Re: [R] Testing for Inequality à la select case
On Sun, Mar 15, 2009 at 11:46 PM, diegol diego...@gmail.com wrote:
...Steve, ...
Actually Stavros (ΣΤΑΥΡΟΣ), not Stephen/Steve (ΣΤΕΦΑΝΟΣ). Both Greek,
but different names.
I still don't understand the analogy. I agree that in this case the R
approach is vectorized. However, your function just as you first proposed it
will not work without a loop.
Approach is vectorized over the range parameter, but not vectorized
over the x parameter. If you want to vectorize over x, you can use
findInterval:
mr -
local({
# Local constants
range= c(0,20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
function(x)
{ idx - findInterval(x,range)
pmax( x*perc[idx], min[idx] )
}
})
And this time, you *do* need pmax. I did refer to cut/split, but only
to say they were unnecessary.
-s
max and pmax are equivalent in this case. I just use pmax as my
default because it acts like other arithmetic operators (+, *, etc.)
which perform pointwise (element-by-element) operations.
It's true. I changed it because I had applied your original version of mr()
to the entire vector x, which gave an incorrect result (perhaps range was
recycled in idx - which(x=range)[1]). If I used max instead of pmax, and
ever happened to use mr() without a loop, the length of the result would be
strange enough for me to realise the error. But then again, I added the if
(length(x) 1) stop(x must have length 1) line, so using max or pmax now
doesn't really make a difference, apart perhaps from run time.
Using cut/split seems like gross overkill here. Among other things,
you don't need to generate labels for all the different ranges.
which(x=range)[1]
seems straightforward enough to me,
I could edit the mr_2() function a little bit to make it more efficient. I
left it mostly unchanged for the thread to be easier to follow. For example
I could replace the last four lines for only:
product - x*percent
ifelse(product minimum, minimum, product)
But I believe you refer to the cut/split functions rather. I agree that
which(x=range)[1] is straighforward, but using such expression will
require a loop to pull the trick, which I don't intend. Am I missing
something?
Regards,
Diego
Stavros Macrakis-2 wrote:
Using cut/split seems like gross overkill here. Among other things,
you don't need to generate labels for all the different ranges.
which(x=range)[1]
seems straightforward enough to me, but you could also use the
built-in function findInterval.
-s
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Universidad de Buenos Aires
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[R] Testing for Inequality à la select case
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 | 2| 65| 0
2 | 10 | 40| 14000
10 | 25 | 30 | 4
25 | 70 | 25 | 75000
70 | 100 | 20 | 175000
100 | inf | -- | 25
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=2, 1,
ifelse(x=10, 2,
ifelse(x=25, 3,
ifelse(x=70, 4,
ifelse(x=100, 5,6)
percent = c(0.65, 0.4, 0.3, 0.25, 0.2, 0)
minimum = c(0, 14000, 4, 75000, 175000, 25)
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 = 2
MyRange = 0.65 * x
Case Is = 10
RCJuiProfDet = IIf(0.40 * x 14000, 14000, 0.4 * x)
Case Is = 25
RCJuiProfDet = IIf(0.3 * x 4, 4, 0.3 * x)
Case Is = 70
RCJuiProfDet = IIf(0.25 * x 75000, 75000, 0.25 * x)
Case Is = 100
RCJuiProfDet = IIf(0.2 * x 175000, 175000, 0.2 * x)
Case Else
' This is always 25. I left it this way so it is analogous
to the R
function
RCJuiProfDet = IIf(0 * x 25, 25, 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
--
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Re: [R] Testing for Inequality à la select case
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 | 2| 65| 0
2 | 10 | 40| 14000
10 | 25 | 30 | 4
25 | 70 | 25 | 75000
70 | 100 | 20 | 175000
100 | inf | -- | 25
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=2, 1,
ifelse(x=10, 2,
ifelse(x=25, 3,
ifelse(x=70, 4,
ifelse(x=100, 5,6)
percent = c(0.65, 0.4, 0.3, 0.25, 0.2, 0)
minimum = c(0, 14000, 4, 75000, 175000, 25)
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 = 2
MyRange = 0.65 * x
Case Is = 10
RCJuiProfDet = IIf(0.40 * x 14000, 14000, 0.4 * x)
Case Is = 25
RCJuiProfDet = IIf(0.3 * x 4, 4, 0.3 * x)
Case Is = 70
RCJuiProfDet = IIf(0.25 * x 75000, 75000, 0.25 * x)
Case Is = 100
RCJuiProfDet = IIf(0.2 * x 175000, 175000, 0.2 * x)
Case Else
' This is always 25. I left it this way so it is analogous to
the R
function
RCJuiProfDet = IIf(0 * x 25, 25, 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
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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
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Testing for Inequality à la select case
On Sun, Mar 15, 2009 at 4:12 PM, diegol diego...@gmail.com wrote:
...This could be done in Excel much tidier in my opinion (especially the
range_aux part), element by element (cell by cell)...
If you'd do it element-by-element in Excel, why not do it
element-by-element in R?
Create a table with the limits of the ranges
range= c(20,100,250,700,1000,Inf)*1000
and then find the index of the appropriate case using something like
idx - which(x=range)[1]
Then the formula becomes simply
pmax( x*perc[idx], min[idx] )
Putting it all together:
mr -
local({
# Local constants
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
function(x)
{ idx - which(x=range)[1]
pmax( x*perc[idx], min[idx] )
}
})
Make sense?
-s
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Re: [R] Testing for Inequality à la select case
Hello Stavros,
If you'd do it element-by-element in Excel, why not do it
element-by-element in R?
Well, actually I was hoping for a vectorized solution so as to avoid
looping. I need to use this formula on rather lengthy vectors and I wanted
to put R's efficiency to some good use. In any case, I had not come up with
your solution. For now, I'd stick to my ugly version.
Make sense?
Perfectly.
x - 1:150 * 1
y - numeric(150)
for (i in 1:150) y[i] - mr(x[i])
identical(MyRange(x), y)
TRUE
I would however use max instead of pmax, since the argument for mr() must be
a vector of length 1. The final version looks like this (also added a line
to avoid vectors of length 1):
mr -
local({
# Local constants
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
function(x)
{if (length(x) 1) stop(x must have length 1)
idx - which(x=range)[1]
max( x*perc[idx], min[idx] )
}
})
Thank you very much for your help.
Diego
Stavros Macrakis-2 wrote:
On Sun, Mar 15, 2009 at 4:12 PM, diegol diego...@gmail.com wrote:
...This could be done in Excel much tidier in my opinion (especially the
range_aux part), element by element (cell by cell)...
If you'd do it element-by-element in Excel, why not do it
element-by-element in R?
Create a table with the limits of the ranges
range= c(20,100,250,700,1000,Inf)*1000
and then find the index of the appropriate case using something like
idx - which(x=range)[1]
Then the formula becomes simply
pmax( x*perc[idx], min[idx] )
Putting it all together:
mr -
local({
# Local constants
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
function(x)
{ idx - which(x=range)[1]
pmax( x*perc[idx], min[idx] )
}
})
Make sense?
-s
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
-
~~
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-tp22527465p22529091.html
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Re: [R] Testing for Inequality à la select case
Hello Baptiste,
I am not very sure how I'd go about that. Taking the range, perc and min
vectors from Stavros' response:
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
For range to work as the breaks argument to cut, I think an additional
first element is needed:
range = c(0, range)
Now I create a dummy vector x and apply cut to create a factor z:
x - 1:150 * 1
z - cut(x = x, breaks = range)
The thing is, I cannot seem to figure out how to use this z factor to create
vectors of the same length as x with the corresponding elements of percent
and min defined above. Admittedly I have never felt very comfortable with
factors. Could you please give me some advice?
Thank you very much.
baptiste auguie-2 wrote:
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 | 2| 65| 0
2 | 10 | 40| 14000
10 | 25 | 30 | 4
25 | 70 | 25 | 75000
70 | 100 | 20 | 175000
100 | inf | -- | 25
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=2, 1,
ifelse(x=10, 2,
ifelse(x=25, 3,
ifelse(x=70, 4,
ifelse(x=100, 5,6)
percent = c(0.65, 0.4, 0.3, 0.25, 0.2, 0)
minimum = c(0, 14000, 4, 75000, 175000, 25)
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 = 2
MyRange = 0.65 * x
Case Is = 10
RCJuiProfDet = IIf(0.40 * x 14000, 14000, 0.4 * x)
Case Is = 25
RCJuiProfDet = IIf(0.3 * x 4, 4, 0.3 * x)
Case Is = 70
RCJuiProfDet = IIf(0.25 * x 75000, 75000, 0.25 * x)
Case Is = 100
RCJuiProfDet = IIf(0.2 * x 175000, 175000, 0.2 * x)
Case Else
' This is always 25. I left it this way so it is analogous
to
the R
function
RCJuiProfDet = IIf(0 * x 25, 25, 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.
-
~~
Diego Mazzeo
Actuarial Science Student
Facultad de Ciencias Económicas
Universidad de Buenos Aires
Buenos Aires, Argentina
--
View this message in context:
Re: [R] Testing for Inequality à la select case
Hi,
I don't use ?cut and ?split very much either, so this may not be good
advice. From what I understood of your problem, I would try something
along those lines,
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
range = c(0, range)
x - 1:150 * 1
percent - x
minimum - x
z - cut(x = x, breaks = range)
levs - levels(z)
split(percent, z, drop = FALSE) - perc
split(minimum, z, drop = FALSE) - min
mydf - data.frame(x, range= z, percent, minimum)
mydf - within(mydf, product - x * percent)
mydf$result - with(mydf, ifelse(product minimum, minimum, product))
str(mydf)
head(mydf)
but it's getting late here so i may well be missing an important thing.
Hope this helps,
baptiste
On 15 Mar 2009, at 23:19, diegol wrote:
Hello Baptiste,
I am not very sure how I'd go about that. Taking the range, perc and
min
vectors from Stavros' response:
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
For range to work as the breaks argument to cut, I think an
additional
first element is needed:
range = c(0, range)
Now I create a dummy vector x and apply cut to create a factor z:
x - 1:150 * 1
z - cut(x = x, breaks = range)
The thing is, I cannot seem to figure out how to use this z factor
to create
vectors of the same length as x with the corresponding elements of
percent
and min defined above. Admittedly I have never felt very
comfortable with
factors. Could you please give me some advice?
Thank you very much.
baptiste auguie-2 wrote:
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 | 2| 65| 0
2 | 10 | 40| 14000
10 | 25 | 30 | 4
25 | 70 | 25 | 75000
70 | 100 | 20 | 175000
100 | inf | -- | 25
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=2, 1,
ifelse(x=10, 2,
ifelse(x=25, 3,
ifelse(x=70, 4,
ifelse(x=100, 5,6)
percent = c(0.65, 0.4, 0.3, 0.25, 0.2, 0)
minimum = c(0, 14000, 4, 75000, 175000, 25)
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 = 2
MyRange = 0.65 * x
Case Is = 10
RCJuiProfDet = IIf(0.40 * x 14000, 14000, 0.4 * x)
Case Is = 25
RCJuiProfDet = IIf(0.3 * x 4, 4, 0.3 * x)
Case Is = 70
RCJuiProfDet = IIf(0.25 * x 75000, 75000, 0.25 * x)
Case Is = 100
RCJuiProfDet = IIf(0.2 * x 175000, 175000, 0.2 * x)
Case Else
' This is always 25. I left it this way so it is analogous
to
the R
function
RCJuiProfDet = IIf(0 * x 25, 25, 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
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PLEASE do read the posting guide
Re: [R] Testing for Inequality à la select case
Hello Baptiste,
Thanks so much for your help. This function which is basically your input
wrapped with curly brackets seems to work alright:
mr_2 - function(x){
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
range = c(0, range)
percent - x
minimum - x
z - cut(x = x, breaks = range)
levs - levels(z)
split(percent, z, drop = FALSE) - perc
split(minimum, z, drop = FALSE) - min
mydf - data.frame(x, range= z, percent, minimum)
mydf - within(mydf, product - x * percent)
mydf$result - with(mydf, ifelse(product minimum, minimum,
product))
mydf$result
}
# Basic Test
x - 1:150 * 1
identical(MyRange(x), mr_2(x))
[1] TRUE
# Yet another test
# (I will have a more in depth look at split, with and within to
feel more comfortable)
x - 150:1 * 1
identical(TramosAutos(x), mr_2(x))
[1] TRUE
Again, thank you very much to both of you.
Have a great week.
Diego
baptiste auguie-2 wrote:
Hi,
I don't use ?cut and ?split very much either, so this may not be good
advice. From what I understood of your problem, I would try something
along those lines,
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
range = c(0, range)
x - 1:150 * 1
percent - x
minimum - x
z - cut(x = x, breaks = range)
levs - levels(z)
split(percent, z, drop = FALSE) - perc
split(minimum, z, drop = FALSE) - min
mydf - data.frame(x, range= z, percent, minimum)
mydf - within(mydf, product - x * percent)
mydf$result - with(mydf, ifelse(product minimum, minimum, product))
str(mydf)
head(mydf)
but it's getting late here so i may well be missing an important thing.
Hope this helps,
baptiste
On 15 Mar 2009, at 23:19, diegol wrote:
Hello Baptiste,
I am not very sure how I'd go about that. Taking the range, perc and
min
vectors from Stavros' response:
range= c(20,100,250,700,1000,Inf)*1000
perc = c(65,40,30,25,20,0)/100
min = c(0,14,40,75,175,250)*1000
For range to work as the breaks argument to cut, I think an
additional
first element is needed:
range = c(0, range)
Now I create a dummy vector x and apply cut to create a factor z:
x - 1:150 * 1
z - cut(x = x, breaks = range)
The thing is, I cannot seem to figure out how to use this z factor
to create
vectors of the same length as x with the corresponding elements of
percent
and min defined above. Admittedly I have never felt very
comfortable with
factors. Could you please give me some advice?
Thank you very much.
baptiste auguie-2 wrote:
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 | 2| 65| 0
2 | 10 | 40| 14000
10 | 25 | 30 | 4
25 | 70 | 25 | 75000
70 | 100 | 20 | 175000
100 | inf | -- | 25
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=2, 1,
ifelse(x=10, 2,
ifelse(x=25, 3,
ifelse(x=70, 4,
ifelse(x=100, 5,6)
percent = c(0.65, 0.4, 0.3, 0.25, 0.2, 0)
minimum = c(0, 14000, 4, 75000, 175000, 25)
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 = 2
MyRange = 0.65 * x
Case Is = 10
RCJuiProfDet = IIf(0.40 * x 14000, 14000, 0.4 * x)
Case Is = 25
RCJuiProfDet = IIf(0.3 * x 4, 4, 0.3 * x)
Case Is =
Re: [R] Testing for Inequality à la select case
Using cut/split seems like gross overkill here. Among other things, you don't need to generate labels for all the different ranges. which(x=range)[1] seems straightforward enough to me, but you could also use the built-in function findInterval. -s __ 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.
Re: [R] Testing for Inequality à la select case
That's what I meant by element-by -element. A vector in R corresponds to a row or a column in Excel, and a vector operation in R corresponds to a row or column of formulae, e.g. Excel A B C 1) 5 10 a1+b1 (= 15) 2) 3 2 a2+b2 (= 5) etc. R A - c(5,3) B - c(10,2) C - A + B Steve, I still don't understand the analogy. I agree that in this case the R approach is vectorized. However, your function just as you first proposed it will not work without a loop. max and pmax are equivalent in this case. I just use pmax as my default because it acts like other arithmetic operators (+, *, etc.) which perform pointwise (element-by-element) operations. It's true. I changed it because I had applied your original version of mr() to the entire vector x, which gave an incorrect result (perhaps range was recycled in idx - which(x=range)[1]). If I used max instead of pmax, and ever happened to use mr() without a loop, the length of the result would be strange enough for me to realise the error. But then again, I added the if (length(x) 1) stop(x must have length 1) line, so using max or pmax now doesn't really make a difference, apart perhaps from run time. Using cut/split seems like gross overkill here. Among other things, you don't need to generate labels for all the different ranges. which(x=range)[1] seems straightforward enough to me, I could edit the mr_2() function a little bit to make it more efficient. I left it mostly unchanged for the thread to be easier to follow. For example I could replace the last four lines for only: product - x*percent ifelse(product minimum, minimum, product) But I believe you refer to the cut/split functions rather. I agree that which(x=range)[1] is straighforward, but using such expression will require a loop to pull the trick, which I don't intend. Am I missing something? Regards, Diego Stavros Macrakis-2 wrote: Using cut/split seems like gross overkill here. Among other things, you don't need to generate labels for all the different ranges. which(x=range)[1] seems straightforward enough to me, but you could also use the built-in function findInterval. -s __ 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. - ~~ 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-tp22527465p22531513.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.
