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How do I sort items into objects via timestamps
How do you get a timestamp in JavaScript?How do I check if an array includes an object in JavaScript?Should I use the datetime or timestamp data type in MySQL?How to insert an item into an array at a specific index (JavaScript)?How do I sort a dictionary by value?Sorting an array of JavaScript objects by propertySort array of objects by string property valueHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?How to pair socks from a pile efficiently?
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I have two arrays of objects, which I want to combine into one array in order of the timestamps that are a part of each object. The first array is an exact second, and the second array is a range of an hour. I want to input the 'humidity' values in the array with the exact timestamps into the one with the hourly ranges, according to where they fit in, time wise. How do I do this in typescript? Both arrays/objects are sorted by time.
For example:
[id: 1, timestampUtc: "2019-02-22T08:24:00Z", humidity: 74,
id: 2, timestampUtc: "2019-02-24T06:20:00Z", humidity: 39,
id: 3, timestampUtc: "2019-02-26T020:03:00Z", humidity: 35]
and
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z",endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12]
to look like:
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03, humidity: 74,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3, humidity: 39,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z", endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12, , humidity: 35]
arrays typescript sorting timestamp
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
add a comment |
I have two arrays of objects, which I want to combine into one array in order of the timestamps that are a part of each object. The first array is an exact second, and the second array is a range of an hour. I want to input the 'humidity' values in the array with the exact timestamps into the one with the hourly ranges, according to where they fit in, time wise. How do I do this in typescript? Both arrays/objects are sorted by time.
For example:
[id: 1, timestampUtc: "2019-02-22T08:24:00Z", humidity: 74,
id: 2, timestampUtc: "2019-02-24T06:20:00Z", humidity: 39,
id: 3, timestampUtc: "2019-02-26T020:03:00Z", humidity: 35]
and
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z",endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12]
to look like:
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03, humidity: 74,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3, humidity: 39,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z", endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12, , humidity: 35]
arrays typescript sorting timestamp
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
add a comment |
I have two arrays of objects, which I want to combine into one array in order of the timestamps that are a part of each object. The first array is an exact second, and the second array is a range of an hour. I want to input the 'humidity' values in the array with the exact timestamps into the one with the hourly ranges, according to where they fit in, time wise. How do I do this in typescript? Both arrays/objects are sorted by time.
For example:
[id: 1, timestampUtc: "2019-02-22T08:24:00Z", humidity: 74,
id: 2, timestampUtc: "2019-02-24T06:20:00Z", humidity: 39,
id: 3, timestampUtc: "2019-02-26T020:03:00Z", humidity: 35]
and
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z",endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12]
to look like:
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03, humidity: 74,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3, humidity: 39,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z", endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12, , humidity: 35]
arrays typescript sorting timestamp
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
I have two arrays of objects, which I want to combine into one array in order of the timestamps that are a part of each object. The first array is an exact second, and the second array is a range of an hour. I want to input the 'humidity' values in the array with the exact timestamps into the one with the hourly ranges, according to where they fit in, time wise. How do I do this in typescript? Both arrays/objects are sorted by time.
For example:
[id: 1, timestampUtc: "2019-02-22T08:24:00Z", humidity: 74,
id: 2, timestampUtc: "2019-02-24T06:20:00Z", humidity: 39,
id: 3, timestampUtc: "2019-02-26T020:03:00Z", humidity: 35]
and
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z",endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12]
to look like:
[id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03, humidity: 74,
id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3, humidity: 39,
id: 6, starttimestampUtc: "2019-02-26T020:00:00Z", endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12, , humidity: 35]
arrays typescript sorting timestamp
arrays typescript sorting timestamp
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
edited Mar 25 at 19:53
LaurenAH
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
asked Mar 25 at 19:47
LaurenAHLaurenAH
2332 silver badges11 bronze badges
2332 silver badges11 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I was able to do this by converting the UTC timestamps into date objects. The following code seemed to work for me, where a and b are your two arrays. I also noticed that there is an extra 0 in the timestamps of the final element of each array.
b.map((element) =>
let entry = a.find((e) =>
let d = new Date(e.timestampUtc)
let start = new Date(element.starttimestampUtc)
let end = new Date(element.endtimestampUtc)
return (d >= start && d <= end)
)
if(entry !== undefined)
element.humidity = entry.humidity
return element
)
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
I was able to do this by converting the UTC timestamps into date objects. The following code seemed to work for me, where a and b are your two arrays. I also noticed that there is an extra 0 in the timestamps of the final element of each array.
b.map((element) =>
let entry = a.find((e) =>
let d = new Date(e.timestampUtc)
let start = new Date(element.starttimestampUtc)
let end = new Date(element.endtimestampUtc)
return (d >= start && d <= end)
)
if(entry !== undefined)
element.humidity = entry.humidity
return element
)
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
add a comment |
I was able to do this by converting the UTC timestamps into date objects. The following code seemed to work for me, where a and b are your two arrays. I also noticed that there is an extra 0 in the timestamps of the final element of each array.
b.map((element) =>
let entry = a.find((e) =>
let d = new Date(e.timestampUtc)
let start = new Date(element.starttimestampUtc)
let end = new Date(element.endtimestampUtc)
return (d >= start && d <= end)
)
if(entry !== undefined)
element.humidity = entry.humidity
return element
)
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
add a comment |
I was able to do this by converting the UTC timestamps into date objects. The following code seemed to work for me, where a and b are your two arrays. I also noticed that there is an extra 0 in the timestamps of the final element of each array.
b.map((element) =>
let entry = a.find((e) =>
let d = new Date(e.timestampUtc)
let start = new Date(element.starttimestampUtc)
let end = new Date(element.endtimestampUtc)
return (d >= start && d <= end)
)
if(entry !== undefined)
element.humidity = entry.humidity
return element
)
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
I was able to do this by converting the UTC timestamps into date objects. The following code seemed to work for me, where a and b are your two arrays. I also noticed that there is an extra 0 in the timestamps of the final element of each array.
b.map((element) =>
let entry = a.find((e) =>
let d = new Date(e.timestampUtc)
let start = new Date(element.starttimestampUtc)
let end = new Date(element.endtimestampUtc)
return (d >= start && d <= end)
)
if(entry !== undefined)
element.humidity = entry.humidity
return element
)
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
edited Mar 25 at 20:24
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
answered Mar 25 at 20:13
Cody DeckardCody Deckard
11 bronze badge
11 bronze badge
add a comment |
add a comment |
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