Path Mapper List Divide

Replies to This Discussion

Permalink Reply by Danny Boyes on January 31, 2014 at 9:21am

I think this is what you are after:

{A;B;C}(i) --> {A;B;C;i\2}

note "\" is backslah divide or integer divide

Permalink Reply by Andrew Lyon on January 31, 2014 at 9:41am

worked like a charm. Can we use most operators in the path mapper?

Permalink Reply by David Rutten on January 31, 2014 at 9:43am

Yes, path mapper accepts all expressions.

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David Rutten

david@mcneel.com