I have created one example for you to understand the problem.
let arr = [];
arr.push({
module_id: 41,
name: 'first'
}, {
module_id: 41,
name: 'second',
important: true,
}, {
module_id: 43,
name: 'third'
});
const lookup = arr.reduce((a, e) => {
a[e.module_id] = ++a[e.module_id] || 0;
return a;
}, {});
console.log('lookup is', lookup);
let unique = [];
arr.filter((e) => {
if (lookup[e.module_id] === 0) {
unique.push(e)
}
})
console.log('unique', unique)first, I have an empty array arr and I am pushing 3 objects in it.
Notice, the module_name. There are two repeatings and I'd like to use the second one with the property name important.
I have used reduce here for finding out which one is repeating based on module_name. it'll return 1 with the key of 41 and 0 for 43. I want to use both but I don't want to duplicate into my unique array. Currently, I'll push only those elements which are unique, in our case module_id: 43.
Now how can I get the duplicate value (with important property)?
You could take a Map and get only distinct first values or the ones with important flag.
const
array = [{ module_id: 41, name: 'first' }, { module_id: 41, name: 'second', important: true }, { module_id: 43, name: 'third' }],
result = Array.from(array
.reduce(
(m, o) => m.set(o.module_id, !o.important && m.get(o.module_id) || o),
new Map
)
.values()
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }try
let unique = [];
arr.filter((e) => {
if (lookup[e.module_id] === 0) {
unique.push(e)
}
else if (e.important) {
unique.push(e)
}
})
let arr = [];
arr.push(
{
module_id: 41,
name: 'first',
},
{
module_id: 41,
name: 'second',
important: true,
},
{
module_id: 43,
name: 'third',
}
);
const result = arr.reduce(
(acc, curr) =>
acc.find((v) => v.module_id === curr.module_id) ? acc : [...acc, curr],
[]
);
console.log(result);