Develop Reference

Javascript check if any of the values exist on array of objects [duplicate]

I have one array like this one:
array1=[{value:1, label:'value1'},{value:2, label:'value2'}, {value:3, label:'value3'}]
I have a second array of integer :
array2=[1,3]
I would like to obtain this array without a loop for :
arrayResult = ['value1', 'value3']
Does someone know how to do it with javascript ?
Thanks in advance

Map the elements in array2 to the label property of the element in array1 with the corresponding value:
array2 // Take array2 and
.map( // map
function(n) { // each element n in it to
return array1 // the result of taking array1
.find( // and finding
function(e) { // elements
return // for which
e.value // the value property
=== // is the same as
n; // the element from array2
}
)
.label // and taking the label property of that elt
;
}
)
;
Without comments, and in ES6:
array.map(n => array1.find(e => e.value === n).label);

You can use .filter and .map, like this
var array1 = [
{value:1, label:'value1'},{value:2, label:'value2'}, {value:3, label:'value3'}
];
var array2 = [1, 3];
var arrayResult = array1.filter(function (el) {
return array2.indexOf(el.value) >= 0;
}).map(function (el) {
return el.label;
});
console.log(arrayResult);

A simple for-loop should suffice for this. In the future you should seriously post some code to show what you have tried.
var array1=[{value:1, label:'value1'},{value:2, label:'value2'}, {value:3, label:'value3'}];
var array2=[1,3];
var result = [];
for (var i = 0; i < array2.length; i++){
result.push(array1[array2[i]-1].label);
}
console.log(result); //["value1", "value3"]
JSBIN

Good answers all. If I may suggest one more alternative using Map as this seems to be suited to a key:value pair solution.
var arr1 = [ {value:1, label:'value1'}, {value:2, label:'value2'}, {value:3, label:'value3'} ];
var arr2 = [1, 3];
// create a Map of the first array making value the key.
var map = new Map( arr1.map ( a => [a.value, a.label] ) );
// map second array with the values of the matching keys
var result = arr2.map( n => map.get ( n ) );
Of course this supposes that the key:value structure of the first array will not become more complex, and could be written in the simpler form of.
var arr1 = [[1,'value1'], [2,'value2'], [3,'value3']]; // no need for names
var arr2 = [1, 3];
var map = new Map( arr1 ); // no need to arr1.map ;
var result = arr2.map( n => map.get ( n ) );

Just index the first array using the _.indexBy function:
var indexedArray1 = _.indexBy(array1, "value");
_.map(array2, function(x) { return indexedArray1[x].label; });

Checking multiple variables with the same "rules"

I have to construct an array of objects. I can do it "long hand," but I'm hoping to find a way to iterate through some variables and check each at "push" them into the right spot in the array.
I have this:
//this is the starting array...I'm going to update these objects
operationTime = [
{"isActive":false,"timeFrom":null,"timeTill":null},//Monday which is operationTime[0]
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null}
];
//I get the below via an API call
var monHours = placeHours.mon_open_close;
var tueHours = placeHours.tue_open_close;
var wedHours = placeHours.wed_open_close;
var thuHours = placeHours.thu_open_close;
var friHours = placeHours.fri_open_close;
var satHours = placeHours.sat_open_close;
var sunHours = placeHours.sun_open_close;
var sunHours = placeHours.sun_open_close;
//here's where I'm stuck.
if (monHours.length>0){
var arr = monHours[0].split("-");
operationTime[0].isActive= true;
operationTime[0].timeFrom= arr[0];
operationTime[0].timeTill= arr[1];
}
else {
operationTime[0].isActive= false;
}
My if/else works perfectly in the above example using Monday, but I don't want to write this for seven days of the week making it unnecessarily complicated. How could I condense this into a single "function" that'll test each variable and push it into the array object in the correct position?

I guess you can put the keys in an array, and then use forEach loop through operationTime and update the object based on the index:
operationTime = [
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null}
];
// make an array of keys that has the same order of the operationTime
var keys = ['mon_open_close', 'tue_open_close', 'wed_open_close', 'thu_open_close', 'fri_open_close', 'sat_open_close', 'sun_open_close'];
var placeHours = {'mon_open_close': ['08:00-17:00'], 'tue_open_close':[], 'wed_open_close':[], 'thu_open_close':[], 'fri_open_close':[], 'sat_open_close':[], 'sun_open_close':['10:20-15:30']}
operationTime.forEach( (obj, index) => {
var dayHours = placeHours[keys[index]];
if(dayHours.length > 0) {
var arr = dayHours[0].split("-");
obj.isActive= true;
obj.timeFrom= arr[0];
obj.timeTill= arr[1];
}
})
console.log(operationTime);

You can try this way with foreach all days's hour,
$all_hours = [monHours, tueHours , wedHours , thuHours , friHours , satHours ,sunHours];
foreach($all_hours as $k=>$hours){
if ($hours.length>0){
$arr = $hours[k].split("-");
operationTime[$k].isActive= true;
operationTime[$k].timeFrom= $arr[0];
operationTime[$k].timeTill= $arr[1];
}
else {
operationTime[$k].isActive = false;
}
}

You can use Object.entries() to iterate properties and values of an object as an array, .map() to define and include index of iteration in block of for..of or other loop. The index is utilized to reference object at index of operationTime array
for (let
[key, prop, index]
of
Object.entries(placeHours)
.map(([key, prop], index) => [key, prop, index]))) {
if (prop.length > 0 ) {
let [arr] = prop.split("-");
operationTime[index].isActive = true;
operationTime[index].timeFrom = arr[0];
operationTime[index].timeTill = arr[1];
}
}

Find common objects in multiple arrays by object ID

I've searched SO for a way to do this but most questions only support two arrays (I need a solution for multiple arrays).
I don't want to compare exact objects, I want to compare objects by their ID, as their other parameters may differ.
So here's the example data:
data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}, etc.]
data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}, etc.]
data3 = [{'id':'20','name':'steve'},{'id':'22','name':'john'}, etc.]
...
I'd like to return all objects whose ID appears in all arrays, and I don't mind which of the set of matched objects is returned.
So, from the data above, I'd expect to return any one of the following:
{'id':'22','name':'andrew'}
{'id':'22','name':'mary'}
{'id':'22','name':'john'}
Thanks

First, you really need an array of arrays - using a numeric suffix is not extensible:
let data = [ data1, data2, ... ];
Since you've confirmed that the IDs are unique within each sub array, you can simplify the problem by merging the arrays, and then finding out which elements occur n times, where n is the original number of sub arrays:
let flattened = data.reduce((a, b) => a.concat(b), []);
let counts = flattened.reduce(
(map, { id }) => map.set(id, (map.get(id) || 0) + 1), new Map()
);
and then you can pick out those objects that did appear n times, in this simple version they'll all come from the first sub array:
let found = data[0].filter(({ id }) => counts.get(id) === data.length);
Picking an arbitrary (unique) match from each sub array would be somewhat difficult, although picking just one row of data and picking the items from that would be relatively easy. Either would satisfy the constraint from the question.

If you want the unique object by Name
data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'mary'}]
data2 = [{'id':'26','name':'mary'},{'id':'85','name':'bill'}]
data3 = [{'id':'29','name':'sophie'},{'id':'22','name':'john'}]
flattened = [ ...data1, ...data2, ...data3 ];
counts = flattened.reduce(
(map, { name }) => map.set(name, (map.get(name) || 0) + 1), new Map()
);
names = []
found = flattened.filter(({ name }) => {
if ((counts.get(name) > 1) && (!names.includes(name))) {
names.push(name);
return true
}
return false
});

its too many loops but , if u can find the common id which is present in all the arrays then it would make your finding easier i think .you can have one array value as reference to find the common id
var global = [];
for(var i = 0;i<data1.length;i++){
var presence = true;
for(var j=0;j<arrays.length;j++){
var temp = arrays[j].find(function(value){
return data1[i].id == value.id;
});
if(!temp){
presence = false;
break;
}
}
if(presence){
global.push(data1[i].id)
}
}
console.log(global);
var data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}];
var data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}];
var data3 = [{'id':'20','name':'steve'},{'id':'22','name':'john'}];
var arrays = [data1, data2, data3];
var global = [];
for(var i = 0;i<data1.length;i++){
var presence = true;
for(var j=0;j<arrays.length;j++){
var temp = arrays[j].find(function(value){
return data1[i].id == value.id;
});
if(!temp){
presence = false;
break;
}
}
if(presence){
global.push(data1[i].id)
}
}
console.log(global);

There's mention you you need n arrays, but also, given that you can:
put all the arrays into an array called data
you can:
combine your arrays
get a list of duplicated IDs (via sort by ID)
make that list unique (unique list of IDs)
find entries in the combined list that match the unique IDs
where the count of those items match the original number of arrays
Sample code:
// Original data
var data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}]
var data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}]
var data3 = [{'id':'13','name':'steve'},{'id':'22','name':'john'}]
var arraycount = 3;
// Combine data into a single array
// This might be done by .pushing to an array of arrays and then using .length
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort?v=control
var data = [].concat(data1).concat(data2).concat(data3);
//console.log(data)
// Sort array by ID
// http://stackoverflow.com/questions/840781/easiest-way-to-find-duplicate-values-in-a-javascript-array
var sorted_arr = data.slice().sort(function(a, b) {
return a.id - b.id;
});
//console.log(sorted_arr)
// Find duplicate IDs
var duplicate_arr = [];
for (var i = 0; i < data.length - 1; i++) {
if (sorted_arr[i + 1].id == sorted_arr[i].id) {
duplicate_arr.push(sorted_arr[i].id);
}
}
// Find unique IDs
// http://stackoverflow.com/questions/1960473/unique-values-in-an-array
var unique = duplicate_arr.filter(function(value, index, self) {
return self.indexOf(value) === index;
});
//console.log(unique);
// Get values back from data
//https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter?v=control
var matches = [];
for (var i = 0; i < unique.length; ++i) {
var id = unique[i];
matches.push(data.filter(function(e) {
return e.id == id;
}))
}
//console.log(matches)
// for data set this will be 13 and 22
// Where they match all the arrays
var result = matches.filter(function(value, index, self) {
return value.length == arraycount;
})
//console.log("Result:")
console.log(result)
Note: There's very likely to be more efficient methods.. I've left this in the hope part of it might help someone

var arr1 = ["558", "s1", "10"];
var arr2 = ["55", "s1", "103"];
var arr3 = ["55", "s1", "104"];
var arr = [arr1, arr2, arr3];
console.log(arr.reduce((p, c) => p.filter(e => c.includes(e))));
// output ["s1"]

Match two multidimensional array in Javascript

I have two multidimensional array and i want to create a third multidimensional array:
var reports = [
[48.98,153.48],
[12.3,-61.64]
];
var vulc = [
["ciccio",48.98,153.48],
["cicci",12.3,-61.64],
["intruso",59.9,99.9]
];
And i want to create a new multidimensional array
var nuovarray= [];
for (i=0; i<= reports.length; i++) {
var attivi= reports[i];
var attlat= attivi[0];
var attlng= attivi[1];
for (s=0; s<=vulc.length; s++){
var vulca= vulc[s];
var vulcanam= vulca[0];
var vulcalat= vulca[1];
var vulcalng= vulca[2];
if ((vulcalat==attlat) && (vulcalng==attlng){
var stato= "A";
nuovarray.push([vulcanam,vulcalat,vulcalng,stato]);
}
else{
var stato= "N";
nuovaarray.push([vulcanam,vulcalat,vulcalng,stato]);
}
}
}
i would like to have
var nuovarray= [
["ciccio",48.98,153.48,"N"],
["cicci",12.3,-61.64,"N"],
["intruso",59.9,99.9,"A"]
];
But i don't know if this code is good :/

As I said in the comment, in the for loop, use < not <= (array of length N has indexes 0 ... N-1) ... and swap the outer loop with the inner loop, and only push with value 'N' before the end of the outer loop if the inner loop hasn't pushed with value 'A'
var reports = [
[48.98,153.48],
[12.3,-61.64]
];
var vulc = [
["ciccio",48.98,153.48],
["cicci",12.3,-61.64],
["intruso",59.9,99.9]
];
var nuovarray= [];
for(var s = 0; s < vulc.length; s++) {
var vulca = vulc[s];
var stato= "A"; // default, no match
var vulcanam= vulca[0];
var vulcalat= vulca[1];
var vulcalng= vulca[2];
for(var i = 0; i < reports.length; i++) {
var attivi = reports[i];
var attlat= attivi[0];
var attlng= attivi[1];
if ((vulcalat==attlat) && (vulcalng==attlng)) {
stato = "N";
break; // we've found a match, so set stato = N and stop looping
}
}
nuovarray.push([vulcanam,vulcalat,vulcalng,stato]);
}
document.getElementById('result').innerHTML = (nuovarray).toSource();
<div id='result'></div>

I believe the code will not work the way it is written. At least, it will not give you the expected output. You are iterating through the vulc array inside the loop which iterates through reports. And you are pushing to the nuovarray inside the inner loop. So I would expect 6 elements in nuovarray, not the 3 elements you are expecting.
Did you try running it? That's the easiest way to prove incorrectness.

var reports = [
[48.98,153.48],
[12.3,-61.64]
];
var vulc = [
["ciccio",48.98,153.48],
["cicci",12.3,-61.64],
["intruso",59.9,99.9]
];
var nuovarray = [];
vulc.forEach(function(item, indx){
var bN = 'undefined' !== typeof reports[indx];
bN = bN && item[1] == reports[indx][0] && item[2] == reports[indx][1];
item.push(bN ? 'N' : 'A');
nuovarray.push(item);
});
console.log(nuovarray);

The code maps the given vulc to nuovarray and add the wanted flag to it. The flag is selected by a search over reports and if found, an 'N' is applied, otherwise an 'A' is applied.
var reports = [
[48.98, 153.48],
[12.3, -61.64]
],
vulc = [
["ciccio", 48.98, 153.48],
["cicci", 12.3, -61.64],
["intruso", 59.9, 99.9]
],
nuovarray = vulc.map(function (a) {
a.push(reports.some(function (b) {
return a[1] === b[0] && a[2] === b[1];
}) ? 'N' : 'A')
return a;
});
document.getElementById('out').innerHTML = JSON.stringify(nuovarray, null, 4);
<pre id="out"></pre>

The map() method creates a new array with the results of calling a provided function on every element in this array.
Array.prototype.map()
The push() method adds one or more elements to the end of an array and returns the new length of the array.
Array.prototype.push()
The some() method tests whether some element in the array passes the test implemented by the provided function.
Array.prototype.some()
var reports = [
[48.98,153.48],
[12.3,-61.64]
];
var vulc = [
["ciccio",48.98,153.48],
["cicci",12.3,-61.64],
["intruso",59.9,99.9]
];
console.log(vulc.map(function (item, index) {
item.push(reports.some(function (report) {
return report[0] == item[1] && report[1] == item[2];
})?"N":"A");
return item;
}));

If performance matters, you should use something better than O(n^2):
var existingPoints = {};
reports.forEach(function (row) {
existingPoints[row.join()] = true;
});
var nuovarray = vulc.map(function (row) {
var point = row.slice(1, 3).join();
var flag = existingPoints[point] ? 'A' : 'N';
return row.concat([flag]);
});

How can I find matching values in two arrays? [duplicate]

This question already has answers here:
Simplest code for array intersection in javascript
(40 answers)
Closed 3 years ago.
I have two arrays, and I want to be able to compare the two and only return the values that match. For example both arrays have the value cat so that is what will be returned. I haven't found anything like this. What would be the best way to return similarities?
var array1 = ["cat", "sum","fun", "run"];
var array2 = ["bat", "cat","dog","sun", "hut", "gut"];
//if value in array1 is equal to value in array2 then return match: cat

You can use :
const intersection = array1.filter(element => array2.includes(element));

Naturally, my approach was to loop through the first array once and check the index of each value in the second array. If the index is > -1, then push it onto the returned array.
​Array.prototype.diff = function(arr2) {
var ret = [];
for(var i in this) {
if(arr2.indexOf(this[i]) > -1){
ret.push(this[i]);
}
}
return ret;
};
​
My solution doesn't use two loops like others do so it may run a bit faster. If you want to avoid using for..in, you can sort both arrays first to reindex all their values:
Array.prototype.diff = function(arr2) {
var ret = [];
this.sort();
arr2.sort();
for(var i = 0; i < this.length; i += 1) {
if(arr2.indexOf(this[i]) > -1){
ret.push(this[i]);
}
}
return ret;
};
Usage would look like:
var array1 = ["cat", "sum","fun", "run", "hut"];
var array2 = ["bat", "cat","dog","sun", "hut", "gut"];
console.log(array1.diff(array2));
If you have an issue/problem with extending the Array prototype, you could easily change this to a function.
var diff = function(arr, arr2) {
And you'd change anywhere where the func originally said this to arr2.

I found a slight alteration on what #jota3 suggested worked perfectly for me.
var intersections = array1.filter(e => array2.indexOf(e) !== -1);
Hope this helps!

This function runs in O(n log(n) + m log(m)) compared to O(n*m) (as seen in the other solutions with loops/indexOf) which can be useful if you are dealing with lots of values.
However, because neither "a" > 1 nor "a" < 1, this only works for elements of the same type.
function intersect_arrays(a, b) {
var sorted_a = a.concat().sort();
var sorted_b = b.concat().sort();
var common = [];
var a_i = 0;
var b_i = 0;
while (a_i < a.length
&& b_i < b.length)
{
if (sorted_a[a_i] === sorted_b[b_i]) {
common.push(sorted_a[a_i]);
a_i++;
b_i++;
}
else if(sorted_a[a_i] < sorted_b[b_i]) {
a_i++;
}
else {
b_i++;
}
}
return common;
}
Example:
var array1 = ["cat", "sum", "fun", "hut"], //modified for additional match
array2 = ["bat", "cat", "dog", "sun", "hut", "gut"];
intersect_arrays(array1, array2);
>> ["cat", "hut"]

Loop through the second array each time you iterate over an element in the first array, then check for matches.
var array1 = ["cat", "sum", "fun", "run"],
array2 = ["bat", "cat", "dog", "sun", "hut", "gut"];
function getMatch(a, b) {
var matches = [];
for ( var i = 0; i < a.length; i++ ) {
for ( var e = 0; e < b.length; e++ ) {
if ( a[i] === b[e] ) matches.push( a[i] );
}
}
return matches;
}
getMatch(array1, array2); // ["cat"]

var array1 = [1, 2, 3, 4, 5, 6];
var array2 = [1, 2, 3, 4, 5, 6, 7, 8, 9];
var array3 = array2.filter(function(obj) {
return array1.indexOf(obj) !== -1;
});

You can use javascript function .find()
As it says in MDN, it will return the first value that is true. If such an element is found, find immediately returns the value of that element. Otherwise, find returns undefined.
var array1 = ["cat", "sum", "fun", "run", "cat"];
var array2 = ["bat", "cat", "dog", "sun", "hut", "gut"];
found = array1.find((val, index) => {
console.log('index', index) // Stops at 0
return array2.includes(val)
})
console.log(found)
Or use .filter(), which loops through every elements first, then give back the result to you.
var array1 = ["cat", "sum", "fun", "run", "cat"];
var array2 = ["bat", "cat", "dog", "sun", "hut", "gut"];
found = array1.filter((val, index) => {
console.log('index', index) // Stops at array1.length - 1
return array2.includes(val)
})
console.log(found)

use lodash
GLOBAL.utils = require('lodash')
var arr1 = ['first' , 'second'];
var arr2 = ['second '];
var result = utils.difference(arr1 , arr2);
console.log ( "result :" + result );

Libraries like underscore and lodash have a utility method called intersection to find matches in arrays passed in. Take a look at: http://underscorejs.org/#intersection

Done as a answer so I can do formatting...
This is the the process you need to go through. Looping through an array for the specifics.
create an empty array
loop through array1, element by element. {
loop through array2, element by element {
if array1.element == array2.element {
add to your new array
}
}
}

If your values are non-null strings or numbers, you can use an object as a dictionary:
var map = {}, result = [], i;
for (i = 0; i < array1.length; ++i) {
map[array1[i]] = 1;
}
for (i = 0; i < array2.length; ++i) {
if (map[array2[i]] === 1) {
result.push(array2[i]);
// avoid returning a value twice if it appears twice in array 2
map[array2[i]] = 0;
}
}
return result;

With some ES6:
let sortedArray = [];
firstArr.map((first) => {
sortedArray[defaultArray.findIndex(def => def === first)] = first;
});
sortedArray = sortedArray.filter(v => v);
This snippet also sorts the firstArr based on the order of the defaultArray
like:
let firstArr = ['apple', 'kiwi', 'banana'];
let defaultArray = ['kiwi', 'apple', 'pear'];
...
console.log(sortedArray);
// ['kiwi', 'apple'];

Iterate on array1 and find the indexof element present in array2.
var array1 = ["cat", "sum","fun", "run"];
var array2 = ["bat", "cat","sun", "hut", "gut"];
var str='';
for(var i=0;i<array1.length;i++){
if(array2.indexOf(array1[i]) != -1){
str+=array1[i]+' ';
};
}
console.log(str)