Creating new arrays from an existing array that holds values with dates

Question

I have an array with values that have dates attached to them.

-x[0].value = 5,    x[0].time = "Mon 24 April 2012"
-x[1].value = 12,   x[1].time = "Mon 24 April 2012"
-x[2].value = 11,   x[2].time = "Mon 23 April 2012"
-x[3].value = 2,    x[3].time = "Mon 20 April 2012"
-x[4].value = 11,   x[4].time = "Mon 20 April 2012"
-x[5].value = 7,    x[5].time = "Mon 20 April 2012"
-x[6].value = 7,    x[6].time = "Mon 20 April 2012"

How do i create many arrays according to similar dates from this array. E.g. end of the day i want.

data1 array will contain:

-x[0].value = 5,    x[0].time = "Mon 24 April 2012"
-x[1].value = 12,   x[1].time = "Mon 24 April 2012"

data2 array will contain:

-x[2].value = 11,   x[2].time = "Mon 23 April 2012"

data3 array will contain:

-x[3].value = 2,    x[3].time = "Mon 20 April 2012"
-x[4].value = 11,   x[4].time = "Mon 20 April 2012"
-x[5].value = 7,    x[5].time = "Mon 20 April 2012"
-x[6].value = 7,    x[6].time = "Mon 20 April 2012"

Your help will be appreciated.

This doesn't make any sense. You speak of data1 array and data2 array, but they are all in the same x array? — jfriend00
– jfriend00, Commented Oct 22, 2012 at 15:03
Alex, no the data cant be sorted correctly always. However in this case the dates are already sorted so thats why they remain the same. But all is fixed thanks. — user1469174
– user1469174, Commented Oct 24, 2012 at 12:36

T.J. Crowder · Accepted Answer · 2012-10-22 15:07:38Z

1

If I'm understanding you correctly, I'd probably loop through maintaining a temporary map keyed by the time values, and sort at the end:

var index;
var data;
var rentry;
var entry;
var map;

data = [];
map = {};
for (index = 0; index < x.length; ++index) { // Or forEach on ES5 systems
    entry = x[index];
    rentry = map[entry.time];
    if (!rentry) {
        rentry = map[entry.time] = [];
        data.push(rentry);
        rentry.time = entry.time;
    }
    rentry.push(entry);
}
map = undefined;
data.sort(function(a, b) {
    if (a.time < b.time) {
        return -1;
    }
    if (a.time > b.time) {
       return 1;
    }
    return 0;
});

Now data[0] has an array of entries with the lowest time value, data[1] the next highest time value, etc.

answered Oct 22, 2012 at 15:07

T.J. Crowder

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2 Comments

user1469174 Over a year ago

Sorry for late response. But thanks guys this worked out quite well. Exactly what i wanted.

T.J. Crowder Over a year ago

@user1469174: Good deal, glad that helped!

dfsq · Accepted Answer · 2012-10-22 15:08:41Z

1

What about this approach:

var arrs = {};
for (var i = 0; i < data.length; i++) {
    if (!arrs[data[i].time]) arrs[data[i].time] = [];
    arrs[data[i].time].push(data[i])
}

So for test data:

var data = [
    {value: 5, time: 'Mon 24 April 2012'},
    {value: 12, time: 'Mon 24 April 2012'},
    {value: 11, time: 'Mon 23 April 2012'},
    {value: 2, time: 'Mon 20 April 2012'},
    {value: 11, time: 'Mon 20 April 2012'},
    {value: 7, time: 'Mon 20 April 2012'},
    {value: 7, time: 'Mon 20 April 2012'},
];

It will create an object (not array) of the next structure:

arrs = {
    "Mon 24 April 2012": [
        {"value": 5, "time": "Mon 24 April 2012"},
        {"value": 12, "time": "Mon 24 April 2012"}
    ],
    "Mon 23 April 2012": [
        {"value": 11, "time": "Mon 23 April 2012"}
    ],
    "Mon 20 April 2012": [
        {"value": 2, "time": "Mon 20 April 2012"},
        {"value": 11, "time": "Mon 20 April 2012"},
        {"value": 7, "time": "Mon 20 April 2012"},
        {"value": 7, "time": "Mon 20 April 2012"}
    ]
}

answered Oct 22, 2012 at 15:08

dfsq

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1 Comment

user1469174 Over a year ago

Worked Great. At first i was getting no values, but is fixed now. thanks

I Hate Lazy · Accepted Answer · 2012-10-22 15:26:33Z

1

This will give you an Array of Arrays.

var dataGroups = data.sort(function(a, b) {
    return a.time.localeCompare(b.time);
}).reduce(function(result, obj) {
    if (result.length && obj.time === result[0][0].time)
        result[0].push(obj);
    else
        result.unshift([obj]);
    return result;
}, []);

I assume you didn't actually want separate incrementing identifiers for each Array, since that's usually not very useful.

http://jsfiddle.net/w6qE9/

The result:

[
    [
        {
            "value": 5,
            "time": "Mon 24 April 2012"
        },
        {
            "value": 12,
            "time": "Mon 24 April 2012"
        }
    ],
    [
        {
            "value": 11,
            "time": "Mon 23 April 2012"
        }
    ],
    [
        {
            "value": 2,
            "time": "Mon 20 April 2012"
        },
        {
            "value": 11,
            "time": "Mon 20 April 2012"
        },
        {
            "value": 7,
            "time": "Mon 20 April 2012"
        },
        {
            "value": 7,
            "time": "Mon 20 April 2012"
        }
    ]
]

edited Oct 22, 2012 at 15:26

answered Oct 22, 2012 at 15:11

I Hate Lazy

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1 Comment

user1469174 Over a year ago

Thanks for your immediate help. It helped a lot because i wanted to create something like this for easy calling to a chart.

jfriend00 · Accepted Answer · 2012-10-22 15:08:03Z

0

It is unclear from your question exactly what kind of result you are trying to achieve, but here is a function that will give you an array of arrays, one array for each time value:

var map = {}, i, time, temp, results;
for (i = 0; i < x.length; i++) {
    time = x[i].time;
    if (time in map) {
        // add this item to the array we already have for this time
        map[time].push(x[i]);
    } else {
        // create a new array for this time and put it in the map
        temp = [];
        temp.push(x[i]);
        map[time] = temp;
        results.push(temp);
    }
}

// here the variable results is an array of arrays, one for each time value

answered Oct 22, 2012 at 15:08

jfriend00

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2 Comments

user1469174 Over a year ago

I dont really care about dates. I wanted the values just sorted into dates so i can use them. Thanks, this was quite helpful.

user1469174 Over a year ago

Just thought i should say, reworked my code and this works like a charm thanks.

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