Forum Discussion
Whitney
Helper II
Helper IIValue furthest from zero
Hi, I have data that captures the best (as a postive number e.g. 3) and worst (as a negative number e.g. -5) position for a workshop in any given fortnight. I'm trying to write a formula to r...
Yes!!... You can use this as a measure
Furthest = if(ABS(MAX(Hoja1[column]))>ABS(MIN(Hoja1[column]));MAX(Hoja1[column]);MIN(Hoja1[column]))Regards!!
Hey there,
To achieve this, you need to write couple of calculated columns,
1) To calculate the ABS value,
absolute = IF('Table'[Position] <0, 'Table'[Position] * -1, 'Table'[Position])2) To calculate the max in the set along (along with it's negative sign, if it's negative),Max =VAR calc = CALCULATE(MAX('Table'[absolute]), FILTER('Table','Table'[absolute]))Return If(calc = 'Table'[absolute],'Table'[Position])I've created a sample report. Please find the attached. Let me know if you need more help.sample
If You use min(table[columnName]) or max(table[columnName]) solve your problem??
WhitneyHi Anonymous,
Using min() or max() neglects the other side of zero. I.e. max takes the largest positive number and min takes the largest negative number, and I am looking for the value furthest from zero which could be positive or negative. Unless there is a way to use min() and max() together in a function to return the value I'm looking for? I'm not sure.
Yes!!... You can use this as a measure
Furthest = if(ABS(MAX(Hoja1[column]))>ABS(MIN(Hoja1[column]));MAX(Hoja1[column]);MIN(Hoja1[column]))Regards!!