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Hi,
Is there any equivalent for Excel SLOPE function.
https://support.office.com/en-us/article/SLOPE-function-11fb8f97-3117-4813-98aa-61d7e01276b9
Regards
Łukasz
LOAD * INLINE [
Month, Value
1,94%
2,99%
3,99%
];
In the frontend, create a text box with expression
=LINEST_M( Value, Month)
It returns 0.025, i.e. 2.5% which seems reasonable to me.
Look into the linest_m function:
linest_m ([{set_expression}][ distinct ] [total [<fld {,fld}>] ]y-expression, x-expression[, y0 [, x0 ]])
returns the aggregated m value (slope) of a linear regression defined by the equation y=mx+b for a series of coordinates represented by paired numbers in x-expression and y-expression iterated over the chart dimension(s). Text values, null values and missing values in any or both pieces of a data-pair will result in the entire data-pair to be disregarded.
An optional valye y0 may be stated forcing the regression line to pass through the y-axis at a given point. By stating both y0 and x0 it is possible to force the regression line to pass through a single fixed coordinate.
Unless both y0 and x0 are stated, the function requires at least two valid data-pairs to calculate. If y0 and x0 are stated, a single data pair will do.
This function has the same limitations for nested aggregation as the avg([{set_expression}] [ distinct ] [ total [<fld { , fld } >]] expression) function. The linest_m function supports Set Analysis and the total qualifier in the same way as the avg([{set_expression}] [ distinct ] [ total [<fld { , fld } >]] expression) function.
Examples:
linest_m( Y, X )
linest_m( A/B, X*Y/3 )
linest_m( total Y, X )
linest_m( total <Z> Y, X)
linest_m( Y, X, 0 )
linest_m( Y, X, 1, 1 )
Thank you for the information and examples.
I have a question how to write an expression.
I have a measure "hits%", and I wanna determine the slope for 3 months.
For Example I have values 94.% 99% 99%, and months numbered 1,2,3.
In excel I get slope = 3%
LOAD * INLINE [
Month, Value
1,94%
2,99%
3,99%
];
In the frontend, create a text box with expression
=LINEST_M( Value, Month)
It returns 0.025, i.e. 2.5% which seems reasonable to me.
Thank you for your help.
I have one small problem.
I have a lot of "values" for various periods.
I want to count linest_m for 3 months back
i try to use
LINEST_M( sum({<Month={'>=$(=only(Month)-2)<=$(=only(Month))'}>}hits) , Month)
or
LINEST_M( sum(aggr(rangesum(above(total sum({<Month=>}hits),0,3)),Month)), Month)
but it's doesn't work