MetricsMachine Copy Transformation

Wei Huang
Wei Huang Posts: 98
edited February 2018 in Type Design Software
1.
Hi all, I'm trying to do a simple copy transformation but I have no idea how to do this:

I want to copy all kerning exceptions that oslash has to oslashacute (both have o for left and right kerning group).

I've read the Manual but I can't understand it, I have this setting but the problem is that it actually copies the oslash kerning exceptions into the [o group kerning?:


2.
Also for future reference, this is the image example in the manual

and here's an example text:
One to Many For example, A > A.sc or {A*.sc} or Aacute.sc or {Aacute.sc} will copy the values of all pairs containing A to A.sc, Aacute.sc , the group containing A.sc and the group containing Aacute.sc. You must explicitly de"ne groups in the Pattern or Replacement if you want to use a group.

How do I even enter that into the Copy panel? 

Pattern: 

A

Replacements Left box: 

A.sc or {A*.sc} or Aacute.sc or {Aacute.sc}

Replacements right box: ???

This doesn't work, any help is much appreciated!

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Comments

  • Kent Lew
    Kent Lew Posts: 944
    I’ve never found that panel in MetricsMachine to be very useful. In order to be versatile, it ends up being confusing instead.

    Seems like what you want to do would be more easily accomplished with a simple Python script from within your UFO editor:
    f = CurrentFont()<br><br>for (l, r), value in f.kerning.items():<br>    if l == 'oslash':<br>        f.kerning[('oslashacute', r)] = value<br>        # print l, r, value<br>    if r == 'oslash':<br>        f.kerning[(l, 'oslashacute')] = value<br>        # print l, r, value<br><br>

    Uncomment those 'print' commands if you want to have a report on which oslash pairs were copied.
  • Wei Huang
    Wei Huang Posts: 98
    Answer in the thread: 
  • Kent Lew
    Kent Lew Posts: 944
    Wei Huang said:
    Answer in the thread: 
    ??! Sorry, I’m not on Twitter. If you ask a question here, you should expect answers here. If that doesn’t work for you, please just let me know so I don’t waste time trying to try to help next time.
  • Mark Simonson
    Mark Simonson Posts: 1,739
    I think he means that the question has been answered (by Tal) on Twitter.
  • Kent Lew
    Kent Lew Posts: 944
    Oh, okay. Thanks, Mark. If that’s true, then I’m sorry to be so quick to take offense. Like I said: Not on Twitter.
  • Mark Simonson
    Mark Simonson Posts: 1,739
    I respect that.  :)
  • Wei Huang
    Wei Huang Posts: 98
    Thank you for your answer @Kent Lew, I hadn't seen your answer before, meant that everyone can check Twitter for some answers including from Tal.

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