图形化自动缩放

Question

我有以下代码来画一个单位圆

open System
open Microsoft.FSharp.Collections
open Microsoft.FSharp.Math
open System.Drawing
open System.Windows.Forms

let make_point (x:float) (y:float) = (fun bit -> if bit = 0.0 then x else y)
let x_of (point:float->float) = point 0.0
let y_of (point:float->float) = point 1.0

let unit_circle (t:float) = 
    make_point (sin <| 2.0 * Math.PI * t)
               (cos <| 2.0 * Math.PI * t)
let draw_connected (curve:float->float->float) (values: float list)=
    let form = new Form(Text = "Curve")
    let drawCurve (g:Graphics) = 
        for t in values do
            let p = curve t        
            g.DrawEllipse(Pens.Red, 
                          float32 (x_of p * 50.0 + (float)form.ClientSize.Width / 2.0), 
                          float32 (y_of p * 50.0 + (float)form.ClientSize.Height / 2.0), 
                          float32 1, 
                          float32 1)
    form.Paint.Add(fun e -> drawCurve e.Graphics)    
    form.Show()

draw_connected unit_circle ([0.0 .. 0.01 .. 1.0])

我并不完全满意，因为我必须手动将x和y坐标“缩放”50以使圆可见。有没有办法让F#自动伸缩？

谢谢。

Answer 1

我认为代码将一个二维点表示为一个带有3个参数的函数--一个标志，x& y。该标志指示返回x和y中的哪一个。如果标志是布尔型而不是浮点型，对于开始来说(稍微)更有意义。我猜代码是从另一种只有浮点数的语言转换而来的？

下面是一个更容易理解的版本：

open System
open Microsoft.FSharp.Collections
open Microsoft.FSharp.Math
open System.Drawing
open System.Windows.Forms
open System.Threading

type Point = {x : float; y : float}

let unit_circle (angle : float) = 
    {
        x = (sin <| 2.0 * Math.PI * angle)
        y = (cos <| 2.0 * Math.PI * angle)
    }

let draw_connected (curve : float -> Point) (radius : float) (angles : float list) =
    let form = new Form(Text = "Curve")
    let drawCurve (gfx : Graphics) =
        for angle in angles do
            let p = curve angle        
            gfx.DrawEllipse(Pens.Red, 
                          float32 (p.x * radius + (float)form.ClientSize.Width / 2.0), 
                          float32 (p.y * radius + (float)form.ClientSize.Height / 2.0), 
                          float32 1,
                          float32 1)
    form.Paint.Add (fun pntEvntArgs -> drawCurve pntEvntArgs.Graphics)    
    form.Show ()
    form

let form = draw_connected unit_circle 50.0 ([0.0 .. 0.01 .. 1.0])

while form.Created do
    Thread.Sleep (1)
    Application.DoEvents ()
done

不确定为什么圆呈现为1个像素椭圆的集合。

在任何情况下，正如Tomas所说，要么圆必须缩放，要么坐标系缩放。否则你最终会得到一个1像素的圆圈。

Answer 2

我并没有完全理解您的代码，但也许您可以使用可以指定给Graphics对象的scale转换。这会更改Graphics的坐标系，因此您执行的所有绘图(例如，使用DrawEllipse)都会自动缩放-您可以通过某种方式设置缩放，使单位圆显示为半径为50的圆。

• 要设置转换，请使用ScaleTransfrom方法(请参见MSDN documentation of Graphics instance (代码中的值为g )。

Answer 3

正如Tomas所说，您可以使用缩放转换。如果要使用小曲线绘制圆，可以使用多个DrawCurve调用:)

为此，我对jon的代码做了一些修改：

• 使用System.Drawing.Point类型而不是System.Drawing.Point unit_circle，以便它返回表示坐标x和y的元组。
• 将角度列表转换为角度序列。这将非常有用，因为我们可以为曲线(基数样条曲线)提供可变数量的节点，由常数N
• implemented a splitRepeatEvery方法表示，例如：

Seq.splitRepeatEvery 3 { 1 .. 10 }返回seq [seq [1; 2; 3]; seq [3; 4; 5]; seq [5; 6; 7]; seq [7; 8; 9]; seq [9; 10]]

代码如下：

module Seq =
    /// Split a sequence into pieces, each n items long
    /// repeating elements between start and end of subsequences.
    let splitRepeatEvery (count : int) (source : seq<'T>) = 
        if not (count > 1) then failwith "count must be superior to 1"
        seq { use e = source.GetEnumerator()
              let hasNext = ref (e.MoveNext())
              while !hasNext do
                 let (block:option<'T>[]) = Array.create count None

                 for i in 0 .. count - 1 do
                     do block.[i] <- if !hasNext then Some(e.Current) else None
                     if (i <> count - 1) then do hasNext := e.MoveNext()

                 yield seq { yield! block }
                 |> Seq.filter (fun x -> x.IsSome)
                 |> Seq.map (function Some(e) -> e | _ -> failwith "" ) }

let unit_circle (angle : float) = 
    (sin <| 2.0 * Math.PI * angle), (cos <| 2.0 * Math.PI * angle)

let draw_connected curve radius (seqOfAngles : float seq seq) knotsCount =
    let form = new Form(Text = "Curve")

    let computeBoundingBox points =
        let search f acc array =
            Array.fold (fun (x,y) (p:Point) -> f p.X x, f p.Y y) acc array
        let minX, minY = search min (form.ClientSize.Width, form.ClientSize.Height) points
        let maxX, maxY = search max (0,0) points
        new Rectangle(minX, minY, abs(minX-maxX), abs(minY-maxY))

    let drawCurves (gfx : Graphics) =
        // Create a buffer for storing our knots
        let buffer = Array.create knotsCount (new Point())
        let mutable i = 0

        for angles in seqOfAngles do
            for angle in angles do
                let x, y = curve angle
                let X = int(x * radius + (float)form.ClientSize.Width  / 2.0)
                let Y = int(y * radius + (float)form.ClientSize.Height / 2.0)
                let P = new Point(X, Y)
                buffer.[i] <- P
                i <- i + 1

            let knots = buffer.[0..(i-1)]
            // Draw spline only if we have one or more knots
            if knots.Length <> 1 then
                gfx.DrawCurve(Pens.Red, knots)
                // For debug: compute BBox of an array of points and draw it
                let debugRect = computeBoundingBox knots
                gfx.DrawRectangle(Pens.Black, debugRect)

            // Don't forget to reset position in buffer between each spline draw call
            i <- 0

    form.Paint.Add (fun pntEvntArgs -> drawCurves pntEvntArgs.Graphics)    
    form.Show ()
    form

// Define constants    
let STEP = 0.050
let N = 4
// Define a new sequence of sequence of angles
let s = {0.0 .. STEP .. 1.0} |> Seq.splitRepeatEvery N
let form = draw_connected unit_circle 120.0 s N

// For debug: print sequence of sequence of angles
s |> Seq.iter (fun s -> Seq.iter (fun x -> printf "%f " x) s; printfn "")

while form.Created do
    Thread.Sleep (1)
    Application.DoEvents ()
done

您可以使用不同值的N (样条线的结数)和STEP (但请注意，应选择STEP，以便1.0f是STEP的倍数或浮点数，以便最后一个序列的最后一个元素足够接近1.0f，否则最后一个样条线将无法连接到第一个！)。瞧！

alt text http://img818.imageshack.us/img818/9765/circles3.png