similar to: Converting dataframe to matrix

I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should

This is a very simple question but I couldn't form a site search quesry that would return a reasonable result set. Say I have a vector: x <- c(0,2,3,4,5,-1,-2) I want to replace all of the values in 'x' with the log of x. Naturally this runs into problems since some of the values are negative or zero. So how can I replace all of the positive elements of x with the log(x) and the

Hi everyone, I cannot install package "svmpath" on R 2.8.1 on Redhat Linux .In fact I cannot install package externally.I am new to Linux so it would be of great help if you all can help me out. Regards, Subhajit. [[alternative HTML version deleted]]

Hi all, Suppose I want to set the values in a column to the log of the values of another column like so: object$LogDist <- log10(object$Distance) How do I loop through the objects if I have object1, object2, etc to perform this function? object1$LogDist <- log10(object1$Distance) object2$LogDist <- log10(object2$Distance) object3$LogDist <- log10(object3$Distance) I was trying to

I just downloaded and installed R 2.3.0 on my Mac G5 running OS X 10.4.6. I also updated with R.app revision 3114 as recommended. Now, when I attemp to use package rgl I get the error > library(rgl) Error: package 'rgl' is not installed for 'arch=ppc' > I have tried reinstalling from CRAN using both binary and source. The source install fails, The binary install yields

I would expect that several math operations should always return values with a class of numeric. If the input is defined with multiple classes, however, the class attribute is preserved. I would think this may have some unintended side-effects. Here's an example: > sessionInfo()$R.version$version.string [1] "R version 3.4.0 (2017-04-21)" > x <- seq.int(5) >

I am trying to fit a very simple broken stick model using the package "segmented" but I have hit a roadblock. > str(data) 'data.frame': 18 obs. of 2 variables: $ Bin : num 0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 ... $ LnFREQ: num 5.06 4.23 3.50 3.47 2.83 ... I fit the lm easily: > fit.lm<-lm(LnFREQ~Bin, data=id07) But I keep getting an error

Hello, I have two questions. The svmpath package provides a svmpath function: --- fit <- svmpath(xtrain, ytrain, kernel.function = radial.kernel, param.kernel = 0.8) --- 1) How to get the optimal lambda value out of this result? The svmpath package also provides a predict function: --- ytest <- predict(fit, xtest) --- How to get a score (or a probability of belonging to one of the two

How can I get the list of non-zero features from svmpath at any given lambda? All I get is following information and information about what features were selected. In Iris example, we have 4 features and 60 cases. In my own example which is 200cases by 300 features, I can't figure out how to print the list of non-zero (beta) or features from svmpath. Any help? Thanks. ---------------- >

Hello, First time posting to this mail list. I'd like to use R in the most efficient way. I'm accomplishing what I want, but feel there is a more R'ish way to do it. Just learning R. *My goal: get ranks of value across rows with row names and column names intact.* I'm guessing one of the [?]apply functions will do what I need, but I couldn't sort out which one (after a lot

Am I trying to perform multiple linear regressions on each 'VARIABLE2'. I figured out that there are different ways, using the following code: (data is given at the end of this message) reg <- lapply(split(TRY, VARIABLE2), function(X){lm(X2 ~ X3, data=X)}) lapply(reg, summary) Which produces the following: $`1` Call: lm(formula = X2 ~ X3, data = X) Residuals: Min

I think you may be confusing (S3) class and ?mode. > x <- seq.int(1:3) > class(x) [1] "integer" > mode(x) [1] "numeric" > class(x+.5) ## coercion [1] "numeric" > mode(x+.5) [1] "numeric" But note: > y <- as.integer(1) > class(y) [1] "integer" > class(y) <- "foo" > mode(y) [1] "numeric" >

Of course SVM is for supervise learning method (classification or regression). You cannot use a boat to fly man !   Justin BEM BP 1917 Yaoundé Tél (237) 99597295 (237) 22040246 ________________________________ Envoyé le : Mardi, 5 Mai 2009, 16h07mn 10s Objet : Re : [R] Support Vector Machines Thank you. I will look this later but I think i've saw this function and i'm not sure

First I have a matrix called stocks1: > class(stocks1) [1] "matrix" Here are the first 5 rows of the last 4 columns: > stocks1[1:5,2:5] [,1] [,2] [,3] [,4] [1,] 80.73 31.95 25.4 25.69 [2,] 83.66 31.95 27.12 25.2 [3,] 83.27 32.93 28.74 26.29 [4,] 83.9 34.07 29.77 26.6 [5,] 82.74 35.18 30.24 27.41 Now, why can't I convert this into a dataframe?

Thanks Bert, I think we agree on the current behaviour, but I'm still not sure if it's desirable. The mode isn't used for method dispatch. In the following example, I have to write `log.foo` in order for the correct method to be called. > x <- seq.int(5) > class(x) <- c("integer", "foo") > half <- function(x) UseMethod("half") >

Hi R Users, I have been trying to work out how to rename column names using grep, basically I have generated these column names using tapply: [1] "NAME" "X1.1" "X2.1" "X3.1" "X4.1" "X5.1" "X6.1" "X7.1" "X8.1" [10] "X1.2" "X2.2" "X3.2" "X4.2"

Hi List I'm running a series of mixed models using lme, and I wonder if there is a way to sort them by AIC prior to testing using anova (lme1,lme2,lme3,....lme7) other than by hand. My current output looks like this. anova (lme.T97NULL.ml,lme.T97FULL.ml,lme.T97NOINT.ml,lme.T972way.ml,lme.T97fc. ml, lme.T97ns.ml, lme.T97min.ml) Model df AIC BIC logLik

Dear all, I may have overlooked something but the following piece of code causes me trouble: > a_data.frame(1:4) > a X1.4 1 1 2 2 3 3 4 4 > apply(a,2,cat) 1 2 3 4Error in ans[[1]] : subscript out of bounds > apply(a,1,cat) 1234NULL Is this the fact my data.frame has only one dimension ? (and do I forget somewhere (but where ?) to specify not to 'drop'

I am trying to set up a function which processes my data according to the following rules: 1. if (x[i]==0) NA 2. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding it +1)) The data this will apply to include a variety of whole numbers not limited to 1 & 0, a number of which may appear consecutively and not separated by zeros. Below is an example with a detailed

Hello, first of all, thanks to LT for \pkg{codeutils}. I agree that it is indeed very useful to identify errors and also to encourage re-thinking past solutions. My problem: I want to compare different sets of related sub-functions which should be used alternatively by the same top-level function. Sets of related functions should be bound together (as lists) and the workspace should be as clean