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I have a table with 3 columns - x position, y-position and colour. I'd like to plot these points and join them with line segments, where the colour of the line segment is determined by 3rd variable. e.g.


I can plot just the points like this

Graphics[{Hue[#3/Length[T]], Point[{#1, #2}]} & @@@ T, Frame -> True, AspectRatio -> 1]

How can I plot line segments instead?


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It seems to be a duplicate but meanwhile try: {Hue[#1[[ 3]]/Length[T]], Line[{##}[[ ;; , ;; 2]]]} & @@@ Partition[T, 2, 1], –  Kuba Aug 26 '13 at 10:58
There are 100 points and 100 colors, which color you want to apply to line between n and n+1 point? the n.? –  Kuba Aug 26 '13 at 11:35
Possibly a duplicate of this question –  m_goldberg Aug 26 '13 at 15:47

2 Answers 2

up vote 6 down vote accepted
T = Table[{n, n, n}, {n, 0, 10}];  

   T[[;; , ;; 2]],
   {Thickness@.02, {Hue[T[[#[[ 1]], 3]]/len], Line@#} & /@ Partition[Range[len], 2, 1] 
], Frame -> True, AspectRatio -> 1]

enter image description here

So that's another approach, but it will blur your colors:

len=Length @ T;

      T[[ ;; , ;; 2]],
      {Thickness@.02, Line[Range[len], VertexColors -> (Hue[#/len] & /@ T[[ ;; , 3]])]}
        , Frame -> True, AspectRatio -> 1]

enter image description here


GraphicsComplex can be useful but for simple cases like here or the one I've faced today it is not a must:

T = Table[{n, n, n}, {n, 0, 10}];
len = Length@T;

          Line[T[[ All, {1, 2}]], 
               VertexColors -> (Hue[#/len] & /@ T[[;; , 3]])]}
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That's perfect thanks! I prefer the blurred one but I didn't know if it was possible. I don't have enough reputation to upvote the answer... sorry –  octopus Aug 26 '13 at 12:10

If I understood you correctly, - there is a simpler way:

data = Table[{x, Sinc[x]}, {x, 0, 10, .5}];

 ColorFunction -> Hue,
 Joined -> True, 
 PlotStyle -> Thickness[.03], 
 Mesh -> All, 
 MeshStyle -> Directive[PointSize[.05], Opacity[.2]]]

enter image description here

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as I understand, he'd explicilty give the color as 3rd argument of the individual points –  Pinguin Dirk Aug 26 '13 at 11:20
@PinguinDirk Yeah I thought it is a possibility. Still you never know. Additionally this is good to know and ColorFunction can be custom in case he just need to specify a simple data-dependent coloring scheme. –  Vitaliy Kaurov Aug 26 '13 at 11:46
perfectly understand- it is what I'd suggest to use too (I am too lazy to build up with primitives when there are built-in functions one could use) –  Pinguin Dirk Aug 26 '13 at 11:48
Many thanks for your answer. In reality, I need to give the colour as an extra argument. (I have a 2D trajectory and want to use the colour to track the time taken to reach each point) –  octopus Aug 26 '13 at 12:12

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