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I know there are multiple ways to remove duplicates from arrays in javascript, the one i use is
let originalArray = [1, 2, 3, 4, 1, 2, 3, 4]
let uniqueArray = array => [...new Set(array)]
console.log(uniqueArray) -> [1, 2, 3, 4]
what i want is something similar but instead of removing the duplicates, to replace it with whatever string or number i want, like this
console.log(uniqueArray) -> [1, 2, 3, 4, "-", "-", "-", "-"]
this has to work with any order, like
[1, 2, 3, 3, 4, 5, 7, 1, 6]
result -> [1, 2, 3, "-", 4, 5, 7, "-", 6]
i tested this solution
const arr = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1];
const deleteAndInsert = uniqueList => {
const creds = uniqueList.reduce((acc, val, ind, array) => {
let { count, res } = acc;
if (array.lastIndexOf(val) === ind) {
res.push(val);
} else {
count++;
};
return { res, count };
}, { count: 0, res: [] });
const { res, count } = creds;
return res.concat(" ".repeat(count).split(" "));
};
console.log(deleteAndInsert(arr));
but only adds it at the end of the uniques, and also, only works with numbers
i want it to work with strings too, like dates as an example
["2021-02-22", "2021-02-23", "2021-02-22", "2021-02-28"]
You could still use a Set and check if the value is in the set.
const
unique = array => array.map((s => v => !s.has(v) && s.add(v) ? v : '-')(new Set));
console.log(...unique([1, 2, 3, 4, 1, 2, 3, 4]));
console.log(...unique([1, 2, 3, 3, 4, 5, 7, 1, 6]));
Just create new Array, use 1 set to control which element appeared, if one element appears more than 1, push new one character like '-'
let originalArray = [1, 2, 3, 4, 1, 2, 3, 4];
let newArray = [];
let set = new Set();
for (let i = 0; i < originalArray.length; i++) {
if(!set.has(originalArray[i])) {
newArray.push(originalArray[i]);
set.add(originalArray[i]);
} else {
newArray.push('-');
}
}
console.log(newArray);
You could do it with reduce
const dashDupes = array => array.reduce((acc, e) => {
if(acc.idx[e])
acc.arr.push('-')
else{
acc.arr.push(e);
}
acc.idx[e] = true;
return acc;
},{idx:{},arr:[]}).arr
console.log(...dashDupes([1, 2, 3, 4, 1, 2, 3, 4]))
console.log(...dashDupes([1, 2, 3, 3, 4, 5, 7, 1, 6]))
This is a very simple approach to the problem:
function uniqueReplace(arr, rep) {
let res = [];
for (x of arr) {
res.push(res.includes(x) ? rep : x);
}
return res;
}
console.log(...uniqueReplace([1, 2, 3, 4, 1, 2, 3, 4], '-'));
console.log(...uniqueReplace([1, 2, 3, 3, 4, 5, 7, 1, 6], '-'));
const myObj = {
a: [1, 2, 3],
b: [2, 4, 6],
c: [10, 20, 30]
}
Into
const myCollection = [
{a: 1, b: 2, c: 10},
{a: 2, b: 4, c: 20},
{a: 3, b: 6, c: 30}
]
I tried combinations of Object.entries, Object.keys and map but I'm always finding myself iterating twice or more over myObj and I'm not happy with any solution I came up with.
So what is the most efficient (in terms of time complexity) and elegant way that you can think to achieve that?
Just in case you'd need variable length:
const myObj = {
a: [1, 2, 3],
b: [2, 4, 6,8, 10],
c: [10, 20, 30, 40],
};
let myCollection = [];
Object.keys(myObj).forEach((k) => {
for (let i = 0; i < myObj[k].length; i++) {
if (!myCollection[i]) myCollection.push({});
myCollection[i][k] = myObj[k][i];
}
});
console.log(myCollection);
You could reduce the entries and map nested arrays.
const
object = { a: [1, 2, 3], b: [2, 4, 6], c: [10, 20, 30] },
result = Object
.entries(object)
.reduce((r, [k, a]) => a.map((v, i) => ({ ...r[i], [k]: v })), []);
console.log(result);
That can be done using Array.reduce. I have attached the conversion code.
const myObj = {
a: [1, 2, 3],
b: [2, 4, 6],
c: [10, 20, 30]
}
const myCollection = [
{a: 1, b: 2, c: 10},
{a: 2, b: 4, c: 20},
{a: 3, b: 6, c: 30}
]
const maxLength = Math.max(...Object.values(myObj).map(item => item.length));
const myObjKeys = Object.keys(myObj);
const result = [ ...Array(maxLength).keys() ].map((index) => {
return myObjKeys.reduce((acc, curKey) => {
if (myObj[curKey].length > index) {
acc[curKey] = myObj[curKey][index];
}
return acc;
}, {});
});
console.log(result);
Using ramdajs, I could suggest you a short way like below
const myObj = {
a: [1, 2, 3],
b: [2, 4, 6],
c: [10, 20, 30]
}
const res = R.map(
R.zipObj(R.keys(myObj)),
R.values(myObj)
)
console.log(res)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>
In ramdajs homepage, criteria 3 in What's Different, it cited that
The parameters to Ramda functions are arranged to make it convenient for currying. The data to be operated on is generally supplied last.
You can use map() on Object.values and then use reduce() on each value.
const myObj = {
a: [1, 2, 3],
b: [2, 4, 6],
c: [10, 20, 30]
}
let keys = Object.keys(myObj);
const arr = Object.values(myObj).map((a) => a.reduce((ac, a, i) => ({...ac, [keys[i]]:a}), {}));
console.log(arr)
Suppose I have:
const KEYS = ['b', 'a', 'c']
const obj = {
2018: {a: 1, b: 2, c: 3},
2019: {a: 4, b: 5, c: 6},
2020: {a: 7, b: 8, c: 9},
}
This is what I would like to get:
const result = {
2018: {
a: [0, 1, 0],
b: [2, 0, 0],
c: [0, 0, 3]
},
2019: {
a: [0, 4, 0],
b: [5, 0, 0],
c: [0, 0, 6]
},,
2020: {
a: [0, 7, 0],
b: [8, 0, 0],
c: [0, 0, 9]
},
}
result['2018'] object has three keys. Each key value is an array that contains the values in the order that is set by KEYS using 0 as fill value.
How can I do something like this?
This is what I tried but obviously is more complicated than this:
const reshaped = Object.entries(obj).map(([key, value]) => {
return { [key]: Object.values(value) }
})
// [
// { 2018: [ 1, 2, 3 ] },
// { 2019: [ 4, 5, 6 ] },
// { 2020: [ 7, 8, 9 ] }
// ]
You could map the wanted keys in order abd build an array for each property.
const
KEYS = ['b', 'a', 'c'],
object = { 2018: { a: 1, b: 2, c: 3 }, 2019: { a: 4, b: 5, c: 6 }, 2020: { a: 7, b: 8, c: 9 } },
result = Object.fromEntries(Object.entries(object).map(([k, o]) => [
k,
Object.fromEntries(Object.entries(o).map(([l, v]) => [
l,
KEYS.map(m => l === m ? v : 0)
]))
]));
console.log(result);
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You could use a combination of Object.entries and Object.fromEntries to map the object and then just create a new array with the length of the KEYS arr.
const KEYS = ['b', 'a', 'c']
const obj = {
2018: {a: 1, b: 2, c: 3},
2019: {a: 4, b: 5, c: 6},
2020: {a: 7, b: 8, c: 9},
}
const result = Object.fromEntries( // Create obj from array of entries
Object.entries(obj).map(([key, value]) => [ // create array of entries from obj and map it
key,
Object.fromEntries( // do the same obj/arr transformation on the value
Object.entries(value).map(([subKey, subValue]) => {
const arr = new Array(KEYS.length).fill(0); // create new array of keys length and fill all zeroes
arr[KEYS.indexOf(subKey)] = subValue; // on the index of the key in the KEYS arr, set the value of the key
return [subKey, arr]; // return subValue
})
)
])
);
console.log(result);
This question already has answers here:
Get all unique values in a JavaScript array (remove duplicates)
(91 answers)
Closed 3 years ago.
I received this question for practice and the wording confused me, as I see 2 results that it might want.
And either way, I'd like to see both solutions.
For example, if I have an array:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
I'm taking this as wanting the final result as either:
let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];
OR:
let finalResult = [1, 9, 10];
The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.
Either way, I'd like to write two functions that does one of each.
This, given by someone else gives my second solution.
let elems = {},
arr2 = arr.filter(function (e) {
if (elems[e] === undefined) {
elems[e] = true;
return true;
}
return false;
});
console.log(arr2);
I'm not sure about a function for the first one (remove all duplicates).
Using Set and Array.from()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));
Alternate using regex
regex explanation here
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(\b,\w+\b)(?=.*\1)/ig, '')
.split(',')
.map(Number);
console.log(res);
Alternate using objects
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc, {
[val]: val
}), {});
console.log(Object.values(obj));
Just use a simple array.filter one-liner:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});
console.log(finalResult);
You could use another filter statement if you wanted the second result:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});
console.log(finalResult);
For the first part you can use Set() and Spread Syntax to remove duplicates.
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)
For the second part you can use reduce()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) => {
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
},{})
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) => {
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
},[])
console.log(res)
You can use closure and Map
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar => {
const mapObj = ar.reduce((acc, e) => {
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
}, new Map())
return function(hasDup = true) {
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
}
}
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))
You can create both arrays in One Go
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) => {
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
}, new Set()));
console.log(Array.from(unique))
console.log(repeated)
You can use Array.prototype.reduce() create a hash object where the keys are the numbers in the array and the values are going to be the the repeated occurrence of numbers in the arr array variable..
Then using Object.keys():
Remove all duplicates Object.keys(hash)
Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence
Code:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);
You could sort the array before and filter the array by checking only one side for duplicates or both sides.
var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)
As many other have said, the first one is just [...new Set(arr)]
For the second, just filter out those that occur more than once:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));
var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
var map = {};
var finalResult = [];
for (var i = 0; i < arr.length; i++) {
if (!map.hasOwnProperty(arr[i])) {
map[arr[i]] = true;
finalResult.push(arr[i]);
}
}
//if you need it sorted otherwise it will be in order
finalResult.sort(function(a, b) {
return a - b
});
console.log(finalResult);
var temparray1 = [[1,3,4],[5,6,7],[8,9,10]];
var final = [];
var obj = {};
for(var temp in temparray1){
for(var test in temparray1[temp]){
obj.b = temparray1[temp][0];
obj.c = temparray1[temp][1];
obj.d = temparray1[temp][2];
}
console.log(obj);
final.push(obj);
}
current output
[{ b: 8, c: 9, d: 10 }
{ b: 8, c: 9, d: 10 }
{ b: 8, c: 9, d: 10 }]
expected Out put:
[{ b: 1, c: 3, d: 4 }
{ b: 5, c: 6, d: 7 }
{ b: 8, c: 9, d: 10 }]
i am running my javascript in node.js -v 8.1.x server
in the console at the end of the for loop it prints the required out put but not in the array push
var temparray = [[1, 3, 4], [5, 6, 7], [8, 9, 10]];
const final = temparray.map(a => ({ b: a[0], c: a[1], d: a[2] }));
console.log(final);
Probably, you set obj outside the for loop, therefore the props are overwritten and you push the same object multiple times into the array. Simply move the obj declaration into the loop. And you probably just need the outer loop.
Btw much shorter:
let final = temparray1.map(
([b,c,d]) => ({b,c,d})
);
You could use Array#map and return an object with the wanted properties.
var array = [[1, 3, 4], [5, 6, 7], [8, 9, 10]],
result = array.map(function (a) {
return { b: a[0], c: a[1], d: a[2] };
});
console.log(result);
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With ES6, you could map the inner arrays as well with an array for the keys with Object.assign and spread syntax ....
var array = [[1, 3, 4], [5, 6, 7], [8, 9, 10]],
keys = ['b', 'c', 'd'],
result = array.map(a => Object.assign(...keys.map((k, i) => ({ [k]: a[i] }))));
console.log(result);
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Here is what you wanted
var temparray1 = [[1,3,4],[5,6,7],[8,9,10]];
var final = [];
for(var temp in temparray1){
var obj = {};
obj['b'] = temparray1[temp][0];
obj['c'] = temparray1[temp][1];
obj['d'] = temparray1[temp][2];
final.push(obj);
}
console.log(final);
Hope this helps!