轴的箭头

Question

使用Mathematica中的Plot命令时如何获取轴的箭头？

感谢您提供任何有用的答案。

Answer 1

对于由Plot生成的2D绘图，以下工作非常有用：

Plot[Sin[x], {x, 0, 10}, AxesStyle -> Arrowheads[0.07]]

或使用自定义箭头：

h = Graphics[Line[{{-1, 1/2}, {0, 0}, {-1, -1/2}}]];
Plot[Sin[x], {x, 0, 10}, 
 AxesStyle -> Arrowheads[{{Automatic, Automatic, h}}]]

Answer 2

以Sjoerd的答案为基础，

如

之类的情节

可以如下获得（例如）：

       Plot[Sin[x], {x, -2\[Pi], 2 \[Pi]},

 AxesStyle-> { 

Directive[{Red,
Arrowheads[{{-0.06,0(*Xleft*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}},
{0.03,.9(*Xright*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}}}]}], 

Directive[{Blue,
Arrowheads[{{-0.05,0(*Ydown*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}},{0.03,.8(*Yup*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}}}]}
]}]

Drawings Tools和Graphics Inspector中提供了很好的箭头示例。获取信息可能有更好的方法，但我annotate a plot带有我喜欢的箭头然后抽象（使用Simon的suggestion）：

Cases["Paste-Graphic_Here", Arrowheads[___], Infinity]

举另一个例子：

代码如下

Plot[Sin[x], {x, -2\[Pi],2 \[Pi]}, 
AxesStyle-> { Directive[{Red,

Arrowheads[{{-0.06,0.1(*Xleft*),
{Graphics[{arrowhead}]/.arrowhead-> arrowhead2,0.98`}},
{0.05,0.95(*Xright*),
{Graphics[{arrowhead}],0.98`}}}]/.arrowhead-> arrowhead4}], 

Directive[{Blue,
Arrowheads[{{-0.05,0(*Ydown*),
{Graphics[{arrowhead}]/.arrowhead-> arrowhead3,0.98`}},{0.03,.8(*Yup*),
{Graphics[{arrowhead}]/.arrowhead-> arrowhead1,0.98`}}}]}

]}]

<强>，其中

arrowhead1=Polygon[{{-1,0.5`},{0,0},{-1,-0.5`}}];

arrowhead2=Polygon[{{-1.5833333333333333`,0.4166666666666667`},{-1.5410500000000003`,0.369283333333333`},{-1.448333333333333`,0.255583333333333`},{-1.3991000000000005`,0.18721666666666673`},{-1.3564666666666663`,0.11826666666666673`},{-1.3268499999999999`,0.05408333333333341`},{-1.3166666666666667`,0.`},{-1.3268499999999999`,-0.048950000000000195`},{-1.3564666666666663`,-0.11228333333333372`},{-1.3991000000000005`,-0.18353333333333333`},{-1.448333333333333`,-0.2562833333333335`},{-1.5410500000000003`,-0.38048333333333345`},{-1.5833333333333333`,-0.43333333333333335`},{0.`,0.`},{-1.5833333333333333`,0.4166666666666667`},{-1.5833333333333333`,0.4166666666666667`}}];

arrowhead3=Polygon[{{-1,0.5`},{0,0},{-1,-0.5`},{-0.6`,0},{-1,0.5`}}];

arrowhead4={{FaceForm[GrayLevel[1]],Polygon[{{-0.6`,0},{-1.`,0.5`},{0.`,0},{-1.`,-0.5`},{-0.6`,0}}],Line[{{-0.6`,0},{-1.`,0.5`},{0.`,0},{-1.`,-0.5`},{-0.6`,0}}]}};

arrowhead5=Polygon[{{-0.6582278481012658`,-0.43037974683544306`},{0.`,0.`},{0.`,0.`},{0.`,0.`},{0.`,0.`},{0.`,0.`},{-0.6455696202531646`,0.43037974683544306`},{-0.4810126582278481`,0.`},{-0.6582278481012658`,-0.43037974683544306`},{-0.6582278481012658`,-0.43037974683544306`}}];

箭头1到5的列表：

Answer 3

Here您在https://math.stackexchange.com/

中发布了解决方案

由于参考文献中的解决方案是针对Plot3D，因此我在Plot []修改（但没有改进）它：

axes[x_, y_, f_, a_] := 
 Graphics[Join[{Arrowheads[a]}, 
   Arrow[{{0, 0}, #}] & /@ {{x, 0}, {0, y}}, 
   {Text[Style["x", FontSize -> Scaled[f]], {0.9*x, 0.1*y}], 
    Text[Style["y", FontSize -> Scaled[f]], {0.1 x, 0.95*y}]
   }]]

Show[Plot[Exp[-x^2], {x, -2, 2}, 
       Axes -> None, 
       PlotRange -> {{-2.1, 2.1}, {-.1, 1.1}}], 
     axes[2, 1, 0.05, 0.02]
]