similar to: Obtaining the value of x at a given value of y in a smooth.spline object

Dear R helpers Suppose I have a data frame as given below mydat = data.frame(x = c(1,1,1, 2, 2, 2, 2, 2, 5, 5, 6), y = c(10, 10, 10, 8, 8, 8, 7, 7, 2, 2, 4)) mydat         x     y 1      1     10 2      1     10 3      1     10 4      2       8 5      2       8 6      2       8 7      2       7 8      2       7 9      5       2 10    5       2 11    6       4 unique(mydat$x) will give me 1,

Hi, I have a data frame that looks like this: >x x1 x2 x3 A 1 1.5 B 2 0.9 B 3 2.7 C 7 1.8 D 7 1.3 I want to "group" by the x1 column and in the case of multiple x$x1 values (e.g., "B")d, return rows that have the smallest values of x2. In the case of rows with only one value of x1 (e.g., "A"), return the row as is. How can I do that?

Hi, I would like to do the following operations: variable.df is a character string that contains the name of the data frame that I want to do the following operations on: variable.df <- data.frame(); # I can do the above command using assign( variable.df, data.frame() ) How can I perform the assignment statements below ? colnames(variable.df) = colnames(some.other.df) variable.df =

Hi, I have a set of (x,y) coordinate pairs that are stored as a list > my_list $x [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 $y [1] -8.0866819 -7.3876052 -6.6849311 -5.9837693 -5.2967432 -4.6525466 [7] -4.0999453 -3.6556190 -3.3076102 -3.0360780 -2.8220465 -2.6532085 [13] -2.5192816 -2.4086241 -2.3072977 -2.1969611 -2.0574250 -1.8737694 [19] -1.6357864

Hi all, I would like to invoke a function that takes multiple arguments (some of which are specified columns in the data frame, and others that are independent of the data frame) on split parts of a data frame, how do I do this? For example, let's say I have a data frame >fitness_data name height weight country rob 5.8 200 usa nancy 5.5 140 germany jen

Dear R People: Could someone recommend a small example for spline regression, please? I would like to show this to my students, step by step. Thanks in advance, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at gmail.com

I'm using R 1.7.0 on linux. With this version of R the package modreg is automatically loaded at start of session. However attempting to use predict.smooth.spline() produces Error: couldn't find function predict.smooth.spline. The function smooth.spline() is OK. What am I missing? ====================================== I.White ICAPB, University of Edinburgh Ashworth Laboratories, West

Hi, I want to use the result from smooth.spline outside R. I take my data ,which is 180 point stored in x and y s <- smooth(x,y) I can know use to e.g. find the interpolated value at e.g. x=500 predict (s,500) My problem is, that i don't know how to implement the predict function. I have looked at literature, but i cannot connect the output of the smooth.spline() to an actual spline

I have noticed a slightly puzzling behaviour exhibited by smooth.spline(). If I do sss <- smooth.spline(x,y) for a certain pair of data vectors x and y, and then do length(sss$x) I get the result ``18''. However if I do length(unique(x)) I get ``27''. Trying to force smooth.spline() to use more knots I tried sss <- smooth.spline(x,y,all.knots=TRUE) but again

Hello, I've noticed that SPLUS seems to have a function for evaluating derivative matrices of splines. I've found the R function that evaluates matrices from 'smooth.spline'; maybe someone has written something to do the same with smooth.basis? regards, s

Dear list, if I do smooth.spline(tmpSec, tmpT, all.knots=T) with the attached data, I get this error-message: Error in smooth.spline(tmpSec, tmpT, all.knots = T) : smoothing parameter value too small If I do smooth.spline(tmpSec[-single arbitrary number], tmpT[-single arbitrary number], all.knots=T) it works! I just don't see it. It works for hundrets other datasets, but not for

Hey, R-listers I was recommended to try using smooth.spline function for estimating 2-Dimensinal curve given a data set. So will you please tell me where to get this R function? Or which package provides this function? Thanks for your point. Fred

Hello. I'm getting an unexpected result when running smooth.spline(). Here is a simple example that replicates the error I'm getting: > aa <- c(1, 2, 3, 8, 8, 8, 8, 8, 8, 8, 8, 8, 12, 13, 14) > bb <- 1:length(aa) > plot(aa, bb) > smooth.spline(aa, bb) Error in smooth.spline(aa, bb) : need at least four unique 'x' values As you can see from the example, my

Can anyone tell me the trick for obtaining the smoother matrix from smooth.spline when there are non-unique values for x. I have the following code but, of course, it only works when all values of x are unique. ## get the smoother matrix (x having unique values smooth.matrix = function(x, df){ n = length(x); A = matrix(0, n, n); for(i in 1:n){ y = rep(0, n); y[i]=1; yi =

Is there any way to identify or infer the inflection points in a smooth spline object? I am doing a comparison of various methods of time-series analysis (polynomial regression, spline smoothing, recursive partitioning) and I am specifically interested in obtaining the julian dates associated with the inflection points inferred by the various models. Tyler e.g.

I like what smooth.spline does but I am unclear on the output. I can see from the documentation that there are fit.coef but I am unclear what those coeficients are applied to.With spline I understand the "noraml" coefficients applied to a cubic polynomial. But these coefficients I am not sure how to interpret. If I had a description of the algorithm maybe I could figure it out but as it

Hi, I have three original curves as follows, n<-seq(20,200,by=10) t<-c(0.1138, 0.1639, 0.2051, 0.2473, 0.2890, 0.3304, 0.3827, 0.4075, 0.4618, 0.4944, 0.5209, 0.5562, 0.5935, 0.6197, 0.6523, 0.6771, 0.6984, 0.7209, 0.7453) es<-c(0.3682, 0.4268, 0.5585, 0.6095, 0.7023, 0.7534, 0.8225, 0.8471, 0.8964, 0.9098, 0.9371, 0.9514, 0.9685, 0.9747, 0.9812, 0.9859, 0.9905, 0.9923, 0.9940)

Hola, Esta es una forma de hacerlo... Mira que lo primero que he modificado es el fichero "x.csv" para sustituir los espacios en los nombres por "_". Y también he quitado los acentos y las eñes... He utilizado el paquete RNNS y la función "mlp()" para ajustar la red. #------------------------------------------- > x <- read.csv("x.csv",

Hello I have installed R-0.90.1 on my Linux (Redhat 6.2) machine, unfortunately I am not able to use a number of commands like e.g. smooth.spline and predict.smooth.spline. The error messages being given by is: Error: Object "smooth.spline" not found With the command library() I have checked or the libraries for the smoothing functions are there, as shown below. -------- >

Hi, forget about the below details. It is not related to the fact that the function is returned from a function. Sorry about that. I've been troubleshooting soo much I've been shoting over the target. Here is a much smaller reproducible example: x <- 1:10 y <- 1:10 + rnorm(length(x)) sp <- smooth.spline(x=x, y=y) ypred <- predict(sp$fit, x) # [1] 2.325181 2.756166 ...