How to convert values in array of objects

Question

Given this array of values:

let values = [
    ["2020-01-01", 1, 3, 2],
    ["2020-02-01", 10, 30, 20],
    ["2020-03-01", 100, 300, 200]
]

How can I convert it in something like this?:

public struct Point {
    var date: String
    var price: Double
}

let expectedResult: [[Point]] = [
    [Point(date: "2020-01-01", price: 1), Point(date: "2020-02-01", price: 10), Point(date: "2020-03-01", price: 100)],
    [Point(date: "2020-01-01", price: 3), Point(date: "2020-02-01", price: 30), Point(date: "2020-03-01", price: 300)],
    [Point(date: "2020-01-01", price: 2), Point(date: "2020-02-01", price: 20), Point(date: "2020-03-01", price: 200)]
]

I expect values to always have 1 String and a different number of Doubles, from that I need an array of Point for each of those Doubles.

I've tried different approaches with .map or nested for loops but I never get to the desired result or I end up with a very hard to read code.

A String is not a Date. You should use Date. And you should also be careful of using Double for currency, as it'll give you perhaps unexpected results. stackoverflow.com/questions/3730019/… — Alexander
– Alexander, Commented May 9, 2021 at 12:55
Yeah I'm not using either String or Double, I just put them here to keep the example simple, the strings in the example could as well be "A", "B", "C" — Ludyem
– Ludyem, Commented May 9, 2021 at 12:58

Esraa Ragab · Accepted Answer · 2021-05-09 12:16:27Z

1

you can do this:

import UIKit

let values = [
    ["2020-01-01", 1.0, 3.0, 2.0],
    ["2020-02-01", 10.0, 30.0, 20.0],
    ["2020-03-01", 100.0, 300.0, 200.0]
]

public struct Point {
    var date: String
    var price: Double
}

var expectedResult: [[Point]] = []

for i in 1..<values[0].count {
    var points: [Point] = []
    for j in 0..<values.count {
        points.append(Point(date: values[j][0] as! String, price: values[j][i] as! Double))
    }
    expectedResult.append(points)
}

print(expectedResult)

answered May 9, 2021 at 12:16

Esraa Ragab

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1 Comment

Alexander Over a year ago

Those explicit for loops can just be replaced by a simple map and flatMap.

Alexander · Accepted Answer · 2021-05-09 12:54:40Z

0

The input is really quite suspicious ([Any] is never fun to work with, you should try to avoid it.), but all you need is two simple maps.

let values = [
    ["2020-01-01", 1, 3, 2],
    ["2020-02-01", 10, 30, 20],
    ["2020-03-01", 100, 300, 200]
]

public struct Point {
    var date: String
    var price: Double
}

let points: [[Point]] = values.map { pointData in
    let date = pointData.first! as! String // You REALLY should parse this into a `Date`.
    let prices = pointData.dropFirst()
    
    return prices.map { Point(date: date, price: Double($0 as! Int)) }
}

points.forEach { print($0) }

answered May 9, 2021 at 12:54

Alexander

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5 Comments

Ludyem Over a year ago

This doesn't give the expect result, for example the first array end up being: [Point(date: "2020-01-01", price: 1), Point(date: "2020-01-01", price: 3), Point(date: "2020-01-01", price: 2)] instead of [Point(date: "2020-01-01", price: 1), Point(date: "2020-02-01", price: 10), Point(date: "2020-03-01", price: 100)]

Alexander Over a year ago

Sounds like you just need to transpose the result. stackoverflow.com/q/39889568/3141234

Alexander Over a year ago

Now this is even more strange, the groupings you have look like this, imgur.com/a/7VDJT0b What's the pattern?

Ludyem Over a year ago

basically each array in values contains the points needed to make 3 charts. for example in: ["2020-01-01", 1, 3, 2] "2020-01-01" is the point in the X axis and 1 would be the point in the Y axis for the first chart, 3 and 2 are the Y points for other two charts with the same X axis value. That's why I need this to end up being 3 different arrays to make 3 different charts. It's way more complex then what it needed to be but I don't have a say in the way I get these values so I have to stick with this approach

Alexander Over a year ago

Interesting. Welp, transposing will the do the job.

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