The IF statement that I have which has the ISNA is the following
=IF($AT$4>0,IF(ISNA(INDEX(projsls,(MATCH($B18,Prjsheet,0)))),0,(INDEX(projsls,(MATCH($B18,Prjsheet,0))))/1000),IF(ISNA(INDEX(ladder20,(MATCH($L18,Ladders,0)))),0,(INDEX(ladder20,(MATCH($L18,Ladders,0))))))
projsls is a named range
Prjsheet is a named range
Ladders is a named range
Ladder 20 is a named range
Originally I wrote a version of this formula as a vlookup but i switched my formulas to index match since i could set up a named range that only looked up one column as opposed to my vlookup formula that looked up an entire worksheet in my named range. By using smaller named ranges, will this make my workbook calculate faster?
=9.99999999999999E+307
=IF($AT$4>0,
LOOKUP(BigNum,CHOOSE({1,2},0,INDEX(projsls,MATCH($B18,Prjsheet,0))/1000)),
LOOKUP(BigNum,CHOOSE({1,2},0,INDEX(ladder20,MATCH($L18,Ladders,0)))))
Why would I have to setup Big Num? and how does the choose command work? ALso I saw that you used lookup. I cannot sort my data so lookup will still work?
If you look at the set up closely, it avoids computing the same thing twice.
Also, LOOKUP as used here is a different matter: It does not require any sorting.
Would this formula calculate extremely fast? is this faster than using Countif to replace the ISNA?
Is this the fastest way to write my formula? i used index match so my named ranges are just one column long. WHen i had my formulas as vlookups, i had my named ranges are 50 columns long, and then i would vlookup the column i need. IN this case , is using index match faster?
You don't have to.Why would I have to setup Big Num?
What does defining Big Number do?