue)) {
//if the property isn't set yet - copy sourceValue
if (typeof objectValue == 'undefined') {
return sourceValue.slice();
} else if (_.isArray(objectValue)) {
//if array already exists - merge 2 arrays
_.forEach(sourceValue, function(sourceArrayItem) {
//find object with the same ID's
var objectArrayItem = _.find(objectValue, {id: sourceArrayItem.id});
if (objectArrayItem) {
//merge objects
_.assign(objectArrayItem, sourceArrayItem);
} else {
objectValue.push(sourceArrayItem);
}
});
return objectValue;
}
}
//if sourceValue isn't array - simply use it
return sourceValue;
});
See the full demo here.

Try this function:
function mergeArrayObjects (a, b) {
var tmp, // Temporary array that will be returned
// Cache values
i = 0,
max = 0;
// Check if a is an array
if ( typeof a !== 'object' || typeof a.indexOf === 'undefined')
return false;
// Check if b is an array
if ( typeof b !== 'object' || typeof b.indexOf === 'undefined')
return false;
// Populate tmp with a
tmp = a;
// For each item in b, check if a already has it. If not, add it.
for (i = 0, max = b.length; i < max; i++) {
if (tmp.indexOf(b[i]) === -1)
tmp.push(b[i]);
}
// Return the array
return tmp;
}
JsFiddle here
Note: Because I'm anal, I decided to see if this function is faster than the alternative proposed. It is.

Using lodash, I would do something like this:
var first = {
name: 'doo',
arr: [
{ id: 1, ver: 1 },
{ id: 3, ver: 3 }
]
};
var second = {
name: 'moo',
arr: [
{ id: 1, ver: 0 },
{ id: 2, ver: 0 }
]
};
_.merge(first, second, function(a, b) {
if (_.isArray(a)) {
return _.uniq(_.union(a, b), 'id');
} else {
return a;
}
});
// →
// {
// name: 'doo',
// arr: [
// { id: 1, ver: 1 },
// { id: 2, ver: 0 },
// { id: 3, ver: 3 }
// ]
// }
The merge() function let's you specify a customizer callback for things like arrays. So we just need to check it it's an array we're dealing with, and if so, use the uniq() and union() functions to find the unique values by the id property.

jQuery using.Map to build Alphabetical array

Hey Im doing something like this:
temp = $.map(response, function(item) {
return {
label: item.name,
value: item.id
};
});
temp.unshift({label: "", value: 0});temp
Is there a way to make sure the return is built in alphabetical order?

You can always sort the response prior to creating the map or simply sort the resulting mapped array.
var response = [{ name: 'z', id: 1 }, { name: 'a', id: 2 }];
response = response.sort(function (a, b) {
return a.name < b.name? -1 : +(a.name > b.name);
});
response[0].name; //a
Please note that +(a.name > b.name) is used to explicitely return 0 when equal and 1 when greater, since that's what the sort function expects.

temp.sort(function(a, b) {
return a.value > b.value ? 1 : -1;
});
var result = temp.map(function(e){
return list[e.index]
});