Given the array const vals = [1, 2, 3, 4, 5, 6, 7, 8, 9];
How can I filter and return a new array of indexed key/value pair objects for example:
const vals = [1, 2, 3, 4, 5, 6, 7, 8, 9];
// My fail attempt using filter()
let obj = vals.filter((n, i) => {
return new Object({ i: n % 2 });
});
return obj;
// expected result [{1:2}, {3:4}, {5:6}, {7:8}]
I need to keep the index values as I will filter 2 different arrays with different criteria and associated them later.
Update
Second attempt using map() as suggested in the comments
let obj = vals.map((n, i) => {
if (n % 2) {
return { [i]: n };
}
});
Gives me the following:
[{0:1}, undefined, {2:3}, undefined, {4:5}, undefined, {6:7}, undefined, {8:9}]
To get a list of { key: value } objects where key is the index, and the values are only even without the odd values, you can do this:
const vals = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const result = vals.map((v, i) => [i, v])
.filter(([_, v]) => v % 2 == 0)
.map(([i, v]) => ({ [i]: v }));
console.log(result);
With the first map, you make a list of [[0, 1], ...] pairs to save the index for later.
Then you filter your index-value pairs so only even values remain.
Then you pack those pairs into an object in another map.
This can be done more efficiently with a single iteration using reduce:
const vals = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const result = vals.reduce((a, v, i) => {
if (v % 2 == 0) {
a.push({ [i]: v });
}
return a;
}, []);
console.log(result);
Youn can try simple for loop or the reduce function
let arr = [];
for(let i = 0; i<vals.length-1;i += 2)
{
let obj={};
obj[vals[i]]=vals[i+1];
arr.push(obj);
};
i have the following object
let filters = {
first_ids: [1,2,3],
second_ids: [2,4,9]
}
Now, i want to loop over them and then return something like the following.
filters[first_ids][]=1&filters[first_ids][]=2&filters[first_ids][]=3&filters[second_ids][]=2&filters[second_ids][]=4&filters[second_ids][]=9&
What i tried is following, but i know this is not a elegant way.
let filters = {
first_ids: [1,2,3],
second_ids: [2,4,9]
}
let finalFilters = []
let firstFilters = filters.first_ids.map((e) => `filters[first_ids][]=${e}&`)
let secondFilters = filters.second_ids.map((e) => `filters[second_ids][]=${e}&`)
finalFilters.push(firstFilters)
finalFilters.push(secondFilters)
finalFilters.toString().replace(/,/g, "")
Also the filters object can have more keys.
Something like this might work:
let filters = {
first_ids: [1, 2, 3],
second_ids: [2, 4, 9],
};
const output = Object.entries(filters).flatMap(([name, ids]) => {
return ids.map(id => {
return `filters[${encodeURIComponent(name)}]=${encodeURIComponent(id)}`;
});
}).join('&');
console.log(output)
The main idea is to look though each key, value pair in the object fitlers, then loop though all the ids, and construct the query string fragment, and finally join it on &.
Take a look at Object.entries, Array.prototype.flatMap and encodeURIComponent
You could take a nested approach for generating the string.
const
filters = { first_ids: [1, 2, 3], second_ids: [2, 4, 9] },
result = Object
.entries(filters)
.flatMap(([k, a]) => a.map(v => `filters[${k}][]=${v}`))
.join('&');
console.log(result);
reduce into an array first and then use join.
let filters = {
first_ids: [1, 2, 3],
second_ids: [2, 4, 9],
};
const res = Object.entries(filters)
.reduce((r, [k, v]) => (v.forEach((n) => r.push(`filters[${k}][]=${n}`)), r), [])
.join("&");
console.log(res)
Here's some generic solution:
function toPhpQueryString(o, key) {
if (o === null || o === undefined)
return '';
if (typeof o !== 'object')
return key + '=' + encodeURIComponent(o);
if (Array.isArray(o))
return o.map(v =>
toPhpQueryString(v, key + '[]')).join('&');
return Object.entries(o).map(p =>
toPhpQueryString(p[1], key + '[' + p[0] + ']')).join('&');
}
//
let filters = {
first_ids: [1, 2, 3],
second_ids: [2, 4, 9],
deep: {
object: {
abc: [33, 44],
xyz: 'hey',
}
}
}
qs = toPhpQueryString(filters, 'filters');
console.log(qs)
What is the concisest way to create an object from a list of keys, all set to the same value. For example,
const keys = [1, 2, 3, 4]
const value = 0
What is the tersest way to attain the object
{
“1”: 0,
“2”: 0,
“3”: 0,
“4”: 0
}
You can use Object.fromEntries
const keys = [1, 2, 3, 4]
const value = 0
const result = Object.fromEntries(keys.map(k => [k, value]))
console.log(result)
Should probably be something among:
const keys = [1, 2, 3 ,4];
const value = 0;
console.log(
keys.reduce((acc, key) => (acc[key] = value, acc), {})
);
The simplest way I can think of would be to use .reduce();
const keys = [1, 2, 3, 4]
const value = 0
const obj = keys.reduce((carry, item) => {
carry[item] = value;
return carry;
}, {});
console.log(obj);
I have an array of objects and I want to find the sum of length of arrays of a certain property(key).
I have this array of objects like
var myArray =
[{
"a" : 1,
"b" : another Array
},
{
"c" : 2,
"b" : another Array
}
.....
]
Is there any way to simplify this below process?
var lengthOfEachObject = myArray.map(function(Obj){
if(Obj.b){
return Obj.b.length;
}
else{
return 0;
}
});
lengthofEachObject.reduce(function(x,y){
return x+y;
})
Answers can also include use of external libraries.
You can use .reduce without .map, this way you can get the total sum by only iterating once over your array. Furthermore, you can use destructing assignment instead of your if statements to set a default value for your b.length if it doesn't exist.
See working example below:
const arr = [{a: 1, b: [1, 2, 3, 4, 5] }, {c: 2, b: [1, 2, 3]}, {e: 3}],
total = arr.reduce((acc, {b={length:0}}) => acc + b.length, 0);
console.log(total);
You can use lodash's _.sumBy():
var myArray = [{"a":1,"b":[1,2,3]},{"c":2,"b":[4,5,6,7,8]},{"c":2}]
var result = _.sumBy(myArray, 'b.length')
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
You could pull out the length with a closure over the wanted key and map this value.
const
length = k => ({ [k]: { length = 0 } = {} }) => length,
add = (a, b) => a + b,
array = [{ a: 1, b: [1, 2, 3, 4] }, { c: 2, b: [1, 2, 3] }];
console.log(array.map(length('b')).reduce(add));
console.log(array.map(length('a')).reduce(add));
With Jasmine is there a way to test if 2 arrays contain the same elements, but are not necessarily in the same order? ie
array1 = [1,2,3];
array2 = [3,2,1];
expect(array1).toEqualIgnoreOrder(array2);//should be true
Edit
Jasmine 2.8 adds arrayWithExactContents that will succeed if the actual value is an Array that contains all of the elements in the sample in any order.
See keksmasta's answer
Original (outdated) answer
If it's just integers or other primitive values, you can sort() them before comparing.
expect(array1.sort()).toEqual(array2.sort());
If its objects, combine it with the map() function to extract an identifier that will be compared
array1 = [{id:1}, {id:2}, {id:3}];
array2 = [{id:3}, {id:2}, {id:1}];
expect(array1.map(a => a.id).sort()).toEqual(array2.map(a => a.id).sort());
You could use expect.arrayContaining(array) from standard jest:
const expected = ['Alice', 'Bob'];
it('matches even if received contains additional elements', () => {
expect(['Alice', 'Bob', 'Eve']).toEqual(expect.arrayContaining(expected));
});
jasmine version 2.8 and later has
jasmine.arrayWithExactContents()
Which expects that an array contains exactly the elements listed, in any order.
array1 = [1,2,3];
array2 = [3,2,1];
expect(array1).toEqual(jasmine.arrayWithExactContents(array2))
See https://jasmine.github.io/api/3.4/jasmine.html
The jest-extended package provides us few assertions to simplify our tests, it's less verbose and for failing tests the error is more explicit.
For this case we could use toIncludeSameMembers
expect([{foo: "bar"}, {baz: "qux"}]).toIncludeSameMembers([{baz: "qux"}, {foo: "bar"}]);
simple...
array1 = [1,2,3];
array2 = [3,2,1];
expect(array1).toEqual(jasmine.arrayContaining(array2));
// check if every element of array2 is element of array1
// to ensure [1, 1] !== [1, 2]
array2.forEach(x => expect(array1).toContain(x))
// check if every element of array1 is element of array2
// to ensure [1, 2] !== [1, 1]
array1.forEach(x => expect(array2).toContain(x))
// check if they have equal length to ensure [1] !== [1, 1]
expect(array1.length).toBe(array2.length)
//Compare arrays without order
//Example
//a1 = [1, 2, 3, 4, 5]
//a2 = [3, 2, 1, 5, 4]
//isEqual(a1, a2) -> true
//a1 = [1, 2, 3, 4, 5];
//a2 = [3, 2, 1, 5, 4, 6];
//isEqual(a1, a2) -> false
function isInArray(a, e) {
for ( var i = a.length; i--; ) {
if ( a[i] === e ) return true;
}
return false;
}
function isEqArrays(a1, a2) {
if ( a1.length !== a2.length ) {
return false;
}
for ( var i = a1.length; i--; ) {
if ( !isInArray( a2, a1[i] ) ) {
return false;
}
}
return true;
}
There is currenly a matcher for this USE CASE:
https://github.com/jest-community/jest-extended/pull/122/files
test('passes when arrays match in a different order', () => {
expect([1, 2, 3]).toMatchArray([3, 1, 2]);
expect([{ foo: 'bar' }, { baz: 'qux' }]).toMatchArray([{ baz: 'qux' }, { foo: 'bar' }]);
});
function equal(arr1, arr2){
return arr1.length === arr2.length
&&
arr1.every((item)=>{
return arr2.indexOf(item) >-1
})
&&
arr2.every((item)=>{
return arr1.indexOf(item) >-1
})
}
The idea here is to first determine if the length of the two arrays are same, then check if all elements are in the other's array.
Here's a solution that will work for any number or arrays
https://gist.github.com/tvler/cc5b2a3f01543e1658b25ca567c078e4
const areUnsortedArraysEqual = (...arrs) =>
arrs.every((arr, i, [first]) => !i || arr.length === first.length) &&
arrs
.map(arr =>
arr.reduce(
(map, item) => map.set(item, (map.get(item) || 0) + 1),
new Map(),
),
)
.every(
(map, i, [first]) =>
!i ||
[...first, ...map].every(([item]) => first.get(item) === map.get(item)),
);
Some tests (a few answers to this question don't account for arrays with multiple items of the same value, so [1, 2, 2] and [1, 2] would incorrectly return true)
[1, 2] true
[1, 2], [1, 2] true
[1, 2], [1, 2], [1, 2] true
[1, 2], [2, 1] true
[1, 1, 2], [1, 2, 1] true
[1, 2], [1, 2, 3] false
[1, 2, 3, 4], [1, 2, 3], [1, 2] false
[1, 2, 2], [1, 2] false
[1, 1, 2], [1, 2, 2] false
[1, 2, 3], [1, 2], [1, 2, 3] false
This algorithm is great for arrays where each item is unique. If not, you can add in something to check for duplicates...
tests = [
[ [1,0,1] , [0,1,1] ],
[ [1,0,1] , [0,0,1] ], //breaks on this one...
[ [2,3,3] , [2,2,3] ], //breaks on this one also...
[ [1,2,3] , [2,1,3] ],
[ [2,3,1] , [1,2,2] ],
[ [2,2,1] , [1,3,2] ]
]
tests.forEach(function(test) {
console.log('eqArraySets( '+test[0]+' , '+test[1]+' ) = '+eqArraySets( test[0] , test[1] ));
});
function eqArraySets(a, b) {
if ( a.length !== b.length ) { return false; }
for ( var i = a.length; i--; ) {
if ( !(b.indexOf(a[i])>-1) ) { return false; }
if ( !(a.indexOf(b[i])>-1) ) { return false; }
}
return true;
}
This approach has worse theoretical worst-case run-time performance, but, because it does not perform any writes on the array, it might be faster in many circumstances (haven't tested performance yet):
WARNING: As Torben pointed out in the comments, this approach only works if both arrays have unique (non-repeating) elements (just like several of the other answers here).
/**
* Determine whether two arrays contain exactly the same elements, independent of order.
* #see https://stackoverflow.com/questions/32103252/expect-arrays-to-be-equal-ignoring-order/48973444#48973444
*/
function cmpIgnoreOrder(a, b) {
const { every, includes } = _;
return a.length === b.length && every(a, v => includes(b, v));
}
// the following should be all true!
const results = [
!!cmpIgnoreOrder([1,2,3], [3,1,2]),
!!cmpIgnoreOrder([4,1,2,3], [3,4,1,2]),
!!cmpIgnoreOrder([], []),
!cmpIgnoreOrder([1,2,3], [3,4,1,2]),
!cmpIgnoreOrder([1], []),
!cmpIgnoreOrder([1, 3, 4], [3,4,5])
];
console.log('Results: ', results)
console.assert(_.reduce(results, (a, b) => a && b, true), 'Test did not pass!');
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>
I am currently using this helper function (for TypeScript). It makes sure that arrays that have non unique elements are supported as well.
function expectArraysToBeEqualIgnoringOrder<T>(arr1: T[], arr2: T[]) {
while(arr1.length > 0) {
expect(arr1.length).toEqual(arr2.length)
const elementToDelete = arr1[0]
arr1 = arr1.filter(element => element !== elementToDelete)
arr2 = arr2.filter(element => element !== elementToDelete)
}
expect(arr2.length).toEqual(0)
}
Many of the other asnwers do not correctly handle cases like this:
array1: [a, b, b, c]
array2: [a, b, c, c]
Here the number of elements in both arrays is the same and both arrays contain all elements from the other array, yet they are different arrays and the test should fail.
It runs in O(n^2) (precisely (n^2 + n) / 2), so it's not suitable for very large arrays, but it's suitable for arrays that are not easilly sorted and therefore can not be compared in O(n * log(n))