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-tp17498394p17498394.html Sent from the R help mailing list archive at Nabble.com.
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 0.0000000 -1.0986123 -1.0986123 0.0000000 -1.3862944 -1.3862944 -1.3862944 0.0000000 -1.6094379 [11] -1.6094379 -1.6094379 -1.6094379 0.0000000>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 at 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:cberry at tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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 0.0000000 -1.0986123 -1.0986123 0.0000000 -1.3862944 > -1.3862944 -1.3862944 0.0000000 -1.6094379 > [11] -1.6094379 -1.6094379 -1.6094379 0.0000000 >> > > 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 at 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:cberry at tajo.ucsd.edu UC San Diego > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 > > ______________________________________________ > R-help at 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.
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 0.0000000 -1.0986123 -1.0986123 0.0000000 -1.3862944 > -1.3862944 -1.3862944 0.0000000 -1.6094379 > [11] -1.6094379 -1.6094379 -1.6094379 0.0000000 >> > > 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 at 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:cberry at tajo.ucsd.edu UC San Diego > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 > > ______________________________________________ > R-help at 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.
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
>
>
>
>
--
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