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Array formula needed

  1. #1
    ZipCurs
    Guest

    Array formula needed

    I want to perform an sumproduct type of function with on column of data (1 x
    n) and the minimum value in each row of a 2 x n matrix. Is there a simple
    formula that will allow me to do this. I know that I could create a new
    column with the minimums or that I could brute force the math, I am just
    hoping for something simple since I have to apply it a few hundred times.

    Thanks

  2. #2
    B. R.Ramachandran
    Guest

    RE: Array formula needed

    Hi,

    If the (1 x n) range data are in Column A, and the (2 x n) matrix data are
    in Columns B and C, use the following formula:
    =SUMPRODUCT(A1:A100,IF(B1:B100<C1:C100,B1:B100,C1:C100)) and confirm with
    CTRL-SHIFT-ENTER.

    Regards,
    B. R. Ramachandran

    "ZipCurs" wrote:

    > I want to perform an sumproduct type of function with on column of data (1 x
    > n) and the minimum value in each row of a 2 x n matrix. Is there a simple
    > formula that will allow me to do this. I know that I could create a new
    > column with the minimums or that I could brute force the math, I am just
    > hoping for something simple since I have to apply it a few hundred times.
    >
    > Thanks


  3. #3
    ZipCurs
    Guest

    RE: Array formula needed

    Thanks, that did the trick. What if instead of a (2xn) matrix, I have one
    (1xn) matrix and one (nx1) matrix that I want to select the minimum from.
    This doesn't seem to work with that.

    "B. R.Ramachandran" wrote:

    > Hi,
    >
    > If the (1 x n) range data are in Column A, and the (2 x n) matrix data are
    > in Columns B and C, use the following formula:
    > =SUMPRODUCT(A1:A100,IF(B1:B100<C1:C100,B1:B100,C1:C100)) and confirm with
    > CTRL-SHIFT-ENTER.
    >
    > Regards,
    > B. R. Ramachandran
    >
    > "ZipCurs" wrote:
    >
    > > I want to perform an sumproduct type of function with on column of data (1 x
    > > n) and the minimum value in each row of a 2 x n matrix. Is there a simple
    > > formula that will allow me to do this. I know that I could create a new
    > > column with the minimums or that I could brute force the math, I am just
    > > hoping for something simple since I have to apply it a few hundred times.
    > >
    > > Thanks


  4. #4
    Max
    Guest

    Re: Array formula needed

    "ZipCurs" wrote
    > .. What if instead of a (2xn) matrix, I have one (1xn) matrix
    > and one (nx1) matrix that I want to select the minimum from.


    Assume the (1xn) is A1:A10, and the (nx1) is B1:Z1
    Try in say, B2: =SUMPRODUCT(A1:A10*MIN(B1:Z1))
    Normal ENTER will do
    --
    Rgds
    Max
    xl 97
    ---
    Singapore, GMT+8
    xdemechanik
    http://savefile.com/projects/236895
    --



  5. #5
    B. R.Ramachandran
    Guest

    RE: Array formula needed

    Hi,

    If the first 1xn range is in A1:A100, and the second (i.e., nx1 ) range is
    in B1:Z1,

    =SUM(A1:A100)*MIN(B1:Z1)

    Regards,
    B. R. Ramachandran

    "ZipCurs" wrote:

    > Thanks, that did the trick. What if instead of a (2xn) matrix, I have one
    > (1xn) matrix and one (nx1) matrix that I want to select the minimum from.
    > This doesn't seem to work with that.
    >
    > "B. R.Ramachandran" wrote:
    >
    > > Hi,
    > >
    > > If the (1 x n) range data are in Column A, and the (2 x n) matrix data are
    > > in Columns B and C, use the following formula:
    > > =SUMPRODUCT(A1:A100,IF(B1:B100<C1:C100,B1:B100,C1:C100)) and confirm with
    > > CTRL-SHIFT-ENTER.
    > >
    > > Regards,
    > > B. R. Ramachandran
    > >
    > > "ZipCurs" wrote:
    > >
    > > > I want to perform an sumproduct type of function with on column of data (1 x
    > > > n) and the minimum value in each row of a 2 x n matrix. Is there a simple
    > > > formula that will allow me to do this. I know that I could create a new
    > > > column with the minimums or that I could brute force the math, I am just
    > > > hoping for something simple since I have to apply it a few hundred times.
    > > >
    > > > Thanks


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