I am trying to check if object array A includes objects from B.
let A = [
{ name: "Max" },
{ name: "Jhon" },
{ name: "Naton" },
]
let B = [
{ name: "Max" },
{ name: "Naton" },
]
So B has two objects that is in array A. How to check this ?
I am trying to achieve it with includes :
for(let entry of this.b){
if(this.a.includes(entry)){
console.log('includes');
}
}
But I get false on includes.
The method Array.includes() compare the entries of the array with the given value. Because your array entries are objects, it will not match. You have to loop at the array yourself and make the comparison.
Array.some() loops on an array and returns true if you returns true at least one. This method is useful when you want to verify something. In our example, we want to verify if the array a contains the b entry.
const a = [{
name: 'Max',
},
{
name: 'Jhon',
},
{
name: 'Naton',
},
];
const b = [{
name: 'Max',
},
{
name: 'Naton',
},
{
name: 'Daddy',
},
];
console.log(b.map(x => a.some(y => y.name === x.name)));
If I break it down :
const a = [{
name: 'Max',
},
{
name: 'Jhon',
},
{
name: 'Naton',
},
];
const b = [{
name: 'Max',
},
{
name: 'Naton',
},
{
name: 'Daddy',
},
];
// Loop on every entry of the b array
b.forEach((x) => {
// x here represent one entry
// first it will worth { name: 'Max' }, then { name: 'Naton' } ...
// for each value we are going to look at a if we can find a match
const isThereAMatch = a.some((y) => {
// y here is worth one entry of the a array
if (y.name === x.name) return true;
return false;
});
if (isThereAMatch === true) {
console.log(`We have found ${x.name} in a`);
} else {
console.log(`We have not found ${x.name} in a`);
}
});
You have to use another loop, then check the property name:
var a = [
{name: "Max"},
{name: "Jhon"},
{name: "Naton"},
];
var b = [
{name: "Max"},
{name: "Naton"},
];
for(let entry of b){
for(let entry2 of a){
if(entry2.name == entry.name){
console.log('includes', entry.name);
}
}
}
OR: You can use string version of object to check with includes():
var a = [
{name: "Max"},
{name: "Jhon"},
{name: "Naton"},
];
var b = [
{name: "Max"},
{name: "Naton"},
];
var aTemp = a.map(i => JSON.stringify(i));
var bTemp = b.map(i => JSON.stringify(i));
for(let entry of bTemp){
if(aTemp.includes(entry)){
console.log('includes', entry);
}
}
When you use Array#includes() method it will always return false because it's comparing objects which aren't equal because they aren't referencing the same object.
You should compare objects properties and not whole objects, you can do it using Array#some() method like this:
for (let entry of this.b) {
if (this.b.some(x => x.name === entry.name)) {
console.log('includes');
}
}
Demo:
A = [{
name: "Max"
},
{
name: "Jhon"
},
{
name: "Naton"
},
]
B = [{
name: "Max"
},
{
name: "Naton"
},
]
//Filter objects that exists in both arrays
let result = A.filter(el=> B.some(x => x.name === el.name));
console.log(result);
Related
I've tried modifying some of the similar solutions on here but I keep getting stuck, I believe I have part of this figured out however, the main caveat is that:
Some of the objects have extra keys, which renders my object comparison logic useless.
I am trying to compare two arrays of objects. One array is the original array, and the other array contains the items I want deleted from the original array. However there's one extra issue in that the second array contains extra keys, so my comparison logic doesn't work.
An example would make this easier, let's say I have the following two arrays:
const originalArray = [{id: 1, name: "darnell"}, {id: 2, name: "funboi"},
{id: 3, name: "jackson5"}, {id: 4, name: "zelensky"}];
const itemsToBeRemoved = [{id: 2, name: "funboi", extraProperty: "something"},
{id: 4, name: "zelensky", extraProperty: "somethingelse"}];
after running the logic, my final output should be this array:
[{id: 1, name: "darnell"}, {id: 3, name: "jackson5"}]
And here's the current code / logic that I have, which compares but doesn't handle the extra keys. How should I handle this? Thank you in advance.
const prepareArray = (arr) => {
return arr.map((el) => {
if (typeof el === "object" && el !== null) {
return JSON.stringify(el);
} else {
return el;
}
});
};
const convertJSON = (arr) => {
return arr.map((el) => {
return JSON.parse(el);
});
};
const compareArrays = (arr1, arr2) => {
const currentArray = [...prepareArray(arr1)];
const deletedItems = [...prepareArray(arr2)];
const compared = currentArray.filter((el) => deletedItems.indexOf(el) === -1);
return convertJSON(compared);
};
How about using filter and some? You can extend the filter condition on select properties using &&.
const originalArray = [
{ id: 1, name: 'darnell' },
{ id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' },
];
const itemsToBeRemoved = [
{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' },
];
console.log(
originalArray.filter(item => !itemsToBeRemoved.some(itemToBeRemoved => itemToBeRemoved.id === item.id))
)
Or you can generalise it as well.
const originalArray = [
{ id: 1, name: 'darnell' },
{ id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' },
];
const itemsToBeRemoved = [
{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' },
];
function filterIfSubset(originalArray, itemsToBeRemoved) {
const filteredArray = [];
for (let i = 0; i < originalArray.length; i++) {
let isSubset = false;
for (let j = 0; j < itemsToBeRemoved.length; j++) {
// check if whole object is a subset of the object in itemsToBeRemoved
if (Object.keys(originalArray[i]).every(key => originalArray[i][key] === itemsToBeRemoved[j][key])) {
isSubset = true;
}
}
if (!isSubset) {
filteredArray.push(originalArray[i]);
}
}
return filteredArray;
}
console.log(filterIfSubset(originalArray, itemsToBeRemoved));
Another simpler variation of the second approach:
const originalArray = [
{ id: 1, name: 'darnell' },
{ id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' },
];
const itemsToBeRemoved = [
{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' },
];
const removeSubsetObjectsIfExists = (originalArray, itemsToBeRemoved) => {
return originalArray.filter(item => {
const isSubset = itemsToBeRemoved.some(itemToBeRemoved => {
return Object.keys(item).every(key => {
return item[key] === itemToBeRemoved[key];
});
});
return !isSubset;
});
}
console.log(removeSubsetObjectsIfExists(originalArray, itemsToBeRemoved));
The example below is a reusable function, the third parameter is the key to which you compare values from both arrays.
Details are commented in example
const arr=[{id:1,name:"darnell"},{id:2,name:"funboi"},{id:3,name:"jackson5"},{id:4,name:"zelensky"}],del=[{id:2,name:"funboi",extraProperty:"something"},{id:4,name:"zelensky",extraProperty:"somethingelse"}];
/** Compare arrayA vs. delArray by a given key's value.
--- ex. key = 'id'
**/
function deleteByKey(arrayA, delArray, key) {
/* Get an array of only the values of the given key from delArray
--- ex. delList = [1, 2, 3, 4]
*/
const delList = delArray.map(obj => obj[key]);
/* On every object of arrayA compare delList values vs
current object's key's value
--- ex. current obj[id] = 2
--- [1, 2, 3, 4].includes(obj[id])
Any match returns an empty array and non-matches are returned
in it's own array.
--- ex. ? [] : [obj]
The final return is a flattened array of the non-matching objects
*/
return arrayA.flatMap(obj => delList.includes(obj[key]) ? [] : [obj]);
};
console.log(deleteByKey(arr, del, 'id'));
let ff = [{ id: 1, name: 'darnell' }, { id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' }]
let cc = [{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' }]
let ar = []
let out = []
const result = ff.filter(function(i){
ar.push(i.id)
cc.forEach(function(k){
out.push(k.id)
})
if(!out.includes(i.id)){
// console.log(i.id, i)
return i
}
})
console.log(result)
I have two array of objects (array1, array2). I am trying return final array(as shown below) which eliminates duplicates from array2 but not array1. I am giving priority to array1.
array1 =[
{ phone: "07485454", name: "John" },
{ phone: "054554", name: "Ryan" },
]
array2 =[
{ phone: "2144564", name: "John" },
{ phone: "286456", name: "Mike" },
]
This is something I want as a final result. Remove duplicates from array2 only.
Final Array:
[
{ phone: "07485454", name: "John" },
{ phone: "054554", name: "Ryan" },
{ phone: "286456", name: "Mike" },
]
This is something that I have tried:
for(let i = 0; i < array1.length; i++) {
let name = array1[i].name;
for(let a = 0; i < array2.length; a++) {
let newname = array2[a].name;
if(name !== newname) {
array1.push(
{
phone: array2[a].phone,
name: array2[a].name
});
}
console.log(array1);
}
}
This is the error I get.
"errorType": "TypeError",
"errorMessage": "Cannot read property 'name' of undefined",
You were close to the solution, you can create a Set to contain names of your array1 elements and pushing array1 elements to your result array; subsequently add elements of array2 which names are not contained in your set to your result array :
function removeDuplicates(array1, array2) {
const result = [];
const set = new Set();
for (const elem of array1) {
result.push(elem);
set.add(elem.name);
}
for (const elem of array2) {
if (!set.has(elem.name)) {
result.push(elem);
}
}
return result;
}
const array1 =[
{ phone: "07485454", name: "John" },
{ phone: "054554", name: "Ryan" }
];
const array2 =[
{ phone: "2144564", name: "John" },
{ phone: "286456", name: "Mike" }
];
console.log(JSON.stringify(removeDuplicates(array1, array2)));
you have done a typo mistake in inner for loop condition a < array2.length
Changing the array1 length will results to infinite loop
Every time, when the if(name !== newname) the condition is true, inserting a new object in array1 will results to changing the array1 length.
let array1 =[
{ phone: "07485454", name: "John" },
{ phone: "054554", name: "Ryan" },
]
let array2 =[
{ phone: "2144564", name: "John" },
{ phone: "286456", name: "Mike" },
]
let result = array2.reduce((accumulator, currentItem) =>{
let index = accumulator.findIndex(item => item.name === currentItem.name);
if(index === -1){
accumulator.push(currentItem);
}
return accumulator;
},array1);
console.log(result);
I have the following array:
[{
name: 'foo',
values: '10,12'
},
{
name: 'bar',
values: 'red,blue'
}]
Using some javascript logic I would like to output the following array:
[{
option1: 10,
option2: 'red'
},
{
option1: 10,
option2: 'blue'
},
{
option1: 12,
option2: 'red'
},
{
option1: 12,
option2: 'blue'
}]
What is the best and correct way to achieve this using javascript?
Lets say your first array is named arr.
var arr = [{
name: 'foo',
values: '10,12'
},
{
name: 'bar',
values: 'red,blue'
}];
var v1 = arr[0].values.split(',');
var v2 = arr[1].values.split(',');
var res = new Array();
for(i in v1){
for(j in v2){
res.push({'option1':v1[i],'option2':v2[j]});
}
}
console.log(res);
Here's an approach that can handle an arbitrary number of objects.
function valuesCrossProduct(input) {
return input.flatMap((current, index, array) => {
let result = [];
let values = current.values.split(',');
for (let v of values) {
for (let i = 0; i < array.length; i++) {
if (i <= index) {
// Skip creating cross products with self (i.e. == index)
// and with previously visited objects (i.e. < index).
continue;
}
let iValues = array[i].values.split(',');
let currentKey = `option${index}`;
let iKey = `option${i}`;
for (let iv of iValues) {
result.push({
[currentKey]: v,
[iKey]: iv,
});
}
}
}
return result;
});
}
let twoElementArray = [{
name: 'foo',
values: '10,12'
},
{
name: 'bar',
values: 'red,blue',
}];
let threeElementArray = [{
name: 'foo',
values: '10,12'
},
{
name: 'bar',
values: 'red,blue',
},
{
name: 'baz',
values: 'wham,bam',
}];
console.log(valuesCrossProduct(twoElementArray));
console.log(valuesCrossProduct(threeElementArray));
Functional for the win.
Note: as it is, this only works for an array of two objects, with any number of values in each, where the first set of values are numbers and the second set are strings, which is what you described above.
const arr = [{
name: 'foo',
values: '10,12'
},
{
name: 'bar',
values: 'red,blue'
}];
const values = arr
.map(o => o.values.split(','))
.reduce((cur, next) => {
return cur.map(c => {
return next.map(n => {
return {
option1: parseInt(c),
option2: n
};
});
}).flat();
});
console.log(values);
If you need generic approach to get possible options from various values.
const options = data => {
let sets = [[]];
data.forEach(({ values }, i) => {
const new_set = [];
values.split(",").forEach(value => {
new_set.push(
Array.from(sets, set => [...set, [`option${i + 1}`, value]])
);
});
sets = new_set.flatMap(set => set);
});
return sets.map(set => Object.fromEntries(set));
};
const data = [
{
name: "foo",
values: "10,12"
},
{
name: "bar",
values: "red,blue,green"
},
{
name: "test",
values: "top,bottom"
}
];
console.log(options(data));
I have an array of arrays below. With ES6, how can I get a count of each value Good, Excellent & Wow into a new array e.g [{name: Good, count: 4} {name: Excellent, count: 5}, {name:Wow, count:2}] in dynamic style. I am attempting to use Object.assign but I am failing to "unique" out the count of the key plus instead, I need to use an array as I am trying to render this out on the front end. Do I need to use reduce? how?
let k = 0
const stats = {}
const remarks = [
[{name: "Good"}],
[{name: "Good"}, {name: "Excellent"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Excellent"}],
[{name: "Excellent"}]
]
remarks.forEach((arr) => {
arr.map((e) => {
Object.assign(stats, { [e.name]: k = k + 1 })
})
})
console.log(stats);
Output:
stats: {Good: 8, Excellent: 11, Wow: 9}
Which is Incorrect plus I need to use an array.
Expected output:
[{name: Good, count: 4} {name: Excellent, count: 5}, {name:Wow, count:2}]
Flatten the array of arrays and reduce it starting with an object like : { Good: 0, Excellent: 0, Wow: 0}
then .map the Object.entries of the result to transform it to an array :
const remarks = [
[{ name: "Good" }],
[{ name: "Good" }, { name: "Excellent" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Excellent" }],
[{ name: "Excellent" }]
];
const result = Object.entries(
remarks.flat().reduce(
(all, { name }) => {
all[name] += 1;
return all;
},
{ Good: 0, Excellent: 0, Wow: 0 }
)
).map(([name, count]) => ({ name, count }));
console.log(result);
You can try below logic:
var data = [[{name: "Good"}],[{name: "Good"}, {name:"Excellent"}],[{name: "Good"}, {name:"Excellent"}, {name:"Wow"}],[{name: "Good"}, {name:"Excellent"}, {name:"Wow"}],[{name:"Excellent"}],[{name:"Excellent"}]]
var nData = [];
(data || []).forEach( e => {
(e || []).forEach(ei => {
var i = (index = nData.findIndex(d => d.name === ei.name)) >=0 ? index : nData.length;
nData[i] = {
name: ei.name,
count : (nData[i] && nData[i].count ? nData[i].count : 0)+1
}
});
});
console.log(nData);
Hope this helps!
You can use reduce, then convert the result into an array of objects:
const counts = remarks.reduce((result, list) => {
list.forEach(remark => {
result[remark.name] = (result[remark.name] || 0) + 1;
});
}, {});
const finalResult = [];
for (let name in counts) {
finalResult.push({name, count: counts[name]});
}
You could achieve this pretty easily by:
1) Flattening the nested array into 1 single level array.
2) Iterating over the flat array and create a "count map" by using Array.prototype.reduce
For example:
const remarks = [
[{
name: 'Good'
}],
[{
name: 'Good'
}, {
name: 'Excellent'
}],
[{
name: 'Good'
}, {
name: 'Excellent'
}, {
name: 'Wow'
}],
[{
name: 'Good'
}, {
name: 'Excellent'
}, {
name: 'Wow'
}],
[{
name: 'Excellent'
}],
[{
name: 'Excellent'
}]
]
const flatten = arr => arr.reduce((accum, el) => accum.concat(el), [])
const map = flatten(remarks).reduce((accum, el) => {
if (accum[el.name]) {
accum[el.name] += 1;
} else {
accum[el.name] = 1;
}
return accum;
}, {});
console.log(map)
First find the counts using reduce than pass that to another function to get the desired view structure:
const Good = 1,
Excellent = 2,
Wow = 3;
const remarks = [
[{name: Good}],
[{name: Good}, {name:Excellent}],
[{name: Good}, {name:Excellent}, {name:Wow}],
[{name: Good}, {name:Excellent}, {name:Wow}],
[{name:Excellent}],
[{name:Excellent}]
];
/*
[{name: Good, count: 4} {name: Excellent, count: 5}, {name:Wow, count:2}]
*/
function counts(remarks) {
return remarks.flat().reduce((acc, v) => {
const name = v.name;
let count = acc[name] || 0;
return {
...acc,
[name]: count + 1
}
}, {});
}
function view(counts) {
return Object.keys(counts).map(key => {
let count = counts[key];
return { name: key, count };
})
}
console.log(view(counts(remarks)));
Any time you are making a smaller set of data, or transforming data, in JavaScript reduce should be the first method you attempt to use. In this case, you may want to pair it with an indexer (hence preloading with an array of index and an array of result).
This works in one pass without needing to know the name values up front.
const remarks = [
[{name: "Good"}],
[{name: "Good"}, {name: "Excellent"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Excellent"}],
[{name: "Excellent"}]
];
const stats = remarks.reduce((p,c) => (
c.forEach( ({name}) => {
if(!p[0].hasOwnProperty(name)){
p[1].push({name:name,count:0});
p[0][name] = p[1].length - 1;
}
p[1][p[0][name]].count++;
}),p),[{},[]])[1];
console.log(stats);
A slightly more concise and definitely less readable approach (but it's worth to mention) could be:
const remarks = [
[{ name: "Good" }],
[{ name: "Good" }, { name: "Excellent" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Excellent" }],
[{ name: "Excellent" }]
];
const stats = Object.entries(
remarks
.flat()
.reduce((acc, {name}) => (acc[name] = -~acc[name], acc), {})))
).map(([name, count]) => ({ name, count }));
console.log(stats);
It uses the comma operator in the reducer to returns the accumulator; and the bitwise operator NOT to create a counter without the needs to initialize the object upfront with all the names.
const flattenedRemarks = _.flatten(remarks);
const groupedRemarks = _.groupBy(flattenedRemarks, (remark) => remark.name);
const remarkCounts = _.mapValues(groupedRemarks, (group) => group.length);
const data = {
"mchale": {
"classes":["ESJ030", "SCI339"], // get the length
"faculty":["Hardy", "Vikrum"] // get the length
},
"lawerence":{
"classes":["ENG001"], // get the length
"faculty":["Speedman", "Lee", "Lazenhower"] // get the length
}
};
const count = Object.keys(data).map(campusName => {
const campus = data[campusName];
return Object.keys(campus).map(key => campus[key].length).reduce((p, c) => p + c, 0);
}).reduce((p, c) => p + c, 0);
console.log(count);
I have the following two Javascript arrays:
const array1 = [{ id: 1}, { id: 2 }, { id: 3 }, { id: 4}];
const array2 = [{ id: 1}, { id: 3 }];
I now want a new array array3 that contains only the objects that aren't already in array2, so:
const array3 = [{ id: 2}, { id: 4 }];
I have tried the following but it returns all objects, and when I changed the condition to === it returns the objects of array2.
const array3 = array1.filter(entry1 => {
return array2.some(entry2 => entry1.id !== entry2.id);
});
Any idea? ES6 welcome
You could reverse the comparison (equal instead of unqual) and return the negated result of some.
const
array1 = [{ id: 1 }, { id: 2 }, { id: 3 }, { id: 4 }],
array2 = [{ id: 1 }, { id: 3 }],
array3 = array1.filter(entry1 => !array2.some(entry2 => entry1.id === entry2.id));
// ^ ^^^
console.log(array3);
Nina's answer is a good start but will miss any unique elements in array 2.
This extends her answer to get the unique elements from each array and then combine them:
const
array1 = [{ id: 1 }, { id: 2 }, { id: 3 }, { id: 4 }],
array2 = [{ id: 1 }, { id: 3 }, { id: 5 }],
array3 = array1.filter(entry1 => !array2.some(entry2 => entry1.id === entry2.id)),
array4 = array2.filter(entry1 => !array1.some(entry2 => entry1.id === entry2.id)),
array5 = array3.concat(array4);
console.log(array5);