Swapping elements of list

Question

Say I have a directed graph inputted in the following format in Mathematica:

g = {{1, "a", 2}, {2, "b", 1}}

I'd like to plot this graph using GraphPlot in Mathematica, which is the following syntax:

GraphPlot[{{1 -> 2, "a"}, {2 -> 1, "b"}}, VertexLabeling -> True, DirectedEdges -> True, 
    VertexCoordinateRules -> Auto, VertexLabeling -> True]

How would I swap the middle elements and the last elements in any list like g, so that g would look closer to the argument of GraphPlot?

{{1 -> 2, "a"}, {2 -> 1, "b"}}

I have searched the Mathematica documentation but was unable to find a good way of doing this.

All you need is a simple replacement rule: g /. {f_Integer, m_String, l_Integer} :> {f -> l, m} — rm -rf, Commented Oct 28, 2013 at 3:48
Also g[[All, 1]] = Rule @@@ g[[All, {1, 3}]];g[[All, {1, 2}]] — Mike Honeychurch, Commented Oct 28, 2013 at 5:26

rm -rf · Accepted Answer · 2013-10-28 20:58:00Z

2

rm -rf has suggested the simplest solution which matches the elements and swaps them around:

g = {{1, "a", 2}, {2, "b", 1}};
g /. {a_Integer, b_String, c_Integer} :> {a -> c, b}
{{1 -> 2, "a"}, {2 -> 1, "b"}}

rm -rf♦

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answered Oct 28, 2013 at 4:13

bill s

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ubpdqn · Accepted Answer · 2013-10-28 04:26:36Z

3

Is this your intention:

Graph[Labeled[#1 \[DirectedEdge] #3, Style[#2, 20]] & @@@ g]

enter image description here

or with more formatting:

Graph[Labeled[#1 \[DirectedEdge] #3, Style[#2, 20]] & @@@ g, 
 VertexLabels -> Placed["Name", Center], VertexSize -> 0.2, 
 VertexLabelStyle -> 20]

enter image description here

answered Oct 28, 2013 at 4:04

ubpdqn

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$\begingroup$ Thanks for this-- yes, this also works, but alas I can only accept one answer. $\endgroup$
– user10245
Commented Oct 28, 2013 at 5:00

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2 Answers 2