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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 a nested/multi-dimensional array like so:
[ [ 1, 1, a ], [ 1, 1 , b ], [ 2, 2, c ], [ 1 ,1, d ] ]
And I want to filter it so that it returns only unique values of the outer array based on the 1st value of each nested array.
So from the above array, it would return:
[ [1,1,a] [2,2,c] ]
Am trying to do this in vanilla javascript if possible. Thanks for any input! =)
Here is my solution.
const dedup = arr.filter((item, idx) => arr.findIndex(x => x[0] == item[0]) == idx)
It looks simple and also somehow tricky a bit.
I realize there's already three solutions, but I don't like them. My solution is
Generic - you can use unique with any selector function
O(n) - it uses a set, it doesn't run in O(n^2) time
So here it is:
/**
* #param arr - The array to get the unique values of
* #param uniqueBy - Takes the value and selects a criterion by which unique values should be taken
*
* #returns A new array containing the original values
*
* #example unique(["hello", "hElLo", "friend"], s => s.toLowerCase()) // ["hello", "friend"]
*/
function unique(arr, uniqueBy) {
const temp = new Set()
return arr.filter(v => {
const computed = uniqueBy(v)
const isContained = temp.has(computed)
temp.add(computed)
return !isContained
})
}
const arr = [ [ 1, 1, 'a' ], [ 1, 1, 'b' ], [ 2, 2, 'c' ], [ 1, 1, 'd' ] ]
console.log(unique(arr, v => v[0]))
You could filter with a set and given index.
const
uniqueByIndex = (i, s = new Set) => array => !s.has(array[i]) && s.add(array[i]),
data = [[1, 1, 'a'], [1, 1, 'b'], [2, 2, 'c'], [1, 1, 'd']],
result = data.filter(uniqueByIndex(0));
console.log(result);
const input = [[1,1,'a'], [1,1,'b'], [2,2,'c'], [1,1,'d']]
const res = input.reduce((acc, e) => acc.find(x => x[0] === e[0])
? acc
: [...acc, e], [])
console.log(res)
Create the object with keys as first element of array. Iterate over array, check if the first element of array exist in the Object, if not push into the array.
const nestedArr = [ [1,1,"a"], [1,1,"b"], [2,2,"c"], [1,1,"d"] ];
const output = {};
for(let arr of nestedArr) {
if(!output[arr[0]]) {
output[arr[0]] = arr;
}
}
console.log(Object.values(output));
Another solution, would be to maintain the count of first array element and if the count is equal to 1, then push in the final array.
const input = [ [1,1,"a"], [1,1,"b"], [2,2,"c"], [1,1,"d"] ],
count = {},
output = [];
input.forEach(arr => {
count[arr[0]] = (count[arr[0]] || 0) + 1;
if(count[arr[0]] === 1) {
output.push(arr);
}
})
console.log(output);
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)
I have an object containing a bunch of arrays like this:
{
names: [0, 1, 2],
gender: [2, 5, 1],
boolean: [7, 2, 1, 6]
}
How can I get the value which is present in all arrays, in this case, 1?
A way of achieving this is by getting the distinct values for all the key values and then looping through the values, checking the value exists for each key's array.
const obj = {
names: [0, 1, 2],
gender: [2, 5, 1],
boolean: [7, 2, 1, 6]
}
// [1] Get the object's keys
const keys = Object.keys(obj);
// [2] Get distinct values by combining all the key arrays.
const distinctValues = [...new Set(keys.reduce((all, curr) => {
return all.concat(obj[curr]);
}, []))];
// [3] Filter the distinct values by checking each value against
// the key's array
const presentInAll = distinctValues.filter((value) => {
return keys.every((key) => obj[key].includes(value));
});
console.log(presentInAll);
I thought I'd add another solution that does not make any temporary copies of the original data which can be useful if the data is larger or memory management is important.
const data = {
names: [0, 1, 2],
gender: [2, 5, 1],
boolean: [7, 2, 1, 6]
};
function findCommon(...arrays) {
// sort arrays by length so we optimize and iterate first by the shortest array
arrays.sort((a, b) => {
return a.length - b.length;
});
let results = new Set();
// for each item in the first array
for (let item of arrays[0]) {
// look in other arrays for this value
let found = true;
for (let i = 1; i < arrays.length; i++) {
if (!arrays[i].includes(item)) {
found = false;
break;
}
}
if (found) {
results.add(item);
}
}
return results;
}
let results = findCommon(data.names, data.gender, data.boolean);
console.log(Array.from(results));
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))