similar to: feature extraction on time series data

I am currently trying to find MLE of a function with four parameters. My codes run well but i don't get the results. I get the following message: BHHH maximisation Number of iterations: 0 Return code: 100 Initial value out of range. I don't know this is so because of the way i have written my loglikelihood or what. The loglikelihood LogLik<-function(param){ beta_1<-param[1]

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",

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,

A data set(dat),has 2 variables: x and a, and 100 rows. I wanna add 2 variables,and call the new data set dat1: var1:f = a/median(a) var2:x_new = x*f My solution: dat1<-transform(dat,f = a/median(a),x_new = x*f) But gets error reply which says that "f" is not exits since dat has no variables called "f". So I have to do through 2 steps:

Dear all, I have several character strings with a high number of different levels. unique(x) gives me values in the range of 100-200. This creates problems as I would like to use them as predictors in a coxph model. I therefore would like to convert each of these strings to a new string (x_new). x_new should be equal to x for the top n categories (i.e. the top n levels with the highest

Dear all, I am using modified LARS algorithm (ref: The Adaptive Lasso and Its Oracle Properties, Zou 2006) for adaptive lasso penalized linear regression. 1. w(j) <- |beta_ols(j)|^(-gamma) gamma>0 and j = 1,...,p 2. define x_new(j) <- x(j)*w(j) 3. apply LARS to solve modified lasso problem out.adalasso <- lars(X_new,y,type="lasso") or enet(X_new,

Hi, folks, Happy work in weekends >_< My question is how to plot the dependent variable against one of the predictors with other predictors as constant. Not for the original data, but after prediction. It means y is the predicted value of the dependent variables. The constane value of the other predictors may be the average or some fixed value. ####### y=1:10 x=10:1 z=2:11

Hi all, Are there functions in R that could help me do the following? We have a special type of regression which is called Geometric Mean Regression. We have done some search and found the following: https://stat.ethz.ch/pipermail/r-help/2011-July/285022.html The question is: how to do the statistical inference on GMR results? More specifically, we are looking for the prediction interval:

Hi Users, I want to fit the maximums of a data series with a cubic spline. How do I go about this in R. I failed to figure out how I can use the mgcv library to do this. Thanks ---------------------------- ZABLONE

I have some data fit to a smooth.spline object as follows: (x=vector of data for the predictor variable, y=vector of data for the response variable) fit <- smooth.spline(x,y) Now, given a spline fit point y_new, I want to be able to find out what value of x_new yielded this fit value. How to do so? (This problem is the inverse of the predict.smooth.spline function, which takes x_new as input

Hello, I am using R and the smooth.Pspline function in the pspline package to smooth some data by using natural cubic splines. After fitting a sufficiently smooth spline using the following call: (ps=smooth.Pspline(x,y,norder=2,spar=0.8,method=1) [the values of x are age in years from 1 to 100] I tried to check that R in fact had fitted a natural cubic spline by checking that the resulting

re: Cubic splines in package "mgcv" I don't have access to Gu (2002) but clearly the function R(x,z) defined on p126 of Simon Wood's book is piecewise quartic, not piecewise cubic. Like Kunio Takezawa (below) I was puzzled by the word "cubic" on p126. As Simon Wood writes, this basis is not actually used by mgcv when specifying bs="cr". Maybe the point is

I'm looking for a function in R similar to t.test() which was generously pointed out to me yesterday, but which can be used for normally distributed data. To recap yesterday: > x <- scan() 1: 62 52 68 23 34 45 27 42 83 56 40 12: Read 11 items > alpha<- .05 > t.test(x) One Sample t-test data: x t = 8.8696, df = 10, p-value = 4.717e-06 alternative hypothesis: true

Hi! > > > Since it is a PCIe card, it does not have the ability to host hardware > > > devices for networking, storage and console. We provide these devices > > > on X100 coprocessors thus enabling a self-bootable equivalent environment > > > for applications. A key benefit of our solution is that it leverages > > > the standard virtio framework for

On Wed, 2013-08-21 at 10:51 -0700, Sudeep Dutt wrote: > ChangeLog: > ========= > > v2 => v3: > a) Patch 1 data structure cleanups, header file include cleanups, > IDA interface reuse and switching to device_create_with_groups(..) > as per feedback from Greg Kroah-Hartman. > b) Patch 7 signal documentation, sleep workaround removal and sysfs > access API

Hi, all I am trying to get the basis matrix and penalty matrix for natural cubic splines. In the "splines" package of R,"ns" can generate the B-spline basis matrix for a natural cubic spline. How can I get the basis matrix and penalty matrix for natural cubic spline. Thanks a lot! Lee [[alternative HTML version deleted]]

On Thu, 2013-08-15 at 12:14 +0200, Pavel Machek wrote: > Hi! > Hi! > > > > Since it is a PCIe card, it does not have the ability to host hardware > > > > devices for networking, storage and console. We provide these devices > > > > on X100 coprocessors thus enabling a self-bootable equivalent environment > > > > for applications. A key

On Thu, 2013-08-15 at 12:14 +0200, Pavel Machek wrote: > Hi! > Hi! > > > > Since it is a PCIe card, it does not have the ability to host hardware > > > > devices for networking, storage and console. We provide these devices > > > > on X100 coprocessors thus enabling a self-bootable equivalent environment > > > > for applications. A key

Hi, I'm using natural cubic splines from splines::ns() in survival regression (regressing inter-arrival times of patients to a queue on queue size). The queue size fluctuates between 3600 and 3900. I would like to be able to run predict.survreg() for sizes <3600 and >3900 by assuming that the rate for <3600 is the same as for 3600 and that for >4000 it's the same as for

Hi Everyone! I live in Sweden and can not get CallerID to work on analog incoming lines. I m trying to find out if DTMF style CallerID works on a FXO card (X100). I`v seen one solution with a modem attached in parallel with the X100 just to provide the ID on its serial port. It must be much better if this can be implemented in to the X100 driver. Any info about this would be highly appreciated.