Den 23.01.2013 07:29, skreiv Brian Barker:
At 22:57 22/01/2013 +0100, Kolbjørn Stuestøl wrote:
Is there any way to find the first occurrence of a number less (or
greater) than a given number in an unsorted array or a column?
Example: A1:A31 contains a list of temperatures a given month. I
would like to find the first day the temperature is below 0 degrees
(metric).
What happens if there is no day in the month when the temperature was
negative? Let's say you want the value zero for the resulting day
number in this case.
This is a little messy, but it works:
o In B1, enter =A1<0 and fill it down the column to B31. This
generates TRUE for your negative values and FALSE otherwise.
o In your result cell, enter =N(MATCH(TRUE;B1:B31;0)) . The MATCH()
function searches for the first TRUE value in the column B values and
returns the relative position in the array of the first match. (Note
that this is not necessarily the row number, though will be in your
example.) This would return the #N/A error if there is no negative
temperature; the N() function makes no change to real dates but
converts this error to zero.
Thank you.
I was hoping there was an easier way doing it.
My first suggestion was =MATCH(A1:A31<0,A1:A31,0) but this does not works.
Instead I used the array formula {=IF(A1:A31<0;1;0)} in B1. Quite
similar to your =A1<0. (In LibreOffice TRUE=1 and FALSE=0).
The next step is the same as you suggests: =N(MATCH(1;B1:B31;0))
Alas the N() function does not works as supposed. It returns "#N/A" if
there are no numbers below zero. I have to figure out a solution.
I am running version LibreOffice 4.0 in Norwegian on Windows 7 if that
matters.
Perhaps some programmers will add a FIND.FIRST.OCCURRENCE(n<0,A1:A31)
function in the future? :-)
You may choose to hide column B, of course - or you could put its
values away somewhere else on the sheet or even on a different sheet.
Incidentally, you are unlikely to get temperatures "below 0 degrees
(metric)", as the SI temperature scale is kelvin, and its zero is the
absolute zero of temperature! But yes: I appreciate that you mean the
celsius scale. ;^)
Yes, you are right. :-) I am using Celcius, not Kelvin of course.
Kolbjoern
I trust this helps.
Brian Barker
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