javascript using reduce function

Question

I have the below array

 ["0,5,p1", "24,29,p2", "78,83,p2", "78,83,p3", "162,167,p2" ]

i want the output as ["5,p1","10,p2","5,p3"] , so p1..3 are video files paying time with start and end time . 0,5 mean p1 profile played for 5 sec and so on. I want to know what profile take what time in total using ECMA script map,reduce function. Here is what i tried but it doesnt work:

 var ca =   uniqueArray.reduce(function(pval, elem) {
    var spl = elem.split(',');
            var difference = Math.round(spl[1] - spl[0]);
            return difference;
    },elem.split(',')[3]);

Isn't it 15,p2 ?

Serge K.
– Serge K.

2017-07-26 08:10:59 +00:00
Commented Jul 26, 2017 at 8:10 — Serge K.
– Serge K., Commented Jul 26, 2017 at 8:10
yes right , i think i missed it..thanks

Mohit H
– Mohit H

2017-07-26 08:14:25 +00:00
Commented Jul 26, 2017 at 8:14 — Mohit H
– Mohit H, Commented Jul 26, 2017 at 8:14

Jamiec · Accepted Answer · 2017-07-26 08:18:42Z

6

I dont think it can be done in one pass, but I could be wrong. I'd go for a 2 step...

Reduce the array to get unique map of pX values
Map the result back to an array in the required format

var input = ["0,5,p1", "24,29,p2", "78,83,p2", "78,83,p3", "162,167,p2" ] 

var step1 = input.reduce(function(p,c){
    var parts = c.split(",");
    if(!p[parts[2]])
       p[parts[2]] = 0;
    p[parts[2]] += parseInt(parts[1],10) - parseInt(parts[0],10);
    return p;
},{});

var result = Object.keys(step1).map(function(e){
    return step1[e] + "," + e;
});

console.log(result);

answered Jul 26, 2017 at 8:18

Jamiec

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Comments

Ben Taliadoros · Accepted Answer · 2017-07-26 08:28:47Z

1

You could use es6 map:

arrayWithNumbers.map(a => {var spl = a.split(','); return (spl[1] - spl[0]) + "," + spl[2]})

edited Jul 26, 2017 at 8:28

answered Jul 26, 2017 at 8:16

Ben Taliadoros

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3 Comments

prasanth Over a year ago

Its not fully satisfied .Op need to added the same arguments value with in array

Ben Taliadoros Over a year ago

@Jamiec I get ["5,p1", "5,p2", "5,p2", "5,p3", "5,p2"] - Is that not what he wanted?

Jamiec Over a year ago

@BenTaliadoros no. Check the question. He's essentially grouping and summing by pX

Nina Scholz · Accepted Answer · 2017-07-26 08:29:04Z

1

For a single loop approach, you could use a hash table for same third parts, like 'p1'. If a hash is given, then update the value with the actual delta.

var array = ["0,5,p1", "24,29,p2", "78,83,p2", "78,83,p3", "162,167,p2"],
    hash = Object.create(null),
    result = array.reduce(function(r, a) {
        var parts = a.split(','),
            delta = parts[1] - parts[0],
            key = parts[2];

        if (!(key in hash)) {
            hash[key] = r.push([delta, key].join()) - 1;
            return r;
        }
        r[hash[key]] = [+r[hash[key]].split(',')[0] + delta, key].join();
        return r;
    }, []);

console.log(result);

edited Jul 26, 2017 at 8:29

answered Jul 26, 2017 at 8:19

Nina Scholz

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Comments

Kannan J · Accepted Answer · 2017-07-26 08:53:30Z

0

I have updated the code. Please check now.

    var ca =   ["0,5,p1", "24,29,p2", "78,83,p2", "78,83,p3", "162,167,p2" ] .reduce(function(result, elem) {
        var spl = elem.split(',');
        var difference = Math.round(spl[1] - spl[0]);
        var found = false 
        for (var i = 0 ; i < result.length; i++) {
          if (result[i].split(',')[1] == spl[2]) {
            result[i] = parseInt(result[i].split(',')[0]) + difference+","+spl[2]; 
            found = true;
          }
        }
        if (!found) result.push(difference+","+spl[2]);       
        return result;
        },[]);
        
     console.log("modified array",ca);

edited Jul 26, 2017 at 8:53

answered Jul 26, 2017 at 8:27

Kannan J

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2 Comments

Kannan J Over a year ago

Someone put a comment as to where the running code is. You have to click on Run code snippet and the answer comes. In case, the expectation is something else, please post in comment so that I can help.

Serge K. Over a year ago

See OP post for expectations, but you don't meet them :/

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