Hi,
I think you could get a cleaner solution using ?cut to split your data
in given ranges (the break argument), and then using this factor to
give the appropriate percentage.
Hope this helps,
baptiste
On 15 Mar 2009, at 20:12, diegol wrote:
Using R 2.7.0 under WinXP.
I need to write a function that takes a non-negative vector and
returns the
parallell maximum between a percentage of this argument and a fixed
value.
Both the percentages and the fixed values depend on which interval x
falls
in. Intervals are as follows:
From | To | % of x | Minimum
---------------------------------------------------------------
0 | 20000 | 65 | 0
20000 | 100000 | 40 | 14000
100000 | 250000 | 30 | 40000
250000 | 700000 | 25 | 75000
700000 | 1000000 | 20 | 175000
1000000 | inf | -- | 250000
Once the interval is determined, the values in x are multiplied by the
percentages applying to the range in the 3rd column.
If the result is less than the fourth column, then the latter is used.
For values of x falling in the last interval, 250,000 must be used.
My best attempt at it in R:
MyRange <- function(x){
range_aux = ifelse(x<=20000, 1,
ifelse(x<=100000, 2,
ifelse(x<=250000, 3,
ifelse(x<=700000, 4,
ifelse(x<=1000000, 5,6)))))
percent = c(0.65, 0.4, 0.3, 0.25, 0.2, 0)
minimum = c(0, 14000, 40000, 75000, 175000, 250000)
pmax(x * percent[range_aux], minimum[range_aux])
}
This could be done in Excel much tidier in my opinion (especially the
range_aux part), element by element (cell by cell),
with a VBA function as follows:
Function MyRange(x as Double) as Double
Select Case x
Case Is <= 20000
MyRange = 0.65 * x
Case Is <= 100000
RCJuiProfDet = IIf(0.40 * x < 14000, 14000, 0.4 * x)
Case Is <= 250000
RCJuiProfDet = IIf(0.3 * x < 40000, 40000, 0.3 * x)
Case Is <= 700000
RCJuiProfDet = IIf(0.25 * x < 75000, 75000, 0.25 * x)
Case Is <= 1000000
RCJuiProfDet = IIf(0.2 * x < 175000, 175000, 0.2 * x)
Case Else
' This is always 250000. I left it this way so it is analogous to
the R
function
RCJuiProfDet = IIf(0 * x < 250000, 250000, 0 * x)
End Select
End Function
Any way to improve my R function? I have searched the help archive
and the
closest I have found is the switch function, which tests for
equality only.
Thank you in advance for reading this.
-----
~~~~~~~~~~~~~~~~~~~~~~~~~~
Diego Mazzeo
Actuarial Science Student
Facultad de Ciencias Económicas
Universidad de Buenos Aires
Buenos Aires, Argentina
--
View this message in context:
http://www.nabble.com/Testing-for-Inequality-%C3%A0-la-%22select-case%22-tp22527465p22527465.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
_____________________________
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
