search for: kathryn

Hi I was wondering if anyone knew how to work out the number of knots that should be applied to each variable when using gams in the mgcv library? Any help or references would be much appreciated. Thanks Kathryn Baldwin

...0 0 0 0 0 0 0 0 [3,] 0 0 0 0 0 0 0 0 0 0 3 7 11 15 19 0 0 0 0 0 [4,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 8 12 16 20 Any suggestion will be greatly appreciated. Best, Kathryn Lord [[alternative HTML version deleted]]

...to get a value you've assigned within a function, out of the function? For example, function () { testing <-read.table("...",header=TRUE) } I want to be able to use 'testing' outside of the function, so be able to call ls() and testing be there... Any ideas? Thanks! Kathryn

...[[2]] [,1] [,2] [,3] [,4] [,5] [,6] [1,] 7 9 11 0 0 0 [2,] 0 0 0 8 10 12 [[3]] [,1] [,2] [,3] [,4] [,5] [,6] [1,] 13 15 17 0 0 0 [2,] 0 0 0 14 16 18 Any suggestion will be greatly appreciated. Regards, Kathryn Lord [[alternative HTML version deleted]]

...{ imsi = 0 if (i==1) {rem.y = y[i]} else {c = length(rem.y) for (j in 1:c) { if (y[i]==rem.y[j]) imsi=1 } if (imsi==0) rem.y = c(rem.y,y[i])} } rem.y #------------------------------------------------------------------------- Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://www.nabble.com/Remove-repeated-values-tp19885503p19885503.html Sent from the R help mailing list archive at Nabble.com.

...------------------------------- > mass.func $`00` prop 5 1 $`10` prop 6 1 $`11` prop 2 3 0.5 0.5 --------------------------------------------------------- which means that A[1,5] = 1 A[2,6] =1 A[3,2] = A[3,3] = 0.5 otherwise, zero. Any suggestion will be greatly appreciated. Regrads, Kathryn Lord p.s. Here is R code. ------------------------------------------------------------------ > dat <- as.data.frame(matrix( c( 2, 1, 1, 3, 1, 1, 6, 1, 0, 5, 0, 0), 4, 3, byrow=T)) > covar <- apply(dat[,-1],1,paste,collapse='') > sp.dat <- split(dat,covar) > y.cova...

...o change this character vector, "zz", zz <- c("12","56","89") to the following numeric matrix. [,1] [,2] [1,] 1 2 [2,] 5 6 [3,] 8 9 Actually, "zz" vector has a long length. Any comments will be greatly appreciated. Kathryn Lord -- View this message in context: http://www.nabble.com/character-vector--%3E-numeric-matrix----tp24703927p24703927.html Sent from the R help mailing list archive at Nabble.com.

...-------------------------------------------------------------------------- In fact, #column of `a` is so big.. Is there a more efficient way to compute this instead of using "apply" or something? or "apply" is only way..? Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://n4.nabble.com/VERY-SIMPLE-QUESTION-tp2013288p2013288.html Sent from the R help mailing list archive at Nabble.com.

...rror 1.000011 0.151850 6.58553 4.5325e-11 Co <--> Co carbonateError 1.000009 0.156217 6.40142 1.5394e-10 carbonate <--> carbonate MWDError 0.589788 0.092128 6.40185 1.5351e-10 MWD <--> MWD Thanks Kathryn

Dear R users I'd like to generate two sets of random numbers with a fixed correlation coefficient, say .4, using R. Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://r.789695.n4.nabble.com/generate-two-sets-of-random-numbers-that-are-correlated-tp3735695p3735695.html Sent from the R help mailing list archive at Nabble.com.

Dear R users, I have 32 observations in data x. After sorting this, I want to compute means and variances of 3 groups divided by "nr". Actually, the number of groups is flexible. Any suggestion will be greatly appreciated. Kathryn Lord --------------------------------------------------------------------------- x=rnorm(32) y=sort(x) nr=matrix(c(12,11,10,10,10,11),2,3) > nr [,1] [,2] [,3] [1,] 12 10 10 -> sum=32 [2,] 11 10 11 -> sum=32 For the 1st row in "nr", index of y = (1,..,12,...

...---------------------------------- 12 0 0 0 0 #- 3rd matrix-------------------------------------- 8 1 0 1 1 3 1 0 1 1 #- 4th matrix-------------------------------------- 9 0 1 0 0 3 0 1 0 0 Any comments will be greatly appreciated. Kathryn Lord -- View this message in context: http://www.nabble.com/Splitting-matrix-into-several-small-matrices-tp24689585p24689585.html Sent from the R help mailing list archive at Nabble.com.

...%B[,1]) = 73.19647 2*exp(X[2,]%*%B[,1]) = 362.54448 1*exp(X[1,]%*%B[,2]) = 1096.633 1*exp(X[2,]%*%B[,2]) = 59874.142 2*exp(X[1,]%*%B[,2]) = 2193.266 2*exp(X[2,]%*%B[,2]) = 119748.283 ------------------------------------ How could I find only 8 values? Any suggestion will be greatly appreciated. Kathryn Lord -- View this message in context: http://www.nabble.com/results-from-%22do.call%22-function-tp20532136p20532136.html Sent from the R help mailing list archive at Nabble.com.

Hi folks, Does anyone know where I can find downloads for old versions of R, around version 1.0? I found the downloads for unix but not windows. Any help would be greatly appreciated! Thank you, Kathryn

...i=0, 1, 2, 3 if k=4, then i=0, 1, 2 if k=5, then i=0, 1 so i'd like to create a list like below. > list k i 1 1 0 2 2 0 3 2 1 4 3 0 5 3 1 6 3 2 7 3 3 8 4 0 9 4 1 10 4 2 11 5 0 12 5 1 I tried expand.grid, but I can't. Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://r.789695.n4.nabble.com/create-a-list-under-constraints-tp3714191p3714191.html Sent from the R help mailing list archive at Nabble.com.

...trix(E) [[2]][[3]] matrix(F) #--------------------------------------------- I'd like to compute matrix(A)+matrix(B)+matrix(C)+matrix(D)+matrix(E)+matrix(F) In fact, I tried "lapply", "do.call" etc, but failed. Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://www.nabble.com/computation-of-matrices-in-list-of-list-tp25094875p25094875.html Sent from the R help mailing list archive at Nabble.com.

...e (or repeat) ------------ { 1 2 . . . 6 } #----------------------------- For example, if I have an error in the 6th line, then I'd like to go back to the 1st line. I've already tried "try", but it didn't work. Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://www.nabble.com/ignore-an-error-and-go-back-to-....-tp25179265p25179265.html Sent from the R help mailing list archive at Nabble.com.

...g[j]))/sqrt(sig[k]+sig[j])}))})) ------------------------------------------------- In fact, this is a part of optimization. I think if it is changed more efficiently, then the running time could be shortened. How could I change this more efficiently? Any suggestion will be greatly appreciated. Kathryn Lord -- View this message in context: http://www.nabble.com/more-efficient-double-summation...-tp19213599p19213599.html Sent from the R help mailing list archive at Nabble.com.

...<- log( sum(exp(sums)) ) + maxsums lbp[k] <- y1[k]*log(4) + y2[k]*log(5) + lsum } elbp <- exp(lbp) ------------------------------------------------------------------------------------ Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://r.789695.n4.nabble.com/My-R-code-is-not-efficient-tp3712762p3712762.html Sent from the R help mailing list archive at Nabble.com.

Hello I am very new to R.  I have an R task to complete that I have not been able to find a straightforward answer to as of yet.  I have a list of values. I would like to count the number of values that are in one bin, the number that fall in the next bin, etc. For example My input file is:  123 48 342 442 43 232 32 129 191 147 I would like the output to be similar to:  0-100 3 100-200 4