画图框 [英] Plotting Deedle frame

问题描述

我有以下代码:

let mychart = frame.GetAllSeries() |> Seq.iter(fun key value -> Chart.Line(value, Name=key) |> Chart.Combine

其中，frame.GetAllSeries()返回seq<KeyValuePair>.我想将序列直接传递到图表中. 我以为我可以遍历整个序列.问题是我找不到可以直接插入到lambda表达式中的单独访问键和值的方法.

where frame.GetAllSeries() returns a seq<KeyValuePair>. I'd like to pipe the sequence directly in a chart. I thought I could iterate through the sequence. The problem is that I can't find an idomatic way to access the key and value separately that could be plugged in directly in the lambda expression.

谢谢.

编辑

这有效:

let chart = frame.GetAllSeries() |> Seq.map( fun (KeyValue(k,v)) -> Chart.Line(v |> Series.observations, Name=k)) |> Chart.Combine

..它可以变得更简单吗?我的数据集很大，恐怕性能会受到如此多转换的影响.

.. could it get simpler? I have a large dataset and I am afraid the performance gets impacted by so many transformations.

推荐答案

您已经弄清楚，使用Series.observations可以为您提供一系列的键值对序列，然后您可以将其传递给Chart.Line等.

As you already figured out, using Series.observations gives you a sequence of key-value pairs from the series that you can then pass to Chart.Line etc.

这绝对是不需要的，您可以使用扩展方法使代码更简单，该方法可以自动绘制序列:

This is definitely something that should not be needed and you can make the code simpler using extension method that lets you automatically plot a series:

[<AutoOpen>]
module FsLabExtensions =
  type FSharp.Charting.Chart with
    static member Line(data:Series<'K, 'V>, ?Name, ?Title, ?Labels, ?Color, ?XTitle, ?YTitle) =
      Chart.Line(Series.observations data, ?Name=Name, ?Title=Title, ?Labels=Labels, ?Color=Color, ?XTitle=XTitle, ?YTitle=YTitle)

如果包含此内容，则可以直接绘制序列:

If you include this, you can plot series directly:

let s = series [ for x in 0.0 .. 0.1 .. 1.0 -> x, sin x ]
Chart.Line(s)

您还可以参考Deedle&通过包含这些重载的实验软件包"FsLab"进行F#图表绘制(请参见此处)

You can also reference Deedle & F# Charting through our experimental package "FsLab" that includes these overloads (see here)

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