有没有办法在mathematica中绘制一组具有相同原点的线？

Question

我有一个由这个列表给出的一组点:

list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};

我希望Mathematica从原点到上面的所有点绘制一条线.换句话说,从原点(0,0)到上述集合中的所有单个点绘制矢量.有没有办法做到这一点？到目前为止,我已经试过了Filling选项,PlotPoints以及VectorPlot,但他们似乎并不能够做我想做的.

Answer 1

开始容易,然后增加难度:

Graphics[{Arrow[{{0, 0}, #}] & /@ list1}]

Graphics[{Arrow[{{0, 0}, #}] & /@ list1}, Axes -> True]

Needs["PlotLegends`"];
list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};
k = ColorData[22, "ColorList"][[;; Length@list1]];

GraphicsRow[{
    Graphics[Riffle[k, Arrow[{{0, 0}, #}] & /@ #], Axes -> True], 
    Graphics@Legend[Table[{k[[i]], #[[i]]}, {i, Length@#}]]}] &@list1

Needs["PlotLegends`"];
list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};
k = ColorData[22, "ColorList"][[;; Length@list1]];
ls = Sequence[Thick, Line[{{0, 0}, {1, 0}}]];

GraphicsRow[{
    Graphics[Riffle[k, Arrow[{{0, 0}, #}] & /@ #], Axes -> True], 
    Graphics@Legend[MapThread[{Graphics[{#1, ls}], #2} &, {k, #}]]}] &@list1

Needs["PlotLegends`"];
list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};
pr = {Min@#, Max@#} & /@ Transpose@list1;
k = ColorData[22, "ColorList"][[;; Length@list1]];

GraphicsRow[{
    Graphics[r = Riffle[k, {Thick,Arrow[{{0, 0}, #}]} & /@ #], Axes -> True], 
    Graphics@
     Legend[MapThread[
             {Graphics[#1, Axes -> True, Ticks -> None, PlotRange -> pr], 
              Text@Style[#2, 20]} &, 
             {Partition[r, 2], #}]]}] &@list1

你也可以调整ListVectorPlot,虽然我不明白为什么你应该这样做,因为它不打算像这样使用:

list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};
data = Table[{i/2, -i/Norm[i]}, {i, list1}];
ListVectorPlot[data, VectorPoints -> All, 
                     VectorScale  -> {1, 1, Norm[{#1, #2}] &}, 
                     VectorStyle  -> {Arrowheads[{-.05, 0}]}]

Answer 2

Graphics[
 {
  Line[{{0, 0}, #}] & /@ list1
  }
 ]

其中/@是函数的简写中缀表示法Map.

我想知道你为什么试过Filling,Plotpoints而且VectorPlot.我必须假设你根本没有阅读过这些文档,因为即使是一个肤浅的阅读也会告诉你这些命令和选项与你正在寻找的功能无关.