search for: newx

...x,mean,sd))/ntotal } wrong <- scan("wrongrawdata.txt", list(x=0)) wrongfit <- fitdistr(wrong$x, "gamma") wrongmean <- mean(wrong$x) wrongshape <- wrongfit[[1]][1] wrongrate <- wrongfit[[1]][2] good <- scan("rawdata.txt", list(x=0)) xmin = 0 newx = good$x xmean = mean(newx) xmax = max(newx)+0.15 goodhist <- hist(newx, br=seq(from=0,to=xmax,by=0.15), probability=T, col="lightyellow") initmean <- (min(newx)+max(newx))/2 totalx <- length(newx) wrongmeanshift <- wrongmean + 0.2 wrongper <- pgamma(wrongmeanshi...

...integer number) is calculated. I tried myself the formula in the on-line documentation but could not get the printed threshold in the error message. > X1 <- c(.2,-.4,-.6,-.5,-.8,-.4,-.9,0,-.2,.1,-.1,.1,.7,.9,0,.3) > X2 <- c(.2,-.4,-.6,-.5,-.8,-.4,-.9,0,-.2,.1,-.1,.1,-.7,.9,0,.3) > newX <- cbind(X1,X2) > wt <- dwt(newX, n.levels=5, boundary="periodic", fast=FALSE) > wt <- dwt(newX,filter="la8",n.levels=6, boundary="periodic", fast=FALSE) Error in dwt(newX, filter = "la8", n.levels = 6, boundary = "periodic", : Inv...

...t the mahalanobis before plotting the QQ plot, but the following error is prompt: > mah=mahalanobis(data,apply(data,2,mean),var(data)) Error in FUN(x, aperm(array(STATS, dims[perm]), order(perm)), ...) : non-numeric argument to binary operator In addition: Warning messages: 1: In mean.default(newX[, i], ...) : argument is not numeric or logical: returning NA 2: In mean.default(newX[, i], ...) : argument is not numeric or logical: returning NA 3: In mean.default(newX[, i], ...) : argument is not numeric or logical: returning NA 4: In mean.default(newX[, i], ...) : argument is not nume...

...emod<-tree(y~x) where y is a factor and x is a matrix of numeric predictors. They have dimensions: > length(y) [1] 1163 > dim(x) [1] 1163 75 I?ve evaluated the tree model and am happy with the fit. I also have a matrix of cases that I want to use the tree model to classify. Call it newx: > dim(newx) [1] 68842 75 The column names of newx match the column names of x. It seems that prediction should be straightforward. To classify the first 10 values of newx, for example, I think I should use: > predict(treemod, newx[1:10,], type = "class") However, this retu...

...;Correlation:"? At first, I thought that this was describing possible correlations among the predictor variables, but then I saw that it also included the model intercept. What do these correlation value mean? ##More detailed information ##function calls: sppl.i.xx = gls(all.all.rch~l10area+newx, data = gtemp, method="ML") sppl.i.ex = gls(all.all.rch~l10area+newx, data = gtemp, method="ML", correlation = corExp(c(20,.8), form=~x+y|area, nugget=TRUE)) ##model summaries > summary(sppl.i.xx) Generalized least squares fit by maximum likelihood Model:...

Hello, How can I label a secondary axis in R? At the moment it's labelled as c(-100,200). Obviously I would like it to be more sensible. Here is the code I am using newx = -100+37.5*((1:9)-1) axis(4,at=newx,labels=(newx+100)/3750) Thanks, Rob -- View this message in context: http://www.nabble.com/Labelling-a-secondary-axis-in-R-tp16807708p16807708.html Sent from the R help mailing list archive at Nabble.com.

...RGIN <- dimnum.from.dimnamename( X,MARGIN ) # this line was added to apply s.call <- ds[-MARGIN] s.ans <- ds[MARGIN] d.call <- d[-MARGIN] d.ans <- d[MARGIN] dn.call <- dn[-MARGIN] dn.ans <- dn[MARGIN] d2 <- prod(d.ans) if (d2 == 0L) { newX <- array(vector(typeof(X), 1L), dim = c(prod(d.call), 1L)) ans <- FUN(if (length(d.call) < 2L) newX[, 1] else array(newX[, 1L], d.call, dn.call), ...) return(if (is.null(ans)) ans else if (length(d.ans) < 2L) ans[1L][-1L] else...

...f each row in our function but it will take several days to complete. The idea is to see if we can use our function with apply. Basically, data$Results<-apply(data,1, ColChange, LowLim=data[1], HighLim=data[2], Vals=data[3:6], NumConsecOut=2) But we get the following error: ?Error in FUN(newX[, i], ...) : unused argument(s) (newX[, i])? Any idea about this error or an alternative way to obtain the results we look for. Thank you very much, Camilo ################# Our function is: ####################### ColChange <- function(LowLim, HighLim, Vals, NumConsecOut) { cols <...

...9;bottomright' ,legend=rownames(data), pch=rang, col='black', cex=.7) title(main='IUB (abk-cmj)/cmj *100 (XY)', sub=paste('correlation:',cor(serie0,serie1)), cex.sub=0.75, font.sub=3, col.sub='grey30') mylm<-lm(serie1~serie0) abline(mylm,col="red") newx<-seq(min(serie0),max(serie0),length.out=length(serie0)) prd<-predict(mylm,newdata=data.frame(serie0=newx),interval = c("confidence"), level = 0.90,type="response") lines(newx,prd[,3],col="red",lty=2) lines(newx,prd[,2],col="red",lty=2) text(newx[1],pr...

...4000 columns. While running svm function I am ending up with the following error. trainfile <- read.csv('0_train_0016435.csv',head=TRUE,na.strings = "NULL") datatrain <- subset(trainfile,select=c(-Class)) model <- svm(datatrain, kernel="radial") Error in FUN(newX[, i], ...) : 'x' is empty I tried substituting "NULL" strings in the data with some numeric values but still I am ending up with error as: model <- svm(datatrain, kernel="radial") Error in FUN(newX[, i], ...) : missing observations in cov/cor In addition: Warning me...

I am hitting an odd message "Error in FUN(newX[, i], ...) : all arguments must have the same length". I can't supply the data as it's a huge data frame but I think this has enough diagnostic information to show the issue. I am sure I am missing something obvious. I've put some extra comments in but otherwise this is cut and...

...as bs,ns, and poly automatically? I am using R 2.3.0 this is their example: n <- 100 set.seed(86) # For reproducibility of the random numbers x <- sort(runif(n)) y <- sort(runif(n)) fit <- lm(y ~ bs(x, df=5)) plot(x, y,col="blue") lines(x, fitted(fit), col="black") newx <- seq(0, 1, len=n) points(newx, predict(fit, data.frame(x=newx)), type="l", col=red, err=-1) thanks, Spencer [[alternative HTML version deleted]]

Hi Chris, Maybe the na.rm=TRUE is affecting things. Try this apply(datTAF[,75:78],2,function(x){ sum(!is.na(x)) }) HTH, Eric On Tue, Sep 26, 2017 at 9:53 AM, Chris Evans <chrishold at psyctc.org> wrote: > I am hitting an odd message "Error in FUN(newX[, i], ...) : all arguments > must have the same length". I can't supply the data as it's a huge data > frame but I think this has enough diagnostic information to show the > issue. I am sure I am missing something obvious. I've put some extra > comments in but otherw...

Could someone please suggest a more clever solution to the following problem than my loop below? Given X a 2xN matrix X, and I a k-subset of N, Generate the (2^k)xN matrix Y with columns not in I all zero and the other columns with all choices of an entry from the first or second row of X. For example, with X <- matrix(1:8, nrow=2) I <- c(1,3) X is 1 3 5 7 2 4 6 8 and Y should be 1 0 5

Dear all with image I can plot only one set of values in one plot. Do somebody have any insight how to put those 2 matrices into one picture so that in one cell in image picture are both values from mat[1,1] and mat2[1,1]. mat<-matrix(1:4, 2,2) mat2<-matrix(4:1,2,2) x <-1:2 y <-1:2 image(x, y, mat) image(x, y, mat2) The only way I found is to mix x or y for both matrices let

...2 for (ctr in 1:10) { # my problem here the both x and y still show the original values from step 1 # in spite of making changes to the old values of the arrays x and y in the function function (x,y) ??? } step3 output < - function(parX,parY){ Variables for New X and Y newx <- array(parX, dim=c(1,length(parX))) newy <- array(parY, dim=c(1,length(parY))) # make some calculation and updated some arrays element in the newX and # #newY # finally assign the global original values x and y with newX and newY...

Colleagues, I am using "nls" successfully (2.11.1, OS X) but I am having difficulties retrieving part of the output - residual sum of squares. I have assigned the output to FIT: > > FIT > Nonlinear regression model > model: NEWY ~ PMESOR + PAMPLITUDE * cos(2 * pi * (NEWX - POFFSET)/PERIOD) > data: parent.frame() > PMESOR PAMPLITUDE POFFSET > 1153.02 -1183.09 24.58 > residual sum-of-squares: 1815056 > > Number of iterations to convergence: 8 > Achieved convergence tolerance: 1.643e-08 As you can see, the object cont...

Hello R Gurus, I am perplexed by the different results I obtained when I ran code like this: set.seed(100) test1<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200) predict(test1, newdata=cbind(NewBinaryY, NewXs), type="response") and this code: set.seed(100) test2<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200, xtest=NewXs, ytest=NewBinarY) The confusion matrices for the two forests I thought would be the same by virtue of the same seed settings, but they differ as do the...

Hi, I am trying to simulate animal movement in a gridded landscape made up of cells. At each time step (iteration), the animal moves from one cell to another in a random fashion. This is how I am simulating movement, where a and b are the x,y co-ordinates of the animal at the previous time step: for (i in 1:no.of.steps){ direction <- sample(1:8, 1) if(direction == 1){ a <- a b <- b -

...time<-5 # time extent tsteps<-501 # time steps newtime<-seq(0,time,len=tsteps) #### Now the things specific for the dynamics along x lam1<- -beta/2*(1+sqrt(1+4*St)) lam2<- -beta/2*(1-sqrt(1+4*St)) xmin<- -0.5 xmax<-0.5 x0<-0.1 vx0<-x0 nx<-101 ## grid intervals along x newx<-seq(xmin,xmax,len=nx) # grid along x # M1 <- do.call("g", c(list(x = newx), mypar)) mypar<-c(q,lam1,lam2) sig_xx<-do.call("sigma_pos",c(list(t=newtime),mypar)) mypar<-c(lam1,lam2,x0,vx0) exp_x<-do.call("expect_position",c(list(t=newtime),mypar))...