有没有办法在 mathematica 中绘制一组具有相同原点的线? [英] Is there a way to draw a set of lines in mathematica all with the same origin point?

问题描述

我有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画一条线.换句话说，从原点 (0,0) 到上述集合中的所有单个点绘制向量.有没有办法做到这一点?到目前为止，我已经尝试了 Filling 选项、PlotPoints 和 VectorPlot，但它们似乎无法做我想做的事.

解决方案

先简单后难:

Graphics[{Arrow[{{0, 0}, #}] &/@列表1}]

Graphics[{Arrow[{{0, 0}, #}] &/@ list1}，轴 ->真的]

Needs["PlotLegends`"];list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};k = ColorData[22, "ColorList"][[;;长度@list1]];图形行[{图形[Riffle[k, Arrow[{{0, 0}, #}] &/@#]，轴 ->真的]，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"][[;;长度@list1]];ls = 序列[粗, 线[{{0, 0}, {1, 0}}]];图形行[{图形[Riffle[k, Arrow[{{0, 0}, #}] &/@#]，轴 ->真的]，图形@Legend[MapThread[{Graphics[{#1, ls}], #2} &, {k, #}]]}] &@list1

Needs["PlotLegends`"];list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};pr = {Min@#, Max@#} &//@转置@list1;k = ColorData[22, "ColorList"][[;;长度@list1]];图形行[{图形[r = Riffle[k, {Thick,Arrow[{{0, 0}, #}]} &/@#]，轴 ->真的]，图形@图例[地图线程[{图形[#1，轴 ->是的，滴答声 ->无，PlotRange ->pr],Text@Style[#2, 20]} &,{分区[r, 2], #}]]}] &@list1

你也可以调整 ListVectorPlot，虽然我不明白你为什么要这样做，因为它不打算这样使用:

list1 = {{3, 1}, {1, 3}, {-1, 2}, {-1, -1}, {1, -2}};数据 = 表[{i/2, -i/Norm[i]}, {i, list1}];ListVectorPlot[data, VectorPoints ->全部，VectorScale ->{1, 1, 范数[{#1, #2}] &},矢量样式 ->{箭头[{-.05, 0}]}]

I have a set of points given by this list:

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

I would like Mathematica to draw a line from the origin to all the points above. In other words draw vectors from the origin (0,0) to all the individual points in the above set. Is there a way to do this? So far I've tried the Filling option, PlotPoints and VectorPlot but they don't seem to be able to do what I want.

解决方案

Starting easy, and then increasing difficulty:

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

You could also tweak ListVectorPlot, although I don't see why you should do it, as it is not intended to use like this:

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}]}]

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