Create a SwiftUI line graph using arrays and the Swift Charts framework

Question

In a SwiftUI app, I'm trying to pass an array of Y values to the Swift Charts framework to draw a line graph. I would like to create the X values based on the length or count of the Y data. I tried the approach shown below but it doesn't work because Charts wants the data as an array of structs. Is there a way to pass arrays directly to the Charts framework?

import SwiftUI
import Charts

struct ContentView: View {
    
    let ydata = [1, 4.5, 3, 6, 7, 5.2, 9, 12.5]
    let xdata = Array(0..<ydata.count)
    let data = [xdata, ydata]

    var body: some View {
        Chart(data) {
            LineMark(
                x: .value("X values", $0),
                y: .value("Y values", $1)
            )
        }
    }
}

Eero · Accepted Answer · 2023-03-17 12:46:51Z

4

This might help

import SwiftUI
import Charts

struct ContentView: View {
    
    let ydata = [1, 4.5, 3, 6, 7, 5.2, 9, 12.5]

    var body: some View {
        Chart(0..<ydata.count, id: \.self)
        { nr in
            LineMark(
                x: .value("X values", nr),
                y: .value("Y values", ydata[nr])
            )
        }
    }
}

answered Mar 17, 2023 at 12:46

Eero

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3 Comments

wigging Over a year ago

What does the nr represent?

Eero Over a year ago

It is counting array elements.

wigging Over a year ago

This is a cleaner approach than the other answer. Thank you for posting this.

Joakim Danielson · Accepted Answer · 2022-11-18 22:28:53Z

3

You can use an array of tuples, here I am using a computed property for the data

var data: [(Int, Double)] {
    Array(zip(Array(0..<ydata.count), ydata))
}

When using a tuple we need to tell Chart what the id is so the use of data is a little different

var body: some View {
    Chart(data, id: \.0) { tuple in
        LineMark(
            x: .value("X values", tuple.0),
            y: .value("Y values", tuple.1)
        )
    }
}

answered Nov 18, 2022 at 22:28

Joakim Danielson

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1 Comment

wigging Over a year ago

That works. Thank you for the help. You could also use $0 and $1 for the x and y values.

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