I have two arrays:
const array = [
[1, 7, 'AAA'],
[2, 5, 'BBB'],
[3, 2, 'CCC'],
[4, 4, 'DDD'],
[4, 9, 'EEE'],
[4, 2, 'FFF'],
[5, 8, 'GGG'],
[6, 2, 'HHH']];
const names = [
[1, 'Joe'],
[2, 'Dave'],
[3, 'Mike'],
[4, 'Sandra'],
[5, 'Sue'],
[6, 'Mary']];
Based on the value in the first column, I want to sum the values in the array[1] and list the three-character letters. The result I'm trying to get is:
const names = [
[1, 'Joe',7,'AAA'],
[2, 'Dave',5,'BBB'],
[3, 'Mike',2,'CCC'],
[4, 'Sandra',15,'DDD, EEE, FFF'],
[5, 'Sue',8,'GGG'],
[6, 'Mary',2,'HHH']]
I'm not sure of the best approach, I'm fairly new to Javascript. What I've managed to do is get the right result when a value in array[0] isn't repeated, but I can't get a sum or list to work.
const counter = (array,value) => array.filter((v) => (v === value)).length;
const arrayCol = (array,value) => array.map(v => v[value]);
const sum = (prevVal, curVal) => prevVal + curVal;
names.forEach ((p,e) => {
array.forEach ((v,x) => (counter(arrayCol(array,0),v[0])===1) ?
(v[0]===p[0]) && names[e].push(v[1],v[2]) :
(v[0]===p[0]) && names[e].push(array.reduce(sum,0)) );
});
console.log(names);
I'm sure the answer has to do with map or filter but not sure how... any pointers appreciated. Thank you
EDIT: All three answers below (from Michael Haddad, Nina Scholz, and testing_22) work and are interesting.
You could collect all data for each group and then map the result in order of the names array.
const
array = [[1, 7, 'AAA'], [2, 5, 'BBB'], [3, 2, 'CCC'], [4, 4, 'DDD'], [4, 9, 'EEE'], [4, 2, 'FFF'], [5, 8, 'GGG'], [6, 2, 'HHH']],
names = [[1, 'Joe'], [2, 'Dave'], [3, 'Mike'], [4, 'Sandra'], [5, 'Sue'], [6, 'Mary']],
groups = array.reduce((r, [id, value, code]) => {
r[id] ??= [0, ''];
r[id][0] += value;
r[id][1] += (r[id][1] && ', ') + code;
return r;
}, {}),
result = names.map(a => [...a, ...groups[a[0]]]);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }You can use a combination of map and reduce, as in:
const array = [[1, 7, 'AAA'], [2, 5, 'BBB'], [3, 2, 'CCC'],[4, 4, 'DDD'], [4, 9, 'EEE'], [4, 2, 'FFF'], [5, 8, 'GGG'], [6, 2, 'HHH']];
const names = [[1, 'Joe'],[2, 'Dave'],[3, 'Mike'],[4, 'Sandra'],[5, 'Sue'],[6, 'Mary']];
const result = names.map(([id, name]) => {
let vals = [];
let sum = array.reduce((acc, [idx, number, XXX]) =>
(idx === id ? (vals.push(XXX), number) : 0) + acc, 0);
return [
id,
name,
sum,
vals.join(", ")
]
})
console.log(result)A basic approach could be:
const array = [[1, 7, 'AAA'], [2, 5, 'BBB'], [3, 2, 'CCC'], [4, 4, 'DDD'], [4, 9, 'EEE'], [4, 2, 'FFF'], [5, 8, 'GGG'], [6, 2, 'HHH']];
const names = [[1, 'Joe'], [2, 'Dave'], [3, 'Mike'], [4, 'Sandra'], [5, 'Sue'], [6, 'Mary']];
let result = [];
for (let name of names) {
let newValue = [...name, 0];
let matchingItems = array.filter(i => i[0] === name[0]);
let strings = []; // for lack of a better name...
for (let item of matchingItems) {
newValue[2] += item[1];
strings.push(item[2]);
}
newValue.push(strings.join(", "));
result.push(newValue);
}
console.log(result);You could also implement the joining logic yourself (I actually prefer this version for readability reasons):
const array = [[1, 7, 'AAA'], [2, 5, 'BBB'], [3, 2, 'CCC'], [4, 4, 'DDD'], [4, 9, 'EEE'], [4, 2, 'FFF'], [5, 8, 'GGG'], [6, 2, 'HHH']];
const names = [[1, 'Joe'], [2, 'Dave'], [3, 'Mike'], [4, 'Sandra'], [5, 'Sue'], [6, 'Mary']];
let result = [];
for (let name of names) {
let newValue = [...name, 0, ""];
let matchingItems = array.filter(i => i[0] === name[0]);
for (let item of matchingItems) {
newValue[2] += item[1];
newValue[3] += newValue[3] === "" ? item[2] : `, ${item[2]}`;
}
result.push(newValue);
}
console.log(result);