K
kcbannon
I having no luck in substituting cell references for actual operators or
independent variables in a MEDIANIF array. It has worked in simpler
functions like COUNTIF. Am I missing something?
In the table below, I want to find the median of the returns in B4:B7
subject to the values in A4:A7 being less than, greater than, equal to
or not equal to (operator in cell A2) a certain variable (value in cell
B2).
Oper Variable
= 5
Value Return
2.5 15.1%
5.0 12.2%
6.2 9.8%
7.3 21.3%
The way I think the formula should be written is:
=median(if(A3:A7=5,B3:B7,""))
However, I want to be able to manipulate the operator (A2) and the
independent variable (B2) by referring to the cells. This is where is
am having problems; the formula returns #Value!
=median(if(A3:A7,$A$2&$B$2),B3:B7,"")))
Tracing through the second formula, it appears that the comma after A7
marks the end of the Logical Test and "$A$2&$B$2" become the Value if
True.
Any help would be greatly appreciated.
independent variables in a MEDIANIF array. It has worked in simpler
functions like COUNTIF. Am I missing something?
In the table below, I want to find the median of the returns in B4:B7
subject to the values in A4:A7 being less than, greater than, equal to
or not equal to (operator in cell A2) a certain variable (value in cell
B2).
Oper Variable
= 5
Value Return
2.5 15.1%
5.0 12.2%
6.2 9.8%
7.3 21.3%
The way I think the formula should be written is:
=median(if(A3:A7=5,B3:B7,""))
However, I want to be able to manipulate the operator (A2) and the
independent variable (B2) by referring to the cells. This is where is
am having problems; the formula returns #Value!
=median(if(A3:A7,$A$2&$B$2),B3:B7,"")))
Tracing through the second formula, it appears that the comma after A7
marks the end of the Logical Test and "$A$2&$B$2" become the Value if
True.
Any help would be greatly appreciated.