search for: ynew

...t a maximum of 20 values has been removed from variable Y. How can I tell R to put some removed values back in the data set, or to remove more values until a maximum number of removed values has been reached? y <- rnorm(40,1,3) x <- 1+2*y1+ rnorm(40,0,5) #Remove values in Y dependent on X: ynew <- rep(NA,40) for (j in 1:40) { if (x[j] < 0){ynew[j] <- rbinom(1,1,0.50)} if (x[j] > 0){ynew[j] <- rbinom(1,1,0.10)} } -- View this message in context: http://r.789695.n4.nabble.com/Simulating-data-loop-tp3019044p3019044.html Sent from the R help mailing list archive at Nabble.com.

...Mz = Mxx + Myy; Cov_xy = Mxx*Myy - Mxy*Mxy; Mxz2 = Mxz*Mxz; Myz2 = Myz*Myz; A2 = 4*Cov_xy - 3*Mz*Mz - Mzz; A1 = Mzz*Mz + 4*Cov_xy*Mz - Mxz2 - Myz2 - Mz*Mz*Mz; A0 = Mxz2*Myy + Myz2*Mxx - Mzz*Cov_xy - 2*Mxz*Myz*Mxy + Mz*Mz*Cov_xy; A22 = A2 + A2; epsilon=1e-12; ynew=1e+20; IterMax=20; xnew = 0; # Newton's method starting at x=0 epsilon=1e-12; ynew=1e+20; IterMax=20; xnew = 0; iter=1:IterMax for (i in 1:IterMax){ yold = ynew; ynew = A0 + xnew*(A1 + xnew*(A2 + 4.*xnew*xnew)); if (abs(ynew) > ab...

...it just doesnt look as if they match in their scale (the barplot is much wider than the "plot"....even though I tried to put limits on the x-axis). Here is an example of what I did: barplot(y, xaxt="n",yaxt="n",ylim=c(-1,45), xlim=c(1,55)) ... par(new=TRUE) plot(x, ynew, lty=2, type="l", ylim=c(0,15), xlim=c(1,55)) Another question: how can I make sure that the "0"-values from the barchart are displayed as well? Thank you so much! Anne-Katrin -- Psst! Geheimtipp: Online Games kostenlos spielen bei den GMX Free Games! http://games.ent...

Dear experts, I am a beginner of R. I'm looking for experts to guide me how to do programming in R in order to randomly replace 5 observations in X explanatory variable with outliers drawn from U(15,20) in sample size n=100. The replacement subject to y < 15. The ultimate goal of my study is to compare the std of y with and without the presence of outliers based on average of 1000

dear:teacher i have a problem which about the polr()(package "MASS"), if the response must have 3 or more levels? and how to fit the polr() to 2 levels? thank you. turly yours [[alternative HTML version deleted]]

...data series and order represents the level of the process. In other words if order=2 then we have an AR (2) process. Now it is easy to to create the y vector within the function, but I'm not clear on how to create the design matrix. For instance if order=2 then y <- as.matrix(rnorm(100)) ynew <- as.matrix(y[3:nrow(y),1]) x <- as.matrix(cbind(rep(1, nrow(y)-2), y[2:(nrow(y)-1),1], y[1:(nrow(y)-2),1])) ynew and x gives me the response vector and design matrix respectively. however, I'm trying to write a general function which will accomodate any order. Hence given the user inpu...

...on sampled data sets. To speed up calculations, I've tried to use internal functions from the Matrix package (as suggested ealier on the list by Doug Bates). So I did: fm2 <- lmer(resistance ~ ET + position + (1|Grp), Semiconductor,method='ML') simdata<-simulate(fm2,nsim=1) ynew <- simdata[,1] mer <- fm2 .Call("mer_update_y", mer, ynew, PACKAGE = "Matrix") mer1u <- LMEoptimize(mer, lmerControl(mer)) What puzzles me is that this call alters my original model fm2 as some kind of side effect. In fact, after the call fm2 is the same as mer1u....

Hi, I have a problem when I want to add new points (or a new line) to the graph. Some points (or parts of the line) are not shown on the graph because they lie beyond the scale of the axis. Is there a way to overcome this so all points (or the entire line) are shown on the graph? Here's an example of my problem: colors = c("red", "blue") plot(x=rnorm(100,0,1),

I think that this is an easy one... I have a matrix where each row is an (x,y,z) triplet. Given a potential (xnew,ynew,znew) triplet I want to know if the matrix already contains a row with the new values (the space already has that point). I can do it using a for loop, but I would like to know if there is anyway in which I can do it without the for loop. I do it now like this (this algorithm appears to be corr...

...ssuming a constant movement during the time interval). The 'time distance' bewteen WP3 and WPnew is (120-73) 47 seconds and the 'time distance' between WPnew and WP4 is (153-120) 33 seconds (whereas total time interval between WP3 and WP4 is 80 seconds). WPnew (with coordinates Xnew,Ynew) should then be located at 80/33*100=41.25% from WP3: Xnew = X3 + (X4-X3)*41.25% and Ynew= Y3 + (Y4-Y3)*41.25% 2) Calculate the average location (average of x3,y3 and x4,y4), at which to create a new waypoint at 2 minutes. 3) A simpler alternative is that the location of the 'closer wayp...

...Loadings plot. Here it is where my struggle comes from with ggplot2. The Loadings are in this dataset! > str(data2) 'data.frame': 4 obs. of 9 variables: $ plsr1 : num 0.9522 0.2986 -0.0898 0.1561 $ plsr2 : num 0.2287 -0.9686 -0.1017 0.0105 $ xnew : num 0 0 0 0 $ ynew : num 0 0 0 0 Aparently with plain R is fairly simple, although I cannot override to the previous plot. >plot(data2$plsr1,data2$plsr2) >arrows(data2$xnew,data2$ynew,data2$plsr1,data2$plsr2,lwd=2,length = 0.2, angle = 30) How to do that with ggplot2 and override it to the previous score...

...t;", "", "B", "", "C") x <- c(0.5, 0.2, 0.3, 0.1, 0.9, 0.4) The empty lines "" always belong to the firm above. Now I want to collapse the dataset so that each firm (A,B, C, etc) has one line only, using summation. So what I would like is yNew <- c("A", "B", "C") xNew <- c(1, 1, 0.4) The problem I'm having is that each firm has a different number of entries for x, so some like C have just one and others have ten or more, so I have difficulty imagining how to use a loop in this case. I'd be gre...

...fyxv=predict(fityxv,where="data") ######## Marginal distribution of gxv # fit marginal distribution of y fitxv=locfit(~x1+x2,alpha=0.5,deg=1) gxv=predict(fitxv,where="data") ######## Prediction of fyxv and gxv # new data vx1=0.2 vx2=0.7 x1new=rep(vx1,ndat) x2new=rep(vx2,ndat) ynew=y # marginal distribution of gxv for new data newdata=data.frame(x1new,x2new) gxvnew=predict(fitxv,newdata) #bug!!! gave the same values as gxv # This bug can be avoid by setting new values into old variables # then, we will get the new predicted values # for example x1=x1new x2=x2new gxvnew2=p...

...Marginal distribution of gxv > # fit marginal distribution of y > fitxv=locfit(~x1+x2,alpha=0.5,deg=1) > gxv=predict(fitxv,where="data") > > ######## Prediction of fyxv and gxv > # new data > vx1=0.2 > vx2=0.7 > x1new=rep(vx1,ndat) > x2new=rep(vx2,ndat) > ynew=y > > # marginal distribution of gxv for new data > newdata=data.frame(x1new,x2new) > gxvnew=predict(fitxv,newdata) #bug!!! gave the same values as gxv > > # This bug can be avoid by setting new values into old variables > # then, we will get the new predicted values > #...

...acepack library. I Have a question: Is there a way to predict new responses using ACE? What I mean is doing something similar to the following code that uses PPR (Projection Pursuit Regression): library(MASS) x <- runif(20, 0, 1) xnew <- runif(2000, 0, 1) y <- sin(x) a <- ppr(x, y, 2) ynew <- predict(ppr, xnew) Any help would be much appretiated, Thanks in advance, Luis Pineda

...[i] + b[6] * I2[i] + b[7] * I3[i] log(mu[i]) <- max(-20, min(20, eta[i])) PRes[i] <- (Y[i] - mu[i]) / sqrt(mu[i]) #estimate residuals (= observed - expected / sqrt(variance)) #Discrepancy measures (used for checking overdispersion) YNew[i] ~ dpois(mu[i]) #New data PResNew[i] <- (YNew[i] - mu[i]) / sqrt(mu[i]) D[i] <- pow(PRes[i], 2) DNew[i] <- pow(PResNew[i], 2) } # Add up discrepancy measures fit <- sum(D[1:N]) fit.new <- sum(DNew[1:N]) for (i...

Hello all, I have an optimization routine that is giving me good results, but the results are not in the nice "model" format like "lm". How can I get optim results into a model so that I can use the clever 'fitted', 'residuals', and 'summary' functions? Using optim is the only way that I was able to make a model that 1) sums the betas to 1, 2)

...00), ncol = 3), display = "none") > str(sm.density(matrix(rnorm(200),ncol=2),display="none")) List of 10 $ eval.points: num [1:50, 1:2] -2.67 -2.57 -2.47 -2.38 -2.28 ... ..- attr(*, "dimnames")=List of 2 .. ..$ : NULL .. ..$ : chr [1:2] "xnew" "ynew" $ estimate : num [1:50, 1:50] 3.76e-05 5.10e-05 7.57e-05 1.22e-04 2.05e-04 ... $ h : Named num [1:2] 0.414 0.512 ..- attr(*, "names")= chr [1:2] "" "" $ h.weights : num [1:100] 1 1 1 1 1 1 1 1 1 1 ... $ weights : num [1:100] 1 1 1 1 1 1 1 1 1...

This may sound like a very naive question, but... give two lists of coordinate pairs (x,y - Cartesian space) is there any simple way to compute the affine transformation matrix in R. I have a set of data which is offset from where i know it should be. I have coordinates of the current data, and matching coordinates of where the data should be. I need to compute the composition of the affine

Hi, I was hoping to clarify the exact behavior associated with this incantation: validate(fit.ols, method='cross', B=50) Output: index.orig training test optimism index.corrected n R-square 0.5612 0.5613 0.5171 0.0442 0.5170 50 MSE 1.3090 1.3086 1.3547 -0.0462 1.3552 50 Intercept 0.0000 0.0000 -0.0040 0.0040