Google Apps Script: filter one 2D array based off the closest timestamp of another 2D array - google-apps-script

Question

I have two 2D-arrays; both of which contain a timestamp and a float-value. Array1 contains hundreds of timestamp-value pairs, while Array2 will generally only contain ~10 timestamp-value pairs.
I am trying to compare Array1 with Array2, and only want to keep Array1's timestamp-value pairings for the timestamps that are closest to the timestamps in Array2.
Array1 = {[59.696, 2020-12-30T00:00:00.000Z],
[61.381, 2020-12-30T00:15:00.000Z],
[59.25, 2020-12-30T00:30:00.000Z],
[57.313, 2020-12-30T00:45:00.000Z],...}
Array2 = {[78.210, 2020-12-30T00:06:00.000Z],
[116.32, 2020-12-30T00:39:00.000Z],...}
So in these Array examples above, I would want Array1 to be filtered down to:
Array1 = {[59.696, 2020-12-30T00:00:00.000Z],
[57.313, 2020-12-30T00:45:00.000Z],...}
because those timestamps are the closest match for timestamps in Array2.
I have tried implementing the suggested code from Find matches for two columns on two sheets google script but can't get it to work, and I can't otherwise find a neat solution for the timestamp matching.
Any help is much appreciated. Let me know if I need to edit or add to my question with more information.

Answer 1

The goal
Given two arrays:
Array1 = [
[59.696, "2020-12-30T00:00:00.000Z"],
[61.381, "2020-12-30T00:15:00.000Z"],
[59.25, "2020-12-30T00:30:00.000Z"],
[57.313, "2020-12-30T00:45:00.000Z"]
]
Array2 = [
[78.210, "2020-12-30T00:06:00.000Z"],
[116.32, "2020-12-30T00:39:00.000Z"]
]
The goal is for each item in Array2 to be matched with the closest date in Array1. So the resultant array of the above example would be 2 items. If Array2 had 100 items and Array1 had 1000, the resultant array would be of 100 items.
I am assuming that each item in Array1 can only be used once.
I am also assuming that the first item in the array, the float value, is to be ignored in the calculation, but kept together with the date and be included in the output.
The script
function filterClosestTimes(array1, array2) {
// Initializing "maps" to keep track of the differences
// closest, will contain the final pairs used
// differences, is used to decide which pair gets used
// positions used, is so that the same pair in array 1
// doesn't get used twice.
closest = []
differences = []
positionsUsed = []
// For each member of array2
array2.forEach((pair, i) => {
// Initializing a date object
targetDate = new Date(pair[1])
// Initializing the position of the current index in the
// tracking arrays.
closest.push(null)
differences.push(Infinity)
positionsUsed.push(null)
// Going through each member of array 1
array1.forEach((subpair, j) => {
// First checking if the position has already been used
if (positionsUsed.includes(j)) return
//initializing date
dateToTest= new Date(subpair[1])
// checking difference between the currently testing date
// of array 2
difference = Math.abs(targetDate - dateToTest)
// Checking if it is the date with the smallest difference
// if so, setting the position in the tracking arrays.
// These values will likely be overwritten many times until
// the date with the least difference is found.
if (differences[i] > difference) {
differences[i] = difference
closest[i] = subpair
positionsUsed[i] = j
}
})
})
return closest
}
function test(){
Logger.log(filterClosestTimes(Array1, Array2))
}
Running test returns:
[["59.696, 2020-12-30T00:00:00.000Z"], ["57.313, 2020-12-30T00:45:00.000Z"]]
Note
This algorithm, seeing as it involves checking every element of one array with almost every one in another array, can get slow. Though if you are only dealing with hundreds of values and comparing with ~10 in your Array2 then it will be fine. Just be aware that this approach has O(n^2) time complexity. Which will mean that as the number of comparisons go up, the time needed to complete the operation increases exponentially. If you try to compare tens of thousands with tens of thousands, then there will be a noticeable wait!
References
JS Date object
Time Complexity