similar to: Red Neuronal complicada categorías

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 R Helpers, I'm struggling with a data preparation problem. I feel that it is a quite easy task but I don't get it done. I hope you can help me with that. I have a data frame looking like this: ID <- c(1,1,1,2,2,3,3,3,3) T <- c(1,2,3,1,4,3,5,6,8) x <- rep(1,9) df <- data.frame(ID,T,x) >df ID T x 1 1 1 1 2 1 1 3 1 2 1 1 2 4 1 3 3 1 3 5 1 3 6 1 3 8 1 I want to

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

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

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:

Hello; last week the power (energy) of the servers fail and it shutdown all. When the power restart in the check I find that the samba4 AD DC have problems. It work, but the users in some of the container or OU dissapear, I mean, when I check with the RSAT the OU in with the user most be, is empty When I try to run # samba-tool dbcheck this is what I have: [root at gtmad ~]# samba-tool

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please help me get this plugin installed... I have no access to techo-weenie.net due to a firewall blocking the weenie crap in its domain name. more specifically, i''m trying to do this: script/plugin install http://svn.techno-weenie.net/projects/plugins and it bombs. On a machine w/ access to this repository, I did an svn checkout and tar''d up the contents... then

I thought I would try btrfs on a new installation of f14. yes, I know its experimental but stable so it seemed to be a good time to try it. I am not sure if I have missed something out of all my searching but am I correct in thinking that currently: I. it is not possible to boot from a snapshot of the operating system and, in particular, the yum snapshots cannot be used for

Estimado Mauricio Monsalvo R es un lenguaje medio complicado, no es orientado a objetos, aunque hay formas para un trabajo con objetos, por otro lado se puede definir una función o emplear 5 paquetes para lo mismo en dos líneas de código. Usted tiene 1500 líneas de código, no es fácil de comprenderlas, tampoco hay recetas mágicas. Yo ordenaría como si fuesen cajas, una entrada, un proceso

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]

Buenas Tengo que analizar unos datos, donde tengo unas piezas fabricadas y el n?mero de ciclos qie aguantan hasta q se parten. Algunas piezas son de un fabricante, otras de otro,.. En cada ciclo a algunas piezas se le somete a algun tipo de tratamiento termico, a otras a otro, algunas se pintan, otras no... Por ejemplo, para una pieza tengo los siguientes datos: Ciclo. Tratamiento. Pintado.

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,

How do I setup relaying in icecast? I tried editing my icecast.conf file like alias techo http://192.76.151.253:8000 and like alias add techo http://192.76.151.253:8000 but neither seemed to work when I tried to open http://myipaddress:8000/techno <p>Thanks, Aaron -- palott@math.umd.edu http://www.math.umd.edu/~palott LCV: IPST 4364A (301)405-4865 Office: IPST 4364D (301)405-4843

Hola. Estoy intentando crear una función pero no logro que termine de cerrar y mi manejo no permite que el google me ayude... n.reg.dep <- function(x, y) { etiqueta <- str_replace(nombres[Variable == y, Descripcion], "[ ]", "") tabla <- datos[, .N, by = .(etiqueta = A3a)] # tabla <- tabla[, Porc := round(N/sum(N)*100,1)] # % tabla[, PorcAc :=

Hi, This is a practical question and I am sure there are many statisticians can give me a hand. I have 500 time series data (500 rows), each row contains 100 intervals, i.e., on each row, I have X1, X2, ..... X100. I am trying to reduce the dimension of this input because the data at the end of each row does not have significant meaning to the project I am doing. I used cubic splines on ea.

The Financial News, July 2002 Production Mini-plants in mobile containers. Co-investment Program "...Science Network will supply to countries and developing regions the technology and the necessary support for the production in series of Mini-plants in mobile containers (40-foot). The Mini-plant system is designed in such a way that all the production machinery is fixed on the platform of