similar to: Which "apply" function to use?

...if that is possible My task is to find the longest streak of continuous days a user participated in a game. Instead of writing an sql function, I chose to use the R's rle function, to get the longest streaks and then update my db table with the results. The (attached) dataframe is something like this: day user_id 2008/11/01 2001 2008/11/01 2002 2008/11/01 2003

Dear all, I am not fluent in R and am struggling to 1) apply a lm to a weight-size dataset, thus the model has to run separately for each species, each year; 2) extract coefs, r-squared, n, etc. The data look like this: year sps cm w 2009 50 16 22 2009 50 17 42 2009 50 18 45 2009 51 15 45 2009 51 16 53 2009 51 17 73 2010 50 15 22 2010 50 16 41 2010 50 16 21 2010

How can I define environments within a function so that they are visible to calls to a sub-function? I have defined an objective function, ProfileErr = function(params,...) which I would like to optimize using standard routines (optim, nlminb,....) but which contains auxiliary variables which need to be updated along with params. No optimization routine in R that I have found has facilities

Hi all I want to assign a constant to a different column for each row eg: m[1,2] <- 0; m[2,3] <- 0; m[3,1] <- 0; m[4,2] <- 0; m[5,1] <- 0; ... etc... i've tried apply/tapply with no luck and also the following coefs <- rtt.abs[,5:8]; coefs.i <- coefs[] == 1; coefs[coefs.i] <- 0; wich results in "matrix subscripts not allowed in replacement"

#I am trying to understand how R fits models for contrasts in a #simple one-way anova. This is an example, I am not stupid enough to want #to simultaneously apply all of these contrasts to real data. With a few #exceptions, the tests that I would compute by hand (or by other software) #will give the same t or F statistics. It is the contrast estimates that R produces #that I can't seem to

Dear all, I executed svm calculation using e1071 library with a microarray data (http://www.iu.a.u-tokyo.ac.jp/~kadota/R/data_Singh_RMA_3274.txt). Then, I shuffled the data samples and executed svm calculation again. The results of 2 calculation were different (in SV, coefs and weights). I attached the script below. Could please tell me why this happens? If possible please tell me how to make

Hi, I have fit a series of ols() models, by group, in this manner: l <- ols(y ~ rcs(x, 4)) ... where the series of 'x' values in each group is the same, however knots are not always identical between groups. The result is a table of 'coefs' derived from the ols objects, by group: group Intercept top top' top'' 1 6.864 0.01 2.241 -2.65

Often in practical situations a predictor has missing values, so that poly crashes. For instance: > x<-1:10 > y<- x - 3 * x^2 + rnorm(10)/3 > x[3]<-NA > lm( y ~ poly(x,2) ) Error in poly(x, 2) : missing values are not allowed in 'poly' > > lm( y ~ poly(x,2) , subset=!is.na(x)) # This does not help?!? Error in poly(x, 2) : missing values are not allowed in

Hi I'm running the latest R on a presumably up to date Linux server. 'Doing something silly I'm sure, but can't see why my saved PLMset objects come out all wrong. To use an example: Setting up an example PLMset (I have the same problem no matter what example I use) > library(affyPLM) > data(Dilution) # affybatch object > Dilution = updateObject(Dilution)

Dears Sirs During my computational work I encountered unexpected behavior when calling "ar" function, namely # time series x<-ts(c(-0.2052083,-0.3764986,-0.3762448,0.3740089,0.2737568,2.8235722,- 1.7783313,0.2728676,-0.3273164),start=c(1978,3),frequency=4,end=c(1980,3)) # ar function res.ar<-ar(x,aic=TRUE,demean=F) # call "ar" again and ............

Hi ! I would like to understand where do affinity.spline.coefs used in function compute.affinities come from ? library(gcrma) data(affinity.spline.coefs) affinity.spline.coefs X1 X2 X3 X4 X5 X1 -0.55004171 -0.58579091 -0.08870557 -0.47774242 0.23205570 0.58002746 X2 X3 X4 X5 X1 X2

How does one write formatted, say "C style", output to a text file. I want something like "y=mx+b" for my linear regressions but with m and b derived from the fit objects. Also, how does one go about extracting just the slope or intercept from such an object. - Greg [[alternative HTML version deleted]]

Hi: I would like to write a function that takes as its input a formula and outputs normalized coefficients ( coef(x)*sdv(x)/sdv(y) ). now, a formula is an object, and I cannot see how to extract the variables for obtaining sdv's. the intent is to write something like my.print.lm( formula ) { model <- lm(formula); coefs <- (t(summary.lm(model)))[1,]; tvals <-

Dear all, I have just sent an email with my problem, but I think no one can see the red part, beacuse it is black. So, i am writing again the codes: rm(list=ls()) #remove almost everything in the memory set.seed(180185) nsim <- 10 mresultx <- matrix(-99, nrow=1000, ncol=nsim) mresultb <- matrix(-99, nrow=1000, ncol=nsim) N <- 200 I <- 5 taus <- c(0.480:0.520) h <-

Hi, I have a question about running multiple in regressions in R and then storing the coefficients. I have a large dataset with several variables, one of which is a state variable, coded 1-50 for each state. I'd like to run a regression of 28 select variables on the remaining 27 variables of the dataset (there are 55 variables total), and specific for each state, ie run a regression of

Hi, Attached is a patch for cosmetics purpose. Please review. Thanks, Linfeng Zhang -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.xiph.org/pipermail/opus/attachments/20170419/34354707/attachment.html> -------------- next part -------------- A non-text attachment was scrubbed... Name: 0001-cosmetics-silk-correct-input-output-arg-comments.patch

Hi there, I am trying to use linear regression to solve the following equation - y <- c(0.2525, 0.3448, 0.2358, 0.3696, 0.2708, 0.1667, 0.2941, 0.2333, 0.1500, 0.3077, 0.3462, 0.1667, 0.2500, 0.3214, 0.1364) x2 <- c(0.368, 0.537, 0.379, 0.472, 0.401, 0.361, 0.644, 0.444, 0.440, 0.676, 0.679, 0.622, 0.450, 0.379, 0.620) x1 <- 1-x2 # equation lmFit <- lm(y ~ x1 + x2) lmFit Call:

Hola, quiero hacer una regresión lineal lm(y ~ x * grupo, data =datos) y: numérica, x: numérica, grupo: factor con dos niveles (1 y 2) pero no calcula el coeficiente de x:grupo2 a cuenta de una singularidad Coefficients: (1 not defined because of singularities) Estimate Std. Error t value Pr(>|t|) (Intercept) -1.283e+06 2.276e+04 -56.359 < 2e-16 *** x

Dear List, A posting to R-Help exposed this problem with the print method for objects of class Arima: > set.seed(1) > x <- arima.sim(n = 100, list(ar = 0.8897, ma = -0.2279)) > mod <- arima(x, order = c(1,0,1)) > coefs <- coef(mod) > mod2 <- arima(x, order = c(1,0,1), fixed = coefs) > mod2 Call: arima(x = x, order = c(1, 0, 1), fixed = coefs) Coefficients: Error

Hi. How can i simulate a binary data set from a logistic regression model?I need to manipulate parameters and so obtain my set of data. I want to show the improve in analyzing binary data with GLM(binomial) model instead of classical ANOVA or NON-MODELS procedures(relative risk-odds ratio-Pearson test of godness of fit...) Can you say me what is the right function to use? Do you know any