Re: [R] help with simple function
Hello, You are missing a '*' in your first try. (As for the second, in R parenthesis are round, not []) sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1, Y2)^2)) This written as a function becomes fun <- function(Y1, Y2) sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1, Y2)^2)) x1 <- rnorm(10) x2 <- rnorm(10) fun(x1, x2) Hope this helps, Rui Barradas Em 17-03-2013 15:47, Miguel Eduardo Delgado Burbano escreveu: hello all I am writing a quite simple script to study dental wear patterns in humans and I wrote this function sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this error message Error: attempt to apply non-function alternatively I wrote this sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))] but this error message appear [1] NA Warning message: NAs introduced by coercion some suggestion to solve this?? thank you so much for your help Miguel [[alternative HTML version deleted]] __ 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. __ 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] help with simple function
Hi, Y1<- 1:4 Y2<- 5:8 sqrt(var(Y1)+var(Y2)^2)-4*((var(Y1)*(var(Y2)-cov(Y1,Y2)^2))) #[1] 9.515593 A.K. - Original Message - From: Miguel Eduardo Delgado Burbano To: r-help@r-project.org Cc: Sent: Sunday, March 17, 2013 11:47 AM Subject: [R] help with simple function hello all I am writing a quite simple script to study dental wear patterns in humans and I wrote this function sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this error message Error: attempt to apply non-function alternatively I wrote this sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))] but this error message appear [1] NA Warning message: NAs introduced by coercion some suggestion to solve this?? thank you so much for your help Miguel [[alternative HTML version deleted]] __ 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. __ 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] help with simple function
On 17-03-2013, at 16:47, Miguel Eduardo Delgado Burbano wrote: > hello all > > I am writing a quite simple script to study dental wear patterns in humans > and I wrote this function > > sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this > error message > > Error: attempt to apply non-function > Write 4*( instead of 4( assuming you want to multiply. > alternatively I wrote this > sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))] > > but this error message appear > > [1] NA > Warning message: > NAs introduced by coercion > [ is an indexing operator. You are trying to index the number 4. Which is impossible. Berend > some suggestion to solve this?? thank you so much for your help > > Miguel > > [[alternative HTML version deleted]] > > __ > 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. __ 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] help with simple function
Here's an added caveat, with subsequently a more detailed explanation of the
output desired:
The data this will apply to includes a variety of whole numbers not limited
to 1 & 0, a number of which may appear consecutively and not separated by
zeros!
e.g. x<-c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
answer = c(1.098, 0.69, NA, -0.69, NA, -0.41, NA, NA, 1.098, NA, NA, NA, NA,
-0.22, 0)
the 1st element, 3, becomes log(3) = 1.098612
the 2nd element, 2, becomes log(2) = 0.6931472
the 3rd element, 0, becomes NA (cannot log zero).
the 4rd element, 1, becomes log(1/(1(number of consecutive zeros immediately
preceding it) + 1 (constant))) = log(1/2) = -0.6931472
the 5th element, 0, becomes NA
the 6th element, 2, becomes log(2/(1(number of consecutive zeros immediately
preceding it) + 1 (constant))) = log(2/3) = -0.4054651
the 7th and 8th elements, both zeros, become NA
the 9th element, 1, becomes log(1/(2(number of consecutive zeros immediately
preceding it) + 1 (constant))) = log(1/3) = 1.098612
the 10-13th elements, all zeros, each become NA
the 14th element, 4, becomes log(4/(4(number of consecutive zeros
immediately preceding it) + 1 (constant))) = log(4/5) = -0.2231436
the 15th element, 1, becomes log(1) = 0
T.D.Rudolph wrote:
>
> I'm trying to build on Jim's approach to change the parameters in the
> function, with new rules:
>
> 1. if (x[i]==0) NA
> 2. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding
> it +1))
>
> x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
> # i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
> NA, NA, NA, NA, -1.61)
> y <- rle(x)
> # attempting to modify Jim's function:
> result <- lapply(seq_along(y$lengths), function(.indx){
> if (y$values[.indx-1] == 0)
> log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1,
> length=y$lengths[.indx]))
> else rep(log(y$values[.indx]), y$lengths[.indx])
> })
> # but I am clearly missing something!
>
> Does it not work because I haven't addressed what to do with the zeros and
> log(0)=-Inf?
> I've tried adding another "ifelse" but I still get the same result.
> Can someone find the error in my ways?
> Tyler
>
>
> jholtman wrote:
>>
>> Does this do what you want:
>>
>>> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
>>> y <- rle(x)
>>> result <- lapply(seq_along(y$lengths), function(.indx){
>> + if (y$values[.indx] == 0)
>> log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
>> length=y$lengths[.indx]))
>> + else rep(log(y$values[.indx]), y$lengths[.indx])
>> + })
>>> unlist(result)
>> [1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
>> -1.0986123 -0.6931472 0.000
>> [10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>>>
>>>
>>
>>
>> On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
>> wrote:
>>
>>>
>>> In fact x[4,2] should = log(x[5,1]/2]
>>> whereas x[3,2] = log(x[5,1/3])
>>>
>>> i.e. The denominator in the log function equals the number of rows
>>> between
>>> m==0 and m>0 (inclusive, hence the "+1")
>>>
>>> Hope this helps!...
>>>
>>>
>>> Charles C. Berry wrote:
>>> >
>>> > On Tue, 27 May 2008, T.D.Rudolph wrote:
>>> >
>>> >>
>>> >>
>>> >> I have a matrix of frequency counts from 0-160.
>>> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>>> >>
>>> >> I would like to apply a function creating a new column
>>> (x[,2])containing
>>> >> values equal to:
>>> >> a) log(x[m,1]) if x[m,1] > 0; and
>>> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
>>> >> values
>>> >> +1)
>>> >>
>>> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) =
>>> -0.6931472
>>> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>>> >>
>>> >
>>> > If you also intend that x[4,2] == x[3,2] in your example, then this
>>> seems
>>> > what you want:
>>> >
>>> >> rle.x <- rle(x[,1])
>>> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
>>> >> rle.x$values )
>>> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
>>> >> rep(log(num/denom),rle.x$lengths)
>>> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000
>>> -1.3862944
>>> > -1.3862944 -1.3862944 0.000 -1.6094379
>>> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
>>> >>
>>> >
>>> > See
>>> >
>>> > ?rep
>>> > ?rle
>>> > ?tail
>>> >
>>> > HTH,
>>> >
>>> > Chuck
>>> >
>>> >
>>> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not
>>> do
>>> it
>>> >> by
>>> >> hand!
>>> >> Tyler
>>> >>
>>> >>
>>> >>
>>> >> --
>>> >> View this message in context:
>>> >>
>>> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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
Re: [R] help with simple function
Here's an added caveat, with subsequently a more detailed explanation of the
output desired:
The data this will apply to includes a variety of whole numbers not limited
to 1 & 0, a number of which may appear consecutively and not separated by
zeros!
e.g. x<-c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
answer = c(1.098, 0.69, NA, -0.69, NA, -0.41, NA, NA, 1.098, NA, NA, NA, NA,
-0.22, 0)
the 1st element, 3, becomes log(3) = 1.098612
the 2nd element, 2, becomes log(2) = 0.6931472
the 3rd element, 0, becomes NA (cannot log zero).
the 4rd element, 1, becomes log(1/(1(number of consecutive zeros immediately
preceding it) + 1 (constant))) = log(1/2) = -0.6931472
the 5th element, 0, becomes NA
the 6th element, 2, becomes log(2/(1(number of consecutive zeros immediately
preceding it) + 1 (constant))) = log(2/3) = -0.4054651
7-8thth elements = NA
the 9th element, 1, becomes log(1/(2(number of consecutive zeros immediately
preceding it) + 1 (constant))) = log(1/3) = 1.098612
the 10-13th elements, all zeros, each become NA
the 14th element, 4, becomes log(4/(4(number of consecutive zeros
immediately preceding it) + 1 (constant))) = log(4/5) = -0.2231436
the 15th element, 1, becomes log(1) = 0
T.D.Rudolph wrote:
>
> I'm trying to build on Jim's approach to change the parameters in the
> function, with new rules:
>
> 1. if (x[i]==0) NA
> 2. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding
> it +1))
>
> x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
> # i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
> NA, NA, NA, NA, -1.61)
> y <- rle(x)
> # attempting to modify Jim's function:
> result <- lapply(seq_along(y$lengths), function(.indx){
> if (y$values[.indx-1] == 0)
> log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1,
> length=y$lengths[.indx]))
> else rep(log(y$values[.indx]), y$lengths[.indx])
> })
> # but I am clearly missing something!
>
> Does it not work because I haven't addressed what to do with the zeros and
> log(0)=-Inf?
> I've tried adding another "ifelse" but I still get the same result.
> Can someone find the error in my ways?
> Tyler
>
>
> jholtman wrote:
>>
>> Does this do what you want:
>>
>>> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
>>> y <- rle(x)
>>> result <- lapply(seq_along(y$lengths), function(.indx){
>> + if (y$values[.indx] == 0)
>> log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
>> length=y$lengths[.indx]))
>> + else rep(log(y$values[.indx]), y$lengths[.indx])
>> + })
>>> unlist(result)
>> [1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
>> -1.0986123 -0.6931472 0.000
>> [10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>>>
>>>
>>
>>
>> On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
>> wrote:
>>
>>>
>>> In fact x[4,2] should = log(x[5,1]/2]
>>> whereas x[3,2] = log(x[5,1/3])
>>>
>>> i.e. The denominator in the log function equals the number of rows
>>> between
>>> m==0 and m>0 (inclusive, hence the "+1")
>>>
>>> Hope this helps!...
>>>
>>>
>>> Charles C. Berry wrote:
>>> >
>>> > On Tue, 27 May 2008, T.D.Rudolph wrote:
>>> >
>>> >>
>>> >>
>>> >> I have a matrix of frequency counts from 0-160.
>>> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>>> >>
>>> >> I would like to apply a function creating a new column
>>> (x[,2])containing
>>> >> values equal to:
>>> >> a) log(x[m,1]) if x[m,1] > 0; and
>>> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
>>> >> values
>>> >> +1)
>>> >>
>>> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) =
>>> -0.6931472
>>> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>>> >>
>>> >
>>> > If you also intend that x[4,2] == x[3,2] in your example, then this
>>> seems
>>> > what you want:
>>> >
>>> >> rle.x <- rle(x[,1])
>>> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
>>> >> rle.x$values )
>>> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
>>> >> rep(log(num/denom),rle.x$lengths)
>>> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000
>>> -1.3862944
>>> > -1.3862944 -1.3862944 0.000 -1.6094379
>>> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
>>> >>
>>> >
>>> > See
>>> >
>>> > ?rep
>>> > ?rle
>>> > ?tail
>>> >
>>> > HTH,
>>> >
>>> > Chuck
>>> >
>>> >
>>> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not
>>> do
>>> it
>>> >> by
>>> >> hand!
>>> >> Tyler
>>> >>
>>> >>
>>> >>
>>> >> --
>>> >> View this message in context:
>>> >>
>>> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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
Re: [R] help with simple function
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
NA, NA, NA, NA, -1.61)
y <- rle(x)
# attempting to modify Jim's function:
result <- lapply(seq_along(y$lengths), function(.indx){
if (y$values[.indx-1] == 0)
log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1,
length=y$lengths[.indx]))
else rep(log(y$values[.indx]), y$lengths[.indx])
})
# but I am clearly missing something!
Does it not work because I haven't addressed what to do with the zeros and
log(0)=-Inf?
I've tried adding another "ifelse" but I still get the same result.
Can someone find the error in my ways?
Tyler
jholtman wrote:
>
> Does this do what you want:
>
>> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
>> y <- rle(x)
>> result <- lapply(seq_along(y$lengths), function(.indx){
> + if (y$values[.indx] == 0)
> log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
> length=y$lengths[.indx]))
> + else rep(log(y$values[.indx]), y$lengths[.indx])
> + })
>> unlist(result)
> [1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
> -1.0986123 -0.6931472 0.000
> [10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>>
>>
>
>
> On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
> wrote:
>
>>
>> In fact x[4,2] should = log(x[5,1]/2]
>> whereas x[3,2] = log(x[5,1/3])
>>
>> i.e. The denominator in the log function equals the number of rows
>> between
>> m==0 and m>0 (inclusive, hence the "+1")
>>
>> Hope this helps!...
>>
>>
>> Charles C. Berry wrote:
>> >
>> > On Tue, 27 May 2008, T.D.Rudolph wrote:
>> >
>> >>
>> >>
>> >> I have a matrix of frequency counts from 0-160.
>> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>> >>
>> >> I would like to apply a function creating a new column
>> (x[,2])containing
>> >> values equal to:
>> >> a) log(x[m,1]) if x[m,1] > 0; and
>> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
>> >> values
>> >> +1)
>> >>
>> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
>> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>> >>
>> >
>> > If you also intend that x[4,2] == x[3,2] in your example, then this
>> seems
>> > what you want:
>> >
>> >> rle.x <- rle(x[,1])
>> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
>> >> rle.x$values )
>> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
>> >> rep(log(num/denom),rle.x$lengths)
>> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944
>> > -1.3862944 -1.3862944 0.000 -1.6094379
>> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
>> >>
>> >
>> > See
>> >
>> > ?rep
>> > ?rle
>> > ?tail
>> >
>> > HTH,
>> >
>> > Chuck
>> >
>> >
>> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do
>> it
>> >> by
>> >> hand!
>> >> Tyler
>> >>
>> >>
>> >>
>> >> --
>> >> View this message in context:
>> >>
>> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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.
>> >>
>> >
>> > Charles C. Berry(858) 534-2098
>> > Dept of Family/Preventive
>> > Medicine
>> > E mailto:[EMAIL PROTECTED] UC San Diego
>> > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
>> 92093-0901
>> >
>> > __
>> > 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.
>> >
>> >
>>
>> --
>> View this message in context:
>> http://www.nabble.com/help-with-simple-function-tp17498394p17502735.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.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
> [[alternative HTML vers
Re: [R] help with simple function
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]>0) log(x[i]/(number of consecutive zeros preceding it +1))
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
NA, NA, NA, NA, -1.61)
y <- rle(x)
# attempting to modify Jim's function:
result <- lapply(seq_along(y$lengths), function(.indx){
if (y$values[.indx-1] == 0)
log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1,
length=y$lengths[.indx]))
else rep(log(y$values[.indx]), y$lengths[.indx])
})
# but I am clearly missing something!
Does it not work because I haven't addressed what to do with the zeros and
log(0)=-Inf?
I've tried adding another "ifelse" but I still get the same result.
Can someone find the error in my ways?
Tyler
T.D.Rudolph wrote:
>
>
> I have a matrix of frequency counts from 0-160.
> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>
> I would like to apply a function creating a new column (x[,2])containing
> values equal to:
> a) log(x[m,1]) if x[m,1] > 0; and
> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values
> +1)
>
> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>
> I will be applying this to nrow(x)=~70,000 so I would prefer to not do it
> by hand!
> Tyler
>
>
>
>
--
View this message in context:
http://www.nabble.com/help-with-simple-function-tp17498394p17551527.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with simple function
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
2. if (x[i]==0) NA
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
NA, NA, NA, NA, -1.61)
# attempting to modify Jim's function...:
y <- rle(x)
result <- lapply(seq_along(y$lengths), function(.indx){
if (y$values[.indx-1] == 0)
log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1,
length=y$lengths[.indx]))
else rep(log(y$values[.indx]), y$lengths[.indx])
})
#...I am clearly missing something!
Does it not work because I haven't addressed what to do with the zeros and
log(0)=-Inf?
I've tried adding another "ifelse" but I still get the same result.
Can someone find the error in my ways?
Tyler
jholtman wrote:
>
> Does this do what you want:
>
>> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
>> y <- rle(x)
>> result <- lapply(seq_along(y$lengths), function(.indx){
> + if (y$values[.indx] == 0)
> log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
> length=y$lengths[.indx]))
> + else rep(log(y$values[.indx]), y$lengths[.indx])
> + })
>> unlist(result)
> [1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
> -1.0986123 -0.6931472 0.000
> [10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>>
>>
>
>
> On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
> wrote:
>
>>
>> In fact x[4,2] should = log(x[5,1]/2]
>> whereas x[3,2] = log(x[5,1/3])
>>
>> i.e. The denominator in the log function equals the number of rows
>> between
>> m==0 and m>0 (inclusive, hence the "+1")
>>
>> Hope this helps!...
>>
>>
>> Charles C. Berry wrote:
>> >
>> > On Tue, 27 May 2008, T.D.Rudolph wrote:
>> >
>> >>
>> >>
>> >> I have a matrix of frequency counts from 0-160.
>> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>> >>
>> >> I would like to apply a function creating a new column
>> (x[,2])containing
>> >> values equal to:
>> >> a) log(x[m,1]) if x[m,1] > 0; and
>> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
>> >> values
>> >> +1)
>> >>
>> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
>> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>> >>
>> >
>> > If you also intend that x[4,2] == x[3,2] in your example, then this
>> seems
>> > what you want:
>> >
>> >> rle.x <- rle(x[,1])
>> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
>> >> rle.x$values )
>> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
>> >> rep(log(num/denom),rle.x$lengths)
>> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944
>> > -1.3862944 -1.3862944 0.000 -1.6094379
>> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
>> >>
>> >
>> > See
>> >
>> > ?rep
>> > ?rle
>> > ?tail
>> >
>> > HTH,
>> >
>> > Chuck
>> >
>> >
>> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do
>> it
>> >> by
>> >> hand!
>> >> Tyler
>> >>
>> >>
>> >>
>> >> --
>> >> View this message in context:
>> >>
>> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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.
>> >>
>> >
>> > Charles C. Berry(858) 534-2098
>> > Dept of Family/Preventive
>> > Medicine
>> > E mailto:[EMAIL PROTECTED] UC San Diego
>> > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
>> 92093-0901
>> >
>> > __
>> > 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.
>> >
>> >
>>
>> --
>> View this message in context:
>> http://www.nabble.com/help-with-simple-function-tp17498394p17502735.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.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
> [[alternative HTML vers
Re: [R] help with simple function
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]>0) log(x[i]/(number of consecutive zeros preceding it +1))
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
NA, NA, NA, NA, -1.61)
y <- rle(x)
# attempting to modify Jim's function:
result <- lapply(seq_along(y$lengths), function(.indx){
if (y$values[.indx-1] == 0)
log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1,
length=y$lengths[.indx]))
else rep(log(y$values[.indx]), y$lengths[.indx])
})
# but I am clearly missing something!
Does it not work because I haven't addressed what to do with the zeros and
log(0)=-Inf?
I've tried adding another "ifelse" but I still get the same result.
Can someone find the error in my ways?
Tyler
jholtman wrote:
>
> Does this do what you want:
>
>> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
>> y <- rle(x)
>> result <- lapply(seq_along(y$lengths), function(.indx){
> + if (y$values[.indx] == 0)
> log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
> length=y$lengths[.indx]))
> + else rep(log(y$values[.indx]), y$lengths[.indx])
> + })
>> unlist(result)
> [1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
> -1.0986123 -0.6931472 0.000
> [10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>>
>>
>
>
> On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
> wrote:
>
>>
>> In fact x[4,2] should = log(x[5,1]/2]
>> whereas x[3,2] = log(x[5,1/3])
>>
>> i.e. The denominator in the log function equals the number of rows
>> between
>> m==0 and m>0 (inclusive, hence the "+1")
>>
>> Hope this helps!...
>>
>>
>> Charles C. Berry wrote:
>> >
>> > On Tue, 27 May 2008, T.D.Rudolph wrote:
>> >
>> >>
>> >>
>> >> I have a matrix of frequency counts from 0-160.
>> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>> >>
>> >> I would like to apply a function creating a new column
>> (x[,2])containing
>> >> values equal to:
>> >> a) log(x[m,1]) if x[m,1] > 0; and
>> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
>> >> values
>> >> +1)
>> >>
>> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
>> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>> >>
>> >
>> > If you also intend that x[4,2] == x[3,2] in your example, then this
>> seems
>> > what you want:
>> >
>> >> rle.x <- rle(x[,1])
>> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
>> >> rle.x$values )
>> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
>> >> rep(log(num/denom),rle.x$lengths)
>> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944
>> > -1.3862944 -1.3862944 0.000 -1.6094379
>> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
>> >>
>> >
>> > See
>> >
>> > ?rep
>> > ?rle
>> > ?tail
>> >
>> > HTH,
>> >
>> > Chuck
>> >
>> >
>> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do
>> it
>> >> by
>> >> hand!
>> >> Tyler
>> >>
>> >>
>> >>
>> >> --
>> >> View this message in context:
>> >>
>> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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.
>> >>
>> >
>> > Charles C. Berry(858) 534-2098
>> > Dept of Family/Preventive
>> > Medicine
>> > E mailto:[EMAIL PROTECTED] UC San Diego
>> > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
>> 92093-0901
>> >
>> > __
>> > 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.
>> >
>> >
>>
>> --
>> View this message in context:
>> http://www.nabble.com/help-with-simple-function-tp17498394p17502735.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.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
> [[alternative HTML version deleted]]
>
> __
Re: [R] help with simple function
In fact x[4,2] should = log(x[5,1]/2] whereas x[3,2] = log(x[5,1/3]) i.e. The denominator in the log function equals the number of rows between m==0 and m>0 (inclusive, hence the "+1") Hope this helps! Charles C. Berry wrote: > > On Tue, 27 May 2008, T.D.Rudolph wrote: > >> >> >> I have a matrix of frequency counts from 0-160. >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) >> >> I would like to apply a function creating a new column (x[,2])containing >> values equal to: >> a) log(x[m,1]) if x[m,1] > 0; and >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero >> values >> +1) >> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612 >> > > If you also intend that x[4,2] == x[3,2] in your example, then this seems > what you want: > >> rle.x <- rle(x[,1]) >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA), >> rle.x$values ) >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 ) >> rep(log(num/denom),rle.x$lengths) > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944 > -1.3862944 -1.3862944 0.000 -1.6094379 > [11] -1.6094379 -1.6094379 -1.6094379 0.000 >> > > See > > ?rep > ?rle > ?tail > > HTH, > > Chuck > > >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do it >> by >> hand! >> Tyler >> >> >> >> -- >> View this message in context: >> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.html >> Sent from the R help mailing list archive at Nabble.com. >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > Charles C. Berry(858) 534-2098 > Dept of Family/Preventive > Medicine > E mailto:[EMAIL PROTECTED]UC San Diego > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > -- View this message in context: http://www.nabble.com/help-with-simple-function-tp17498394p17500891.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with simple function
Well done Jim.
It always ends up a little more complicated than I originally hoped, but I
certainly couldn't improve on it!
jholtman wrote:
>
> Does this do what you want:
>
>> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
>> y <- rle(x)
>> result <- lapply(seq_along(y$lengths), function(.indx){
> + if (y$values[.indx] == 0)
> log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
> length=y$lengths[.indx]))
> + else rep(log(y$values[.indx]), y$lengths[.indx])
> + })
>> unlist(result)
> [1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
> -1.0986123 -0.6931472 0.000
> [10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>>
>>
>
>
> On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
> wrote:
>
>>
>> In fact x[4,2] should = log(x[5,1]/2]
>> whereas x[3,2] = log(x[5,1/3])
>>
>> i.e. The denominator in the log function equals the number of rows
>> between
>> m==0 and m>0 (inclusive, hence the "+1")
>>
>> Hope this helps!...
>>
>>
>> Charles C. Berry wrote:
>> >
>> > On Tue, 27 May 2008, T.D.Rudolph wrote:
>> >
>> >>
>> >>
>> >> I have a matrix of frequency counts from 0-160.
>> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
>> >>
>> >> I would like to apply a function creating a new column
>> (x[,2])containing
>> >> values equal to:
>> >> a) log(x[m,1]) if x[m,1] > 0; and
>> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
>> >> values
>> >> +1)
>> >>
>> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
>> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
>> >>
>> >
>> > If you also intend that x[4,2] == x[3,2] in your example, then this
>> seems
>> > what you want:
>> >
>> >> rle.x <- rle(x[,1])
>> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
>> >> rle.x$values )
>> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
>> >> rep(log(num/denom),rle.x$lengths)
>> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944
>> > -1.3862944 -1.3862944 0.000 -1.6094379
>> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
>> >>
>> >
>> > See
>> >
>> > ?rep
>> > ?rle
>> > ?tail
>> >
>> > HTH,
>> >
>> > Chuck
>> >
>> >
>> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do
>> it
>> >> by
>> >> hand!
>> >> Tyler
>> >>
>> >>
>> >>
>> >> --
>> >> View this message in context:
>> >>
>> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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.
>> >>
>> >
>> > Charles C. Berry(858) 534-2098
>> > Dept of Family/Preventive
>> > Medicine
>> > E mailto:[EMAIL PROTECTED] UC San Diego
>> > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
>> 92093-0901
>> >
>> > __
>> > 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.
>> >
>> >
>>
>> --
>> View this message in context:
>> http://www.nabble.com/help-with-simple-function-tp17498394p17502735.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.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
> [[alternative HTML version deleted]]
>
> __
> 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.
>
>
--
View this message in context:
http://www.nabble.com/help-with-simple-function-tp17498394p17503599.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.
Re: [R] help with simple function
Does this do what you want:
> x<-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
> y <- rle(x)
> result <- lapply(seq_along(y$lengths), function(.indx){
+ if (y$values[.indx] == 0)
log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
length=y$lengths[.indx]))
+ else rep(log(y$values[.indx]), y$lengths[.indx])
+ })
> unlist(result)
[1] -0.6931472 0.000 -1.0986123 -0.6931472 0.000 -1.3862944
-1.0986123 -0.6931472 0.000
[10] -1.6094379 -1.3862944 -1.0986123 -0.6931472 0.000
>
>
On Tue, May 27, 2008 at 8:04 PM, T.D.Rudolph <[EMAIL PROTECTED]>
wrote:
>
> In fact x[4,2] should = log(x[5,1]/2]
> whereas x[3,2] = log(x[5,1/3])
>
> i.e. The denominator in the log function equals the number of rows between
> m==0 and m>0 (inclusive, hence the "+1")
>
> Hope this helps!...
>
>
> Charles C. Berry wrote:
> >
> > On Tue, 27 May 2008, T.D.Rudolph wrote:
> >
> >>
> >>
> >> I have a matrix of frequency counts from 0-160.
> >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
> >>
> >> I would like to apply a function creating a new column (x[,2])containing
> >> values equal to:
> >> a) log(x[m,1]) if x[m,1] > 0; and
> >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero
> >> values
> >> +1)
> >>
> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
> >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
> >>
> >
> > If you also intend that x[4,2] == x[3,2] in your example, then this seems
> > what you want:
> >
> >> rle.x <- rle(x[,1])
> >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA),
> >> rle.x$values )
> >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 )
> >> rep(log(num/denom),rle.x$lengths)
> > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944
> > -1.3862944 -1.3862944 0.000 -1.6094379
> > [11] -1.6094379 -1.6094379 -1.6094379 0.000
> >>
> >
> > See
> >
> > ?rep
> > ?rle
> > ?tail
> >
> > HTH,
> >
> > Chuck
> >
> >
> >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do
> it
> >> by
> >> hand!
> >> Tyler
> >>
> >>
> >>
> >> --
> >> View this message in context:
> >>
> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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.
> >>
> >
> > Charles C. Berry(858) 534-2098
> > Dept of Family/Preventive
> > Medicine
> > E mailto:[EMAIL PROTECTED] UC San Diego
> > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
> 92093-0901
> >
> > __
> > 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.
> >
> >
>
> --
> View this message in context:
> http://www.nabble.com/help-with-simple-function-tp17498394p17502735.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.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] help with simple function
In fact x[4,2] should = log(x[5,1]/2] whereas x[3,2] = log(x[5,1/3]) i.e. The denominator in the log function equals the number of rows between m==0 and m>0 (inclusive, hence the "+1") Hope this helps!... Charles C. Berry wrote: > > On Tue, 27 May 2008, T.D.Rudolph wrote: > >> >> >> I have a matrix of frequency counts from 0-160. >> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) >> >> I would like to apply a function creating a new column (x[,2])containing >> values equal to: >> a) log(x[m,1]) if x[m,1] > 0; and >> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero >> values >> +1) >> >> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 >> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612 >> > > If you also intend that x[4,2] == x[3,2] in your example, then this seems > what you want: > >> rle.x <- rle(x[,1]) >> num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA), >> rle.x$values ) >> denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 ) >> rep(log(num/denom),rle.x$lengths) > [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944 > -1.3862944 -1.3862944 0.000 -1.6094379 > [11] -1.6094379 -1.6094379 -1.6094379 0.000 >> > > See > > ?rep > ?rle > ?tail > > HTH, > > Chuck > > >> I will be applying this to nrow(x)=~70,000 so I would prefer to not do it >> by >> hand! >> Tyler >> >> >> >> -- >> View this message in context: >> http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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. >> > > Charles C. Berry(858) 534-2098 > Dept of Family/Preventive > Medicine > E mailto:[EMAIL PROTECTED]UC San Diego > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 > > __ > 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. > > -- View this message in context: http://www.nabble.com/help-with-simple-function-tp17498394p17502735.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.
Re: [R] help with simple function
On Tue, 27 May 2008, T.D.Rudolph wrote: I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612 If you also intend that x[4,2] == x[3,2] in your example, then this seems what you want: rle.x <- rle(x[,1]) num <- ifelse(rle.x$values == 0, c(tail(rle.x$values,-1),NA), rle.x$values ) denom <- ifelse(rle.x$values == 0 , rle.x$lengths +1 , 1 ) rep(log(num/denom),rle.x$lengths) [1] -0.6931472 0.000 -1.0986123 -1.0986123 0.000 -1.3862944 -1.3862944 -1.3862944 0.000 -1.6094379 [11] -1.6094379 -1.6094379 -1.6094379 0.000 See ?rep ?rle ?tail HTH, Chuck I will be applying this to nrow(x)=~70,000 so I would prefer to not do it by hand! Tyler -- View this message in context: http://www.nabble.com/help-with-simple-function-tp17498394p17498394.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. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ 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.
