Group Nested array by value and find average

Question

Most other answer shows hot to group array of object but not array of arrays. Looking to group nested arrays and find average of the rest of the items

I am looking to group an array by the url (2nd item in array) and find average of rest of the item. So we should be able to group all url with same path for eg "/url/ranking" which have can be found in 2 sub arrays whereas /different/url in three. I am also looking to find average of rest of the numbers when we group the array based on the url.

[
 [
    'randomstring4',
    "/url/ranking/",
    12.1176,
    76.8978,
    8.8990
 ],
 [
    'randomstring3',
    "/url/ranking/",
    33.76,
    24.8978,
    7.8990
 ],
 [
    'randomstring2',
    "/different/url/",
    42.1176,
    26.8978,
    83.8990
 ],
 [
    'randomstring1',
    "/different/url/",
    33.76,
    24.8978,
    73.8990
 ],
 [
    'randomstring1',
    "/different/url/",
    13.689,
    4.118,
    53.03
 ],

]

In above example we will group by the url and get average for rest of items for eg

[ 'randomstring4', "/url/ranking/", 22.93, // (12.1176+33.76)/2 50.89, 8.39 ]

This would provide a means to get scoring of url from different website and find and average score upon data produced.

Here is the result with one of the array item (rest may need to be calculated accordingly)

[ 
'randomstring4', 
"/url/ranking/", 
22.93, // (12.1176+33.76)/2 50.89, 8.39 ]
 (76.8978+24.8978)/2),
( 8.8990+7.8990)/2
]

My trial here https://codepen.io/anon/pen/ymJbPe

please add the code to the question along with the error or what not work. — Nina Scholz
– Nina Scholz, Commented Jul 25, 2019 at 6:49
It should be the whole array grouped by arr[2] item _(arr).groupBy(row => row[2]) so '/url/ranking/' then /different/url/ and so on — June
– June, Commented Jul 25, 2019 at 7:00

Nina Scholz · Accepted Answer · 2019-07-25 07:06:37Z

2

You could group the items first by taking the value with their sum and count and render the result array with the grouped data.

var data = [['randomstring4', "/url/ranking/", 12.1176, 76.8978, 8.8990], ['randomstring3', "/url/ranking/", 33.76, 24.8978, 7.8990], ['randomstring2', "/different/url/", 42.1176, 26.8978, 83.8990], ['randomstring1', "/different/url/", 33.76, 24.8978, 73.8990], ['randomstring1', "/different/url/", 13.689, 4.118, 53.03]],
    groups = data.reduce((g, array) => {
        var key = array[1];
        g[key] = g[key] || [];
        array.slice(2).forEach((v, i) => {
            g[key][i] = g[key][i] || { sum: 0, count: 0 };
            g[key][i].sum += v;
            g[key][i].count++;
        });
        return g;
    }, {}),
    result = Object
        .entries(groups)
        .map(([group, values]) => (['', group, ...values.map(({ sum, count }) => sum / count)]));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Jul 25, 2019 at 7:06

Nina Scholz

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5 Comments

Kunal Mukherjee Over a year ago

Nice solution, for now its unclear what the first item of the list may be you have taken it as to be empty string ''

Nina Scholz Over a year ago

i am unsure where this part is coming from. actually it could be the first appearance of the array of the group an its first string.

Kunal Mukherjee Over a year ago

the OP hasn't mentioned that situation

June Over a year ago

+1. Nicely made. Is it possible to add key each of the final result score so it is easy to identify them rather than being an array result could be an object ` { url: "/url/ranking/", "site1.com": 22.9388, "site2.com": 50.897800000000004, "site3.com": 8.399 }`

Nina Scholz Over a year ago

@June, then you need another array with the wanted keys.

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