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Jeffrey Martin3
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Q: Using the new Transpose function

It's great to see the new Transpose function, as well as many other improvements in Numbers 09.

I am trying to transpose a 3 x 3 array. Perhaps I am thinking too Excel, but I put the transpose formula in a destination cell or even a range of cells, and it seems to do nothing. In Excel, the result is a 3x3 array, but I'm not at all getting how use this functino in Numbers. Must I also apply an index function along with it in each cell? Seems odd if this is so, but any insight is greatly appreciated.

MacBook Pro, Mac OS X (10.5.6)

Posted on Jan 12, 2009 5:11 PM

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Jeffrey Martin3

Q: Using the new Transpose function

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  • by KOENIG Yvan,

    KOENIG Yvan Jan 25, 2009 12:30 PM in response to Ralph McLaughlin
    Level 8 (41,790 points)
    Jan 25, 2009 12:30 PM in response to Ralph McLaughlin
    Hello

    At last I discovered that there was a bug in the posted script.
    I used two wrong variables so it was always adding new rows and columns.

    This time it is corrected.

    The correct script is available on my iDisk.

    Yvan KOENIG (from FRANCE dimanche 25 janvier 2009 21:30:31)

  • by rcarbaugh,

    rcarbaugh rcarbaugh Feb 24, 2009 9:03 PM in response to KOENIG Yvan
    Level 1 (0 points)
    Feb 24, 2009 9:03 PM in response to KOENIG Yvan
    Mr. Koenig:
    You are a tireless helper with a lot of patience for the rest of us. Thank you for your investment in helping us use Numbers. I just started using Numbers today, being a big Appleworks fan since ClarisWorks3, with a working knowledge of Excel. The transpose function has sent me into this forum, and I have read many of your posts in my research.

    Maybe this explanation might help someone who is having trouble with the syntax in the guide of TRANSPOSE. What was in the guide was really scary for me.

    As to the TRANSPOSE function, it seems like it does not do what we all were hoping. I shortened up your shortcut a little, and simply used the INDEX function, without the TRANSPOSE function. Let's say you have a 6 row by 2 column series, A1:F2, and your target is the series C11:D16. If you put a reference index series in a A11 to A16 of integers 1-6, then in the row B10 to C10 of 1-2, then C11 =INDEX($A$1:$F$2,B$10,$A11). Fill that down and right in all the cells, and you have a transposed array.

    I didn't follow the row()-row(1) that you had, which is probably even easier, but this one works and it is pretty easy, flexible and independent of array size (providing you adjust the indices, of course).
    RC<

  • by KOENIG Yvan,

    KOENIG Yvan Feb 25, 2009 5:46 AM in response to rcarbaugh
    Level 8 (41,790 points)
    Feb 25, 2009 5:46 AM in response to rcarbaugh
    If you are able to build the formulas by yourself, it's perfect.
    The script was written for those which aren't or which have no time to spent with that.
    When you write the formulas by yourself, you may work with references like $B12.
    When writing a script, it is much more efficient to use formulas which doesn't use this kind of reference because it requires that the script calculates every reference one by one.
    With my formula, it builds one formula and apply it to every cells in the range.

    Let me add that formulas using references of the kind B12 are not robust enough to resist to a sort while mine is.

    But I repeat, nobody is forced to use available tools.
    The number of questions asked here when the response is easily available in the Help is a perfect prooh of this statement.

    Yvan KOENIG (from FRANCE mercredi 25 février 2009 14:46:31)
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