El 12/02/12 16:04, Barry Smith escribió:
Thank you for your help, Miguel.
=MATCH(N(A2),'20120208'!B$2:B$1491,0)
That is part of the problem. That says within the WB1 column, that
there is a match on the 122 row, but the actual row is 123... so MATCH
gives a row relative to the top of the vector. That is even better than
I was hoping for (I was thinking that I'd have to subtract a scalar
because the matrix started in Row 2 (or 3, since I'm trying to create
the stats before the draw on 2/8).
So, I dupe the MATCH function for each WB column, and can I then use MIN
to find the first occurrence...
Something like
=MIN( MATCH(N(A2),'20120208'!B$2:B$1491,0),
MATCH(N(A2),'20120208'!C$2:C$1491,0),
MATCH(N(A2),'20120208'!D$2:D$1491,0),
MATCH(N(A2),'20120208'!E$2:E$1491,0),
MATCH(N(A2),'20120208'!F$2:F$1491,0) )
That would give me the lowest row in the matrix where N(A2) is found...
I think.
Thank you for your help, Miguel.
Paz, (escribo español también)
For you can use a matrix fomula:
=ArrayFormula(MIN(IF('20120208'!B$2:H$1490=$A2;ROW('20120208'!B$2:H$1490);999999)))
This is how is view in google docs,
To enter in LibreOffice, intruduce the formula:
=MIN(IF('20120208'.B$2:H$1490=$A2;ROW('20120208'.B$2:H$1490);999999))
and finish with Crtl+Shift+Enter
Take care of sheet separator, if you use Libo or google docs.
Miguel Ángel.
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