Develop Reference

TypeScript - Take object out of array based on attribute value

My array looks like this:
array = [object {id: 1, value: "itemname"}, object {id: 2, value: "itemname"}, ...]
all my objects have the same attibutes, but with different values.
Is there an easy way I can use a WHERE statement for that array?
Take the object where object.id = var
or do I just need to loop over the entire array and check every item? My array has over a 100 entries, so I wanted to know if there was a more efficient way

Use Array.find:
let array = [
{ id: 1, value: "itemname" },
{ id: 2, value: "itemname" }
];
let item1 = array.find(i => i.id === 1);
Array.find at MDN: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/find

I'd use filter or reduce:
let array = [
{ id: 1, value: "itemname" },
{ id: 2, value: "itemname" }
];
let item1 = array.filter(item => item.id === 1)[0];
let item2 = array.reduce((prev, current) => prev || current.id === 1 ? current : null);
console.log(item1); // Object {id: 1, value: "itemname"}
console.log(item2); // Object {id: 1, value: "itemname"}
(code in playground)
If you care about iterating over the entire array then use some:
let item;
array.some(i => {
if (i.id === 1) {
item = i;
return true;
}
return false;
});
(code in playground)

You can search a certain value in array of objects using TypeScript dynamically if you need to search the value from all fields of the object without specifying column
var searchText = 'first';
let items = [
{ id: 1, name: "first", grade: "A" },
{ id: 2, name: "second", grade: "B" }
];
This below code will search for the value
var result = items.filter(item =>
Object.keys(item).some(k => item[k] != null &&
item[k].toString().toLowerCase()
.includes(searchText.toLowerCase()))
);
Same approach can be used to make a Search Filter Pipe in angularjs 4 using TypeScript

I had to declare the type to get it to work in typescript:
let someId = 1
array.find((i: { id: string; }) => i.id === someId)

You'll have to loop over the array, but if you make a hashmap to link each id to an index and save that, you only have to do it once, so you can reference any objeft after that directly:
var idReference = myArray.reduce(function( map, record, index ) {
map[ record.id ] = index;
return map;
}, {});
var objectWithId5 = myArray[ idReference["5"] ];
This does assume all ids are unique though.

How to combine _.map and _.filter in a more efficient way?

I am using Lodash in my Angular project and I was wondering if there is a better way to write the following code:
$scope.new_arr = _.map(arr1, function(item){
return _.assign(item, {new_id: _.find(arr2, {id: item.id})});
});
$scope.new_arr = _.filter($scope.new_arr, function (item) {
return item.new_id !== undefined;
});
I am trying to combine values from one array to same objects in other array, and I want to ignore the objects that not appear in both arrays (it is something like join or left outer join in the sql language).
Here is a fiddle with an example of this code: Click me!

i think is better to use chaining
$scope.new_arr = _.chain(arr1)
.map(function(item) {
return _.merge(
{}, // to avoid mutations
item,
{new_id: _.find(arr2, {id: item.id})}
);
})
.filter('new_id')
.value();

https://jsfiddle.net/3xjdqsjs/6/
try this:
$scope.getItemById = (array, id) => {
return array.find(item => item.id == id);
};
$scope.mergeArrays = () => {
let items_with_ids = arr1.filter(item => !_.isNil($scope.getItemById(arr2,item.id)));
return items_with_ids.map(item => _.assign(item, {new_id: $scope.getItemById(arr2,item.id)}));
};

The answers provided here are all runtime of O(n^2), because they first run an outer loop on the first array, with an inner loop on the second array. You can instead run this in O(n). First, create a hashmap of all the ids in arr2 in a single loop; this will allow us an order 1 lookup. In the second loop on arr1, check this hashmap to determine if those items exist with O(n). Total Complexity is n + n = 2n, which is just O(n).
// provision some test arrays
var arr1 = [
{
id: 2
},
{
id: 4
},
{
id: 6
}
]
var arr2 = [
{
id: 3
},
{
id: 4
},
{
id: 5
},
{
id: 6
}
]
// First, we create a map of the ids of arr2 with the items. Complexity: O(n)
var mapIdsToArr2Items = _.reduce(arr2, function(accumulator, item) {
accumulator[item.id] = item;
return accumulator;
}, {});
// Next, we use reduce (instead of a _.map followed by a _.filter for slightly more performance.
// This is because with reduce, we loop once, whereas with map and filter,
// we loop twice). Complexity: O(n)
var combinedArr = _.reduce(arr1, function(accumulator, item) {
// Complexity: O(1)
if (mapIdsToArr2Items[item.id]) {
// There's a match/intersection! Arr1's item matches an item in arr 2. Include it
accumulator.push(item);
}
return accumulator;
}, []);
console.log(combinedArr)

You could first make a Map with arr1 and then map the items of arr2 with the properties of arr1.
var arr1 = [{ id: 1, title: 'z' }, { id: 2, title: 'y' }, { id: 3, title: 'x' }, { id: 4, title: 'w' }, { id: 5, title: 'v' }],
arr2 = [{ id: 2, name: 'b' }, { id: 3, name: 'c' }, { id: 4, name: 'd' }, { id: 5, name: 'e' }],
map = new Map(arr1.map(a => [a.id, a])),
result = arr2.map(a => Object.assign({}, a, map.get(a.id)));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Create new array from iterating JSON objects and getting only 1 of its inner array

See jsfiddle here: https://jsfiddle.net/remenyLx/2/
I have data that contains objects that each have an array of images. I want only the first image of each object.
var data1 = [
{
id: 1,
images: [
{ name: '1a' },
{ name: '1b' }
]
},
{
id: 2,
images: [
{ name: '2a' },
{ name: '2b' }
]
},
{
id: 3
},
{
id: 4,
images: []
}
];
var filtered = [];
var b = data1.forEach((element, index, array) => {
if(element.images && element.images.length)
filtered.push(element.images[0].name);
});
console.log(filtered);
The output needs to be flat:
['1a', '2a']
How can I make this prettier?
I'm not too familiar with JS map, reduce and filter and I think those would make my code more sensible; the forEach feels unnecessary.

First you can filter out elements without proper images property and then map it to new array:
const filtered = data1
.filter(e => e.images && e.images.length)
.map(e => e.images[0].name)
To do this in one loop you can use reduce function:
const filtered = data1.reduce((r, e) => {
if (e.images && e.images.length) {
r.push(e.images[0].name)
}
return r
}, [])

You can use reduce() to return this result.
var data1 = [{
id: 1,
images: [{
name: '1a'
}, {
name: '1b'
}]
}, {
id: 2,
images: [{
name: '2a'
}, {
name: '2b'
}]
}, {
id: 3
}, {
id: 4,
images: []
}];
var result = data1.reduce(function(r, e) {
if (e.hasOwnProperty('images') && e.images.length) r.push(e.images[0].name);
return r;
}, [])
console.log(result);

All answers are creating NEW arrays before projecting the final result : (filter and map creates a new array each) so basically it's creating twice.
Another approach is only to yield expected values :
Using iterator functions
function* foo(g)
{
for (let i = 0; i < g.length; i++)
{
if (g[i]['images'] && g[i]["images"].length)
yield g[i]['images'][0]["name"];
}
}
var iterator = foo(data1) ;
var result = iterator.next();
while (!result.done)
{
console.log(result.value)
result = iterator.next();
}
This will not create any additional array and only return the expected values !
However if you must return an array , rather than to do something with the actual values , then use other solutions suggested here.
https://jsfiddle.net/remenyLx/7/

Sort array of objects by arbitrary list in JavaScript

Given an array of objects like this:
objects = [
{ id: 'aaaa', description: 'foo' },
{ id: 'bbbb', description: 'bar' },
{ id: 'cccc', description: 'baz' }
];
And an array of strings like this:
order = [ 'bbbb', 'aaaa', 'cccc' ];
How would I sort the first array so that the id attribute matches the order of the second array?

Try this:
objects.sort(function(a, b){
return order.indexOf(a.id) - order.indexOf(b.id)
});
Assuming the variables are like you declared them in the question, this should return:
[
{ id: 'bbbb', description: 'bar' },
{ id: 'aaaa', description: 'foo' },
{ id: 'cccc', description: 'baz' }
];
(It actually modifies the objects variable)

You need a way to translate the string into the position in the array, i.e. an index-of function for an array.
There is one in newer browsers, but to be backwards compatible you need to add it if it's not there:
if (!Array.prototype.indexOf) {
Array.prototype.indexOf = function(str) {
var i;
for (i = 0; i < this.length; i++) if (this[i] == str) return i;
return -1;
}
}
Now you can sort the array by turning the string into an index:
objects.sort(function(x,y){ return order.indexOf(x.id) - order.indexOf(y.id); });
Demo: http://jsfiddle.net/Guffa/u3CQW/

Use a mapping object for (almost) constant access time:
/* Create a mapping object `orderIndex`:
{
"bbbb": 0,
"aaaa": 1,
"cccc": 2
}
*/
const orderIndex = {}
order.forEach((value, index) => orderIndex[value] = index);
// Sort
objects.sort((a, b) => orderIndex[a.id] - orderIndex[b.id]);
// data
const objects = [
{ id: 'aaaa', description: 'foo' },
{ id: 'bbbb', description: 'bar' },
{ id: 'cccc', description: 'baz' }
];
const order = [ 'bbbb', 'aaaa', 'cccc' ];
/* Create a mapping object `orderIndex`:
{
"bbbb": 0,
"aaaa": 1,
"cccc": 2
}
*/
const orderIndex = {}
order.forEach((value, index) => orderIndex[value] = index);
// Sort
objects.sort((a, b) => orderIndex[a.id] - orderIndex[b.id]);
// Log
console.log('orderIndex:', orderIndex);
console.log('objects:', objects);

How can I get a unique array based on object property using underscore

I have an array of objects and I want to get a new array from it that is unique based only on a single property, is there a simple way to achieve this?
Eg.
[ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ]
Would result in 2 objects with name = bill removed once.

Use the uniq function
var destArray = _.uniq(sourceArray, function(x){
return x.name;
});
or single-line version
var destArray = _.uniq(sourceArray, x => x.name);
From the docs:
Produces a duplicate-free version of the array, using === to test object equality. If you know in advance that the array is sorted, passing true for isSorted will run a much faster algorithm. If you want to compute unique items based on a transformation, pass an iterator function.
In the above example, the function uses the objects name in order to determine uniqueness.

If you prefer to do things yourself without Lodash, and without getting verbose, try this uniq filter with optional uniq by property:
const uniqFilterAccordingToProp = function (prop) {
if (prop)
return (ele, i, arr) => arr.map(ele => ele[prop]).indexOf(ele[prop]) === i
else
return (ele, i, arr) => arr.indexOf(ele) === i
}
Then, use it like this:
const obj = [ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ]
obj.filter(uniqFilterAccordingToProp('abc'))
Or for plain arrays, just omit the parameter, while remembering to invoke:
[1,1,2].filter(uniqFilterAccordingToProp())

If you want to check all the properties then
lodash 4 comes with _.uniqWith(sourceArray, _.isEqual)

A better and quick approach
var table = [
{
a:1,
b:2
},
{
a:2,
b:3
},
{
a:1,
b:4
}
];
let result = [...new Set(table.map(item => item.a))];
document.write(JSON.stringify(result));
Found here

You can use the _.uniqBy function
var array = [ { id: 1, name: 'bob' }, { id: 2, name: 'bill' }, { id: 1, name: 'bill' },{ id: 2, name: 'bill' } ];
var filteredArray = _.uniqBy(array,function(x){ return x.id && x.name;});
console.log(filteredArray)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>
In the above example, filtering is based on the uniqueness of combination of properties id & name.
if you have multiple properties for an object.
then to find unique array of objects based on specific properties, you could follow this method of combining properties inside _.uniqBy() method.

I was looking for a solution which didn't require a library, and put this together, so I thought I'd add it here. It may not be ideal, or working in all situations, but it's doing what I require, so could potentially help someone else:
const uniqueBy = (items, reducer, dupeCheck = [], currentResults = []) => {
if (!items || items.length === 0) return currentResults;
const thisValue = reducer(items[0]);
const resultsToPass = dupeCheck.indexOf(thisValue) === -1 ?
[...currentResults, items[0]] : currentResults;
return uniqueBy(
items.slice(1),
reducer,
[...dupeCheck, thisValue],
resultsToPass,
);
}
const testData = [
{text: 'hello', image: 'yes'},
{text: 'he'},
{text: 'hello'},
{text: 'hell'},
{text: 'hello'},
{text: 'hellop'},
];
const results = uniqueBy(
testData,
item => {
return item.text
},
)
console.dir(results)

In case you need pure JavaScript solution:
var uniqueProperties = {};
var notUniqueArray = [ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ];
for(var object in notUniqueArray){
uniqueProperties[notUniqueArray[object]['name']] = notUniqueArray[object]['id'];
}
var uniqiueArray = [];
for(var uniqueName in uniqueProperties){
uniqiueArray.push(
{id:uniqueProperties[uniqueName],name:uniqueName});
}
//uniqiueArray

unique array by id property with ES6:
arr.filter((a, i) => arr.findIndex(b => b.id === a.id) === i); // unique by id
replace b.id === a.id with the relevant comparison for your case