Reputation: 1620
I have an array with the following values (example):
[
1367848800000: true,
1367935200000: true,
1368021600000: true,
1368108000000: true,
1368194400000: true,
1368367200000: true,
1368540000000: true,
1368626400000: true,
1368712800000: true
]
Where the index is a date time. The date time will always be at 12:00:00 on a date.
In this example, the first five dates are consecutive, then one day by itself, and then another group of 3 dates. An example of what I mean is below.
Now, what I am trying to do is find sequential dates and put them into an array as follows:
[
1367848800000,
1367935200000,
1368021600000,
1368108000000,
1368194400000
],
[
1368367200000,
1368540000000,
1368626400000,
],
[
1368712800000Ω
]
So in the end, I have an array, with 3 arrays of all the times. I have tried numerous pieces of code, but everything bugs out and nothing is worth posting on here. Any help would be much appreciated!
Upvotes: 18
Views: 3597
Reputation: 4085
// Preconditions: singleArray contains the input array with each element corresponding to a time index. singleArray is sorted.
var outputArray = new Array();
var stack = new Array();
var stackSize = 0;
var i;
for( i = 0; i < singleArray.length; i++ )
{
// Get the last element on the stack
var lastElement = (stackSize == 0) ? 0 : stack.pop();
// Compare to see if difference is one day
if( singleArray[i] - lastElement == 86400000 ) // 24 * 60 * 60 * 1000
{
// Dates are 1 day apart
if( lastElement != 0 ) stack.push(lastElement);
stack.push(singleArray[i]);
stackSize++;
}
else
{
if( lastElement != 0 ) stack.push(lastElement);
var tempQueue = new Array();
while(stackSize > 0)
{
// Build up a new array containing consecutive days
// using a queue
tempQueue.push(stack.pop());
stackSize--;
}
// Push the consecutive days onto the next place in the output array.
outputArray.push(tempQueue);
// Start a new group of consecutive dates
stack.push(singleArray[i]);
stackSize++;
}
}
Upvotes: 1
Reputation: 4144
tried array sort
with forEach
var dates = [1367848800000, 1367935200000, 1368021600000,
1368108000000, 1368194400000, 1368367200000,
1368540000000, 1368626400000, 1368712800000];
var k = 0 , sorted = [[]];
dates.sort( function ( a, b ){
return +a > +b ? 1 : +a == +b ? 0: -1;
})
.forEach( function( v , i ){
var a = v,b = dates[i+1]||0;
sorted[k].push( +a );
if ( (+b - +a) > 86400000) {
sorted[++k] = []
}
});
Later you can sort them per counts
sorted.sort( function ( a,b ){
return a.length > b.length ? -1: 1;
});
The sorted
array contains desired result jsfiddle
Upvotes: 1
Reputation: 2126
Gotta love these puzzles. Nice answers everyone, here's mine more jQueryish approach.
var datearray = {
1367848800000: true,
1367935200000: true,
1368021600000: true,
1368108000000: true,
1368194400000: true,
1368367200000: true,
1368540000000: true,
1368626400000: true,
1368712800000: true
};
$(function() {
var result = dateSequences(datearray);
}
function dateSequences(array) {
// parse json object to array of keys
var keys = Object.keys(array);
// sort it up
keys = keys.sort();
// convert them to dates
var dates = new Array();
$.each(keys, function(i) {
dates.push(new Date(parseInt(keys[i])));
});
// now we have array of dates, search for sequential dates
var final = new Array();
var prevdate = undefined;
var currentseq = 0;
$.each(dates, function(i, d) {
// undefined?
// first sequence
if (prevdate == undefined) {
final.push(new Array());
final[currentseq].push(d);
}
else {
// compare if difference to current date in loop is greater than a day
var comp=new Date();
comp.setDate(prevdate.getDate()+2);
// Advance sequence if it is
if (comp < d) {
currentseq++;
final[currentseq] = new Array();
}
// Push the date to current sequence
final[currentseq].push(d);
}
// store previous
prevdate = d;
});
return final;
}
Fiddle:
Upvotes: 1
Reputation: 8651
Sth like this could do:
function sequentialize(dArr) {
dArr = Object.keys(dArr).slice().sort();
var last;
var arrs = [[]];
for (var i = 0, l = dArr.length; i < l; i++) {
var cur = new Date();
cur.setTime(dArr[i]);
last = last || cur;
if (isNewSequence(cur, last)) {
arrs.push([]);
}
arrs[arrs.length - 1].push(cur.getTime()); //always push to the last index
last = cur;
}
return arrs;
function isNewSequence(a, b) {
if (a.getTime() - b.getTime() > (24 * 60 * 60 * 1000))
return true;
return false;
}
}
Now if you pass your example Array/Object
to the sequentialize function
var dates = {
1367848800000: true,
1367935200000: true,
1368021600000: true,
1368108000000: true,
1368194400000: true,
1368367200000: true,
1368540000000: true,
1368626400000: true,
1368712800000: true
};
console.log(sequentialize(dates));
This gives the following output
[
[
1367848800000,
1367935200000,
1368021600000,
1368108000000,
1368194400000
],
[
1368367200000
],
[
1368540000000,
1368626400000,
1368712800000
]
]
This simply
creates an array out of the Date keys,
Sorts them
Iterates over them
If the difference of the Current and Last Date is greate than a day
Push a new Array to the Sequence Array
Push the Current Date to the last Array in the Sequence Array
Demo on JSBin
Note: You may have to change the isNewSequence
function to actually fit your needs
Upvotes: 1
Reputation: 145478
The following approach uses array .reduce()
method:
var arr = [1367848800000, 1367935200000, 1368021600000,
1368108000000, 1368194400000, 1368367200000,
1368540000000, 1368626400000, 1368712800000],
i = 0,
result = arr.reduce(function(stack, b) {
var cur = stack[i],
a = cur ? cur[cur.length-1] : 0;
if (b - a > 86400000) {
i++;
}
if (!stack[i])
stack[i] = [];
stack[i].push(b);
return stack;
}, []);
console.log(result);
DEMO: http://jsfiddle.net/gbC8B/1/
Upvotes: 4