5
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I have a list of directed edges:

list={1 \[DirectedEdge] 1, 1 \[DirectedEdge] 10, 2 \[DirectedEdge] 2, 
3 \[DirectedEdge] 3, 3 \[DirectedEdge] 5, 3 \[DirectedEdge] 7, 
3 \[DirectedEdge] 8, 3 \[DirectedEdge] 9, 3 \[DirectedEdge] 11, 
4 \[DirectedEdge] 4, 5 \[DirectedEdge] 2, 5 \[DirectedEdge] 5, 
5 \[DirectedEdge] 7, 5 \[DirectedEdge] 8, 5 \[DirectedEdge] 9, 
5 \[DirectedEdge] 11, 5 \[DirectedEdge] 14, 6 \[DirectedEdge] 1, 
6 \[DirectedEdge] 2, 6 \[DirectedEdge] 4, 6 \[DirectedEdge] 5, 
6 \[DirectedEdge] 6, 6 \[DirectedEdge] 7, 6 \[DirectedEdge] 8, 
6 \[DirectedEdge] 9, 6 \[DirectedEdge] 10, 6 \[DirectedEdge] 11, 
6 \[DirectedEdge] 12, 6 \[DirectedEdge] 13, 6 \[DirectedEdge] 14, 
6 \[DirectedEdge] 15, 6 \[DirectedEdge] 16, 6 \[DirectedEdge] 17, 
7 \[DirectedEdge] 7, 7 \[DirectedEdge] 9, 8 \[DirectedEdge] 8, 
8 \[DirectedEdge] 9, 8 \[DirectedEdge] 11, 8 \[DirectedEdge] 14, 
8 \[DirectedEdge] 16, 8 \[DirectedEdge] 17, 9 \[DirectedEdge] 9, 
10 \[DirectedEdge] 9, 10 \[DirectedEdge] 10, 11 \[DirectedEdge] 2, 
11 \[DirectedEdge] 8, 11 \[DirectedEdge] 9, 11 \[DirectedEdge] 11, 
11 \[DirectedEdge] 15, 11 \[DirectedEdge] 16, 11 \[DirectedEdge] 17, 
12 \[DirectedEdge] 9, 12 \[DirectedEdge] 12, 13 \[DirectedEdge] 9, 
13 \[DirectedEdge] 13, 13 \[DirectedEdge] 16, 13 \[DirectedEdge] 17, 
14 \[DirectedEdge] 14, 15 \[DirectedEdge] 9, 15 \[DirectedEdge] 15, 
16 \[DirectedEdge] 9, 16 \[DirectedEdge] 16, 17 \[DirectedEdge] 17};

I want to draw a histogram in the following structure. The list above consists of In- and Out-links, and I like to construct a histogram combining both In and Out link numbers with different colors in the bars of the histogram. For example, if 5 links go out of vertex 1 and 10 links come in to vertex 1, the bar associated with vertex 1 should have a height of 15. But then, I should be able to see the bars in two colors to visualize the In and Out sample.

I know that Degree, InDegree and OutDegree are the related functions for calculating the data that will be used in constructing the histogram, but I have no idea how I put them in the above structure.

The ideal (limited with my knowledge) solution is to write a Mathematica function, the argument of which is a list of directed edges.

Thank you for your time.

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4 Answers 4

Reset to default
8
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Updated to return chart of loop-free part

You can use BarChart for this (incorporating halmir's suggestions):

inOutGraph[e_] := With[{g = SimpleGraph @ Graph[Sort @ VertexList[e], e]},
    BarChart[
        Transpose @ Through @ {VertexInDegree, VertexOutDegree} @ g,
        ChartLayout -> "Stacked"
    ]
]

inOutGraph[list]

enter image description here

or:

inOutGraph[e_] := With[{g = SimpleGraph @ Graph[Sort @ VertexList[e], e]},
    BarChart[
        Transpose @ Through @ {VertexInDegree, Minus @* VertexOutDegree} @ g,
        ChartLayout -> "Stacked"
    ]
]

inOutGraph[list]

enter image description here

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6
  • $\begingroup$ Very nice. You even chose the colors I was imagining. Thanks. $\endgroup$
    – Tugrul Temel
    Commented May 10, 2019 at 22:59
  • 1
    $\begingroup$ The BarChart does not appear to have the vertices in order. E.g. 3 has 1 in and 6 out but your chart shows a total of 5 for 3 instead if 7.. $\endgroup$
    – Edmund
    Commented May 10, 2019 at 23:05
  • $\begingroup$ @Edmund Thanks for letting me know $\endgroup$
    – Carl Woll
    Commented May 10, 2019 at 23:11
  • $\begingroup$ @Edmund: Yes, I was about to post the same question you raised. $\endgroup$
    – Tugrul Temel
    Commented May 10, 2019 at 23:13
  • 1
    $\begingroup$ @Edmund just define g = Graph[Range[17], list] and use it. $\endgroup$
    – halmir
    Commented May 11, 2019 at 1:44
7
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g1 = SimpleGraph @ Graph[Range[17], list];
{ind, outd} = Through[{VertexInDegree, VertexOutDegree}[g1]];

GraphComputation`GraphPropertyChart

GraphComputation`GraphPropertyChart[g1, Automatic -> {ind, outd}, 
 ChartStyle -> {Blue, Yellow}, VertexShapeFunction -> None, 
 VertexSize -> .75, VertexLabels -> Placed["Name", Center], 
 ChartLegends -> {"VertexInDegree", "VertexOutDegree"}]

enter image description here

Add the options PolarGridLines->False and PolarAxes->False to get

enter image description here

PairedBarChart

PairedBarChart[Style[#, ColorData[97][1]] & /@ ind, 
 Style[#, ColorData[97][2]] & /@ outd, BarSpacing -> {3, 0, 0}, 
 BarOrigin -> Top, ChartLabels -> Placed[Range[17], Axis]]

enter image description here

PairedHistogram + WeightedData

PairedHistogram[Style[WeightedData[Range@17, ind], ColorData[97][1]], 
 Style[WeightedData[Range@17, outd],  ColorData[97][2]], 17, 
 BarSpacing -> 3, BarOrigin -> Top, Axes -> {False, True}, 
 Epilog -> (Text[#, {#, 0}] & /@ Range[17])]

enter image description here

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5
  • 1
    $\begingroup$ It is really excellent. I am amazed with your extended and very useful answers, which answered many other questions of mine. Thank you so much. $\endgroup$
    – Tugrul Temel
    Commented May 11, 2019 at 9:26
  • $\begingroup$ @Tugrul, my pleasure. $\endgroup$
    – kglr
    Commented May 11, 2019 at 9:32
  • $\begingroup$ With n=17 (number of vertices) a single vertex can have max of 32 links, excluding self loops. Can we set the size of the circle in your 1st polar histogram to 2(n-1)=32 so that I can easily identify those sectors performing the best in the given digraph? Thanks. $\endgroup$
    – Tugrul Temel
    Commented May 11, 2019 at 10:53
  • 1
    $\begingroup$ @Tugrul, it should be possible but it seems to be non-trivial (several things I tried with PolarAxes, PolarGridLines ... did not give the desired result.) $\endgroup$
    – kglr
    Commented May 11, 2019 at 12:56
  • $\begingroup$ Thank you for your trials. I appreciate it. Regards. $\endgroup$
    – Tugrul Temel
    Commented May 11, 2019 at 13:15
3
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With ordered vertices

vl = VertexList@list;
degree = Association @@ Thread[{"In", "Out"} -> Through@{VertexInDegree, VertexOutDegree}@list];
degree = #[[Ordering[vl]]] & /@ degree;
vl = vl[[Ordering[vl]]];

BarChart[Transpose@Values@degree,
 ChartLayout -> "Stacked",
 ChartLabels -> {vl, None},
 ChartLegends -> SwatchLegend[Automatic, {"In", "Out"}]]

Mathematica graphics

Hope this helps.

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1
  • $\begingroup$ and @Carl: Would be possible to exclude SelfLoops from the histogram? $\endgroup$
    – Tugrul Temel
    Commented May 10, 2019 at 23:26
2
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An alternative with ordered vertices, and self-loops removed:

listToCounts[list_] := Module[{l2, maxvertex, goingout, goingin},
   l2 = DeleteCases[list, (a_ \[DirectedEdge] a_)];
   maxvertex = Max[{l2[[All, 1]], l2[[All, 2]]}];
   goingout = Table[Count[l2[[All, 1]], vertex], {vertex, maxvertex}];
   goingin = Table[Count[l2[[All, 2]], vertex], {vertex, maxvertex}];
   Transpose[{goingin, goingout}]
   ];

counts = listToCounts[list]

{{1, 1}, {3, 0}, {0, 5}, {1, 0}, {2, 6}, {0, 15}, {3, 1}, {4, 5}, {11,0}, {2, 1}, {4, 6}, {1, 1}, {1, 3}, {3, 0}, {2, 1}, {4, 1}, {4, 0}}

BarChart[counts, ChartLayout -> "Stacked", ChartLegends -> {"in", "out"}, ChartStyle -> {Orange, Blue}, ChartLabels -> {ToString[#] & /@ Range[Max[{list[[All, 1]], list[[All, 2]]}]], None}]

enter image description here

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