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Re-exploring a way to rank all of the same age the same. Rather than { rank=same; <label 1>, <label 2>, <label 3> } as I am using right now, I think a better way would be to create subgraphs (labeled based on age) containing all of a particular age then connect adjacent (with gaps) ages with invisible edges style = invis
This will have the effect of boxing all of the same age together and implicitly sets `rank="same``.
Another route is use of the group attribute where things in the same group are aligned on the perpendicular axis (i.e., LR alignment if rankdir="TB").
I should likely try activating newrank=true which will make rank constraints take precedence over edge constraints.
newrank might also allow my intermediate plan which was the use of both an implicit rank=same via subgraphs based on age AND group=<string> for perhaps aligning on the perpendicular axis.
New rank was tried and decided to be kept as the default as it slightly improved the readability of the graph in testing. When grouping based on presumed lineage the result was not what I had hoped so this complicated behavior was not retained moving forward. Latest work this effect was ef7c251 but other than forcing older individuals to point toward younger individuals, there was not way to force more pedigree formatting constraints to improve the readability further.
Re-exploring a way to rank all of the same age the same. Rather than
{ rank=same; <label 1>, <label 2>, <label 3> }
as I am using right now, I think a better way would be to create subgraphs (labeled based on age) containing all of a particular age then connect adjacent (with gaps) ages with invisible edgesstyle = invis
This will have the effect of boxing all of the same age together and implicitly sets `rank="same``.
Another route is use of the
group
attribute where things in the same group are aligned on the perpendicular axis (i.e., LR alignment ifrankdir="TB"
).Originally posted by @rhagenson in #4 (comment)
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