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Create an array of objects based on an object if one or more properties have multiple values differentiated by a comma

i'm trying to duplicate objects based on two properties that have multiple values differentiated by a comma.
For example:
I have an object
const obj = {
id: 1
date: "2021"
tst1: "111, 222"
tst2: "AAA, BBB"
}
And I would like the result to be an array of 2 objects in this case (because there are 2 values in tst1 OR tst2, these 2 properties will always have the same nr of values differentiated by a comma)
[{
id: 1,
date: "2021",
tst1: "111",
tst2: "AAA",
},
{
id: 1,
date: "2021",
tst1: "222",
tst2: "BBB",
}]
What I tried is this:
I created a temporary object
const tempObject = {
id: obj.id,
date: obj.date,
}
And then I would split and map the property that has multiple values, like this:
cont newObj = obj.tst1.split(",").map(function(value) {
let finalObj = {}
return finalObj = {
id: tempObject.id,
date: tempObject.date,
tst1: value,
})
And now, the newObj is an array of objects and each object contains a value of tst1.
The problem is I still have to do the same for the tst2...
And I was wondering if there is a simpler method to do this...
Thank you!

Here is an example that accepts an array of duplicate keys to differentiate. It first maps them to arrays of entries by splitting on ',' and then trimming the entries, then zips them by index to create sub-arrays of each specified property, finally it returns a result of the original object spread against an Object.fromEntries of the zipped properties.
const mapDuplicateProps = (obj, props) => {
const splitProps = props.map((p) =>
obj[p].split(',').map((s) => [p, s.trim()])
);
// [ [[ 'tst1', '111' ], [ 'tst1', '222' ]], [[ 'tst2', 'AAA' ], [ 'tst2', 'BBB' ]] ]
const dupeEntries = splitProps[0].map((_, i) => splitProps.map((p) => p[i]));
// [ [[ 'tst1', '111' ], [ 'tst2', 'AAA' ]], [[ 'tst1', '222' ], [ 'tst2', 'BBB' ]] ]
return dupeEntries.map((d) => ({ ...obj, ...Object.fromEntries(d) }));
};
const obj = {
id: 1,
date: '2021',
tst1: '111, 222',
tst2: 'AAA, BBB',
};
console.log(mapDuplicateProps(obj, ['tst1', 'tst2']));

Not sure if that's what you're searching for, but I tried making a more general use of what you try to do:
const duplicateProperties = obj => {
const properties = Object.entries(obj);
let acc = [{}];
properties.forEach(([key, value]) => {
if (typeof value === 'string' && value.includes(',')) {
const values = value.split(',');
values.forEach((v, i) => {
if (!acc[i]) {
acc[i] = {};
}
acc[i][key] = v.trim();
});
} else {
acc.forEach(o => o[key] = value);
}
});
return acc;
};
const obj = {
id: 1,
date: '2021',
tst1: '111, 222',
tst2: 'AAA, BBB',
};
console.log(duplicateProperties(obj));

You could start by determining the length of the result using Math.max(), String.split() etc.
Then you'd create an Array using Array.from(), returning the correct object for each value of the output index.
const obj = {
id: 1,
date: "2021",
tst1: "111, 222",
tst2: "AAA, BBB",
}
// Determine the length of our output array...
const length = Math.max(...Object.values(obj).map(s => (s + '').split(',').length))
// Map the object using the relevant index...
const result = Array.from({ length }, (_, idx) => {
return Object.fromEntries(Object.entries(obj).map(([key, value]) => {
const a = (value + '').split(/,\s*/);
return [key, a.length > 1 ? a[idx] : value ]
}))
})
console.log(result)
.as-console-wrapper { max-height: 100% !important; }

Is there any other way to check multiple values are present in array of object?

I have the below array of objects and I want to check if two different users are present in this array .if present i have to run some logic
let result = [{"name": "ABC"},{"name": "CDE"},{"name": "FGH"},{"name": "XYZ"}];
var newArr = [];
var hasMatch = result.filter(function(val) {
if (val.name == "FGH"){
newArr.push(val)
} else if (val.name == "ABC") {
newArr.push(val)
}
});
console.log(newArr)
if (newArr.length == 2) {
//do logic
}
It's working as expected but I'm looking for a different approach for this. could someone advise?

Not optimized for speed, but does the job
let arr = [
{
"name": "ABC"
},
{
"name": "CDE"
},
{
"name": "FGH"
},
{
"name": "XYZ"
}
];
let users = ["ABC", "XYZ"]
let hasAllUsers = users.every(user => arr.some(item => item.name == user))
console.log(hasAllUsers)
// if(hasAllUser) {...}

It's a pretty roundabout way to zero in on the logic you're trying to express. Note how the result in hasMatch is never even used. That's really all you're looking for, does the array "have the values".
There's no need to push values to another array and check if that array has values. Just check of the original array has them.
Which could be as simple as:
let result = [{"name": "ABC"},{"name": "CDE"},{"name": "FGH"},{"name": "XYZ"}];
if (result.filter(r => r.name === "FGH" || r.name === "ABC").length === 2) {
// do logic
}
Or if you want to refactor the condition into a variable:
let result = [{"name": "ABC"},{"name": "CDE"},{"name": "FGH"},{"name": "XYZ"}];
let hasMatch = result.filter(r => r.name === "FGH" || r.name === "ABC").length === 2;
if (hasMatch) {
// do logic
}
Or a bit more verbose for clarity:
let result = [{"name": "ABC"},{"name": "CDE"},{"name": "FGH"},{"name": "XYZ"}];
let filteredResult = result.filter(r => r.name === "FGH" || r.name === "ABC");
let hasMatch = filteredResult.length === 2;
if (hasMatch) {
// do logic
}

You can simply create another array with the valid users and filter your array to match each items that are this array.
This can be done using the Array#includes method
const users = [{"name": "ABC"},{"name": "CDE"},{"name": "FGH"},{"name": "XYZ"}];
const validUsers = ["ABC", "FGH", "AnotherUser"];
const matchUsers = users.filter(user => validUsers.includes(user.name))
console.log(matchUsers)

You could count the wanted names.
const
data = [{ name: "ABC" }, { name: "CDE" }, { name: "FGH" }, { name: "XYZ" }],
names = ['ABC', 'FGH'],
result = data.reduce((t, { name }) => t + names.includes(name), 0);
console.log(result);

Try using a named function and pass in the array, key, and one or more values with the rest operator ...values. Use .flatMap() to filter with
[...values].includes(obj[key])
// ["ABC", "XYZ"].includes(obj.name)
and any non-match returns an empty array []. The final return is an array with a sub-array and the length of said sub-array.
const result = [["ABC", "XYZ"], 2]
// result[0][0] = "ABC"
// result[0][1] = "XYZ"
// result[1] = 2
const arr = [{"name": "ABC"},{"name": "CDE"},{"name": "FGH"},{"name": "XYZ"}];
function hasMatch(array, key, ...values) {
const result = array.flatMap((obj, idx) =>
[...values].includes(obj[key]) ? obj : []);
return [result, result.length];
}
console.log(hasMatch(arr, "name", "ABC", "XYZ"));
console.log(hasMatch(arr, "name", "FGH", "IJK", "LMN", "ABC", "XYZ"));

Filter Multidimensional Array with Another Array Values - JavaScript [duplicate]

I have two array, One of them is like this:
const array1 = [
Object {
"ItemId": 1,
},
Object {
"ItemId": 2,
},
]
other array is :
const array2 = [
Object {
"obj1": Object {
"Id": 4736,
},
"obj2": Object {
"ItemId": 1,
},
}
]
I want to get items in array1, which are not equals in obj2 of array2 .
I tried this but doesnt work
array1.filter(function (item) {
return array2.map((x) => {
return x.obj2 != item;
});

instead of array2.map, you're looking for Array.protype.some or Array.prototype.every. Why?
array1.filter is expecting a truthy value for every item.
array2.map is returning an array (!) with the results of its callback. In your case it's the result of each comparisson x.obj2 != item. It could look something like this: [true, false]. But this array will always evaluate to true.
array2.some() or array2.every() also iterate over the array but will return a boolean for the condition you're using. You can try for yourself and check the code sample afterwards.
const array1 = [
{
"ItemId": 1,
},
{
"ItemId": 2,
},
];
const array2 = [
{
"obj1": {
"ItemId": 4736,
},
"obj2": {
"ItemId": 1,
},
},
];
// errornous filter usage
const test = array1.filter(item => {
const arr = array2.map(x => x.obj2 !== item);
return arr;
});
// suggestion
const filteredResult = array1.filter(item => {
// both ways work
// return array2.every(x => x.obj2.ItemId !== item.ItemId);
return !array2.some(x => x.obj2.ItemId === item.ItemId);
});
console.log('initial result', test);
console.log('correct result', filteredResult);

Try this
array1=array1.filter((item) => {
return array2[0].obj2.ItemId !== item.ItemId;
})

Convert an array of objects to a dictionary by letter [duplicate]

This question already has answers here:
Most efficient method to groupby on an array of objects
(58 answers)
Closed 3 years ago.
I have an array of objects that looks like this:
let stuff = [
{
"id": "48202847",
"name": "Doe"
},
{
"id": "17508",
"name": "Marie"
},
{
"id": "175796",
"name": "Robert"
},
{
"id": "175796",
"name": "Ronald"
},
]
What I want to get is a dictionary looking something like this:
{
"D": [{"id": "48202847", "name": "Doe"}],
"M": [{"id": "17508", "name": "Marie"}],
"R": [{"id": "175796", "name": "Robert"}, {"id": "175796", "name": "Ronald"}]
}
Notice how all the people whose name starts with "R" are listed under one key.
This is my function that creates a dictionary with the person's name as the key:
const byId = (array) =>
array.reduce((obj, item) => {
obj[item.name] = item
return obj
}, {})
But this obviously doesn't do what I want it to. I do have some ideas of how to make this possible, but they are extremely legacy and I would love to know how to do this right.
Any help is appreciated!

You need the first character, uppercase and an array for collecting the objects.
const byId = array =>
array.reduce((obj, item) => {
var key = item.name[0].toUpperCase(); // take first character, uppercase
obj[key] = obj[key] || []; // create array if not exists
obj[key].push(item); // push item
return obj
}, {});
let stuff = [{ id: "48202847", name: "Doe" }, { id: "17508", name: "Marie" }, { id: "175796", name: "Robert" }, { id: "175796", name: "Ronald" }],
result = byId(stuff)
console.log(result);

Here's a solution based on Set, map, reduce and filter:
let stuff = [{"id": "48202847","name": "Doe"},{"id": "17508","name": "Marie"},{"id": "175796","name": "Robert"},{"id": "175796","name": "Ronald"}];
let result = [...new Set(stuff.map(x => x.name[0]))]
.reduce((acc, val) => {
return acc = { ...acc,
[val]: stuff.filter(x => x.name.startsWith(val))
}
}, {});
console.log(result);

Great solution Nina! Could be made a little cleaner by utilizing the spread operator.
const byId = (array) =>
array.reduce((obj, item) => {
var key = item.name[0].toUpperCase();
return {
...obj,
[key]: obj[key] ? [...obj[key], item] : [item],
}
}, {});

Find difference between two arrays

I have following Plunkr which works perfectly.
https://plnkr.co/edit/WDjoEK7bAVpKSJbAmB9D?p=preview
It uses the _.differenceWith() function of lodash, in order two save all array values, which are not contained by the two arrays.
var result = _.differenceWith(data, test, _.isEqual);
Now I have two problems:
1.) In our project we use an older Lodash Version where the function differenceWith is not implemented
2.) I only need to compare one value of the array. This currently compares the complete objects. I only need to compare the id property.

This will find the objects in arr1 that are not in arr2 based on the id attribute.
var arr1 = [ { "id": "1" }, { "id": "2" }, { "id": "3" } ];
var arr2 = [ { "id": "1" }, { "id": "2" } ];
var result = arr1.filter(o1 => arr2.filter(o2 => o2.id === o1.id).length === 0);
console.log(result);
Note that this example does not require lodash.
If you want to use a different comparison instead of id, you can change the o2.id === o1.id part to a different property.
Here is a more generic solution:
var arr1 = [ { "name": "a" }, { "name": "b" }, { "name": "c" } ];
var arr2 = [ { "name": "a" }, { "name": "c" } ];
function differenceWith(a1, a2, prop) {
return a1.filter(o1 => a2.filter(o2 => o2[prop] === o1[prop]).length === 0);
}
var result = differenceWith(arr1, arr2, 'name');
console.log(result);