Mathematica's Plot 中多个函数的检测和样式 [英] Detection and styling of multiple functions in Mathematica's Plot

问题描述

这个问题让我开始思考 Mathematica 如何检测正在绘制的多个函数.我发现我真的不明白这个过程.

This question started me thinking about how Mathematica detects multiple functions being plotted. I find that I really do not understand the process.

考虑:

Plot[{1, Sequence[2, 3], 4}, {x, 0, 1}, PlotRange -> {0, 5}]

我可以理解 Plot 最初在列表中找到三个元素，但是它如何知道"设置 2 和 3 的样式?相同的?就好像有一个关于这两个元素来自起始列表的哪一部分的记忆.这是如何工作的?

I can understand that Plot finds three elements in the list initially, but how does it "know" to style 2 and 3 the same? It is as though there is a memory of what part of the starting list those two elements came from. How does this work?

推荐答案

好吧，它知道只有三个参数:

Well, it knows that there three arguments just so:

In[13]:= Function[x, Length[Unevaluated[x]], HoldAll][{1, 
  Sequence[2, 3], 4}]

Out[13]= 3

如果允许 x 求值，则

If x is allowed to evaluate, then

In[14]:= Function[x, Length[x], HoldAll][{1, Sequence[2, 3], 4}]

Out[14]= 4

:

In[15]:= Hold[{1, Sequence[2, 3], 4}]

Out[15]= Hold[{1, Sequence[2, 3], 4}]

换句话说，序列的扁平化需要评估器.

in other words, flattening of Sequence requires evaluator.

编辑 2: 我显然错过了提出的真正问题，现在将尝试回答.

EDIT 2: I clearly missed the real question posed and will try to answer it now.

Once Plot 确定它构建的参数数量 {{ style1, Line ..}, {style2, Line..}, ... }.在 {1,Sequence[2,3],4} 的情况下，我们得到以下结构:

Once Plot determines the number of argument it builds {{ style1, Line ..}, {style2, Line..}, ... }. In the case of {1,Sequence[2,3],4} we get the following structure:

In[23]:= Cases[
  Plot[{1, Sequence[2, 3], 4}, {x, 0, 1}, 
   PlotRange -> {0, 5}], {_Hue, __Line}, 
  Infinity] /. {x_Line :> Line, _Hue -> Hue}

Out[23]= {{Hue, Line}, {Hue, Line, Line}, {Hue, Line}}

绘制 {1,{2,3},4} 时，我们得到不同的结构:

When plotting {1,{2,3},4} we get a different structure:

In[24]:= Cases[
  Plot[{1, List[2, 3], 4}, {x, 0, 1}, 
   PlotRange -> {0, 5}], {_Hue, __Line}, 
  Infinity] /. {x_Line :> Line, _Hue -> Hue}

Out[24]= {{Hue, Line}, {Hue, Line}, {Hue, Line}, {Hue, Line}}

因为列表会变平，只是不使用评估器.因此，如您所见，出现相同颜色的标记是因为 Sequence[2,3] 被视为一个返回两个元素列表的黑盒函数:

because lists would be flattened, just not using the evaluator. So as you see the tagging in the same color occurs because Sequence[2,3] is treated as a black-box function which returns a list of two elements:

In[25]:= g[x_?NumberQ] := {2, 3}

In[26]:= Cases[
  Plot[{1, g[x], 4}, {x, 0, 1}, PlotRange -> {0, 5}], {_Hue, __Line}, 
  Infinity] /. {x_Line :> Line, _Hue -> Hue}

Out[26]= {{Hue, Line}, {Hue, Line, Line}, {Hue, Line}}

我试图构建一个顶级实现来构建这样的结构，但必须与评估器作斗争.例如:

I was trying to build a top-level implementation which would build such a structure, but one has to fight the evaluator. For example:

In[28]:= Thread /@ Function[x,
   Thread[{Hold @@ {Range[Length[Unevaluated[x]]]}, Hold[x]}, Hold]
   , HoldAll][{1, Sequence[2, 3], 4}]

Out[28]= Hold[Thread[{{1, 2, 3}, {1, Sequence[2, 3], 4}}]]

现在我们必须评估线程而不评估其参数，这将给出{{1, 1}, {2, Sequence[2,3]}, {3, 4}}，其中列表的第一个元素是一个标签，后面的一次是要采样的函数.

Now we have to evaluate the Thread without evaluating its arguments, which would give {{1, 1}, {2, Sequence[2,3]}, {3, 4}}, where the first element of the list is a tag, and the subsequent once are functions to be sampled.

希望这会有所帮助.

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