similar to: set the lower bound of normal distribution to 0 ?

Dear list, I have a dataset containing values obtained from two different instruments (x and y). I want to generate 5 samples from normal distribution for each instrument based on their means and standard deviations. The problem is values from both instruments are non-negative, so if using rnorm I would get some negative values. Is there any options to determine the lower bound of normal

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Dear all, The 'make check' step fails for the pacakge mva on IBM AIX. The tail of the Rout log file looks like: > for(factors in 2:4) print(update(Harman23.FA, factors = factors)) Call: factanal(factors = factors, covmat = Harman23.cor) Uniquenesses: height arm.span forearm lower.leg weight 0.170 0.107 0.166

I currently work on a draft of an aquatic bioassessment. The conditions tested are the following: ER river water T dechlorinated water control 0.5 + 0.5mg / L of malate T + 1 dechlorinated water control + 1g / L of malate T ED dechlorinated water control SED + ER + river water sediment SED ED + sediment + water dechlorinated. It is the result of AChE in muscle (fillet of fish). The production of

Dear all,I am a new user of WinBUGS and need your help. After running the following code, I got parameters of beta0 through beta4 (stats, density), but I don't know how to get the prediction of the last value of h, the variable I set to NA and want to model it using the following code.Does anyone can given me a hint? Any advice would be greatly appreciated.Best

Good morning, suppose the following: m <- round(matrix(rnorm(16), ncol=4), 3) a <- rev(c(0.01, 0.025, 0.05, 0.1)) rownames(m) <- a colnames(m) <- c("0.25,0.75", "0.4,0.6", "0.1,0.9", "0.4,0.9") m 0.25,0.75 0.4,0.6 0.1,0.9 0.4,0.9 0.1 1.034 -0.119 -1.213 0.619 0.05 0.035 1.074 0.525 1.671 0.025 -1.687 0.960

Hello Everyone, My name is Thomas and I have been using R for one week. I recently found your site and have been able to search the archives of posts. This has given me some great information that has allowed me to craft an initial design to an inquiry I would like to make into the breakdown of McNemar's test. I have read an intro to R manual and the posting guides and hope I am not violating

Hi, is there a possibilty to get my function output with "cat " as data.frame.rows with variables ? Var1---------------- 8 15 1 3 Var2---------------- 0.170 0.319 0.0213 0.0638 Var3---------------- 83.8 88.6 90 75 Var4---------------- 84.3 84.3 100 83.3 Var5---------------- 62.5 56 20 53.3 function(data.frame) { .... ....

Hello all: I am confused about the output from a lm() model with an incomplete design/missing level. I have two categorical predictors and a continuous covariate (day) that I am using to model larval mass (l.mass): leaf.species has three levels - map, syc, and oak cond.time has two levels - 30 and 150. There are no response values for Map-150, so that entire, two-way, level is missing.

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

I have the following code, which tests the split on a data.frame and the split on each column (as vector) separately. The runtimes are of 10 time difference. When m and k increase, the difference become even bigger. I'm wondering why the performance on data.frame is so bad. Is it a bug in R? Can it be improved? > system.time(split(as.data.frame(x),f)) user system elapsed 1.700

Hello, I am trying to filter a data set like below so that the peaks in the Phase value are more obvious and can be identified by a peak finding function following the useful advise of Carl Witthoft. I have written the following for(i in length(data$Phase)){ newphase=if(abs(data$Phase[i+1]-data$Phase[i])>6){ data$Phase[i+1] }else{data$Phase[i] } } I get the following error which I have not

Hi,? I have two big data set.? data _1 :? > dim(data_1) [1] 15820 5 > head(data_1) ? ?Chromosome ?????Start????????End????????Feature GroupA_3 1: ? ? ? ????????chr1 521369 ?750000 ????chr1-0001 ? ?????0.170 2: ? ? ? ????????chr1 750001 ?800000 ????chr1-0002 ? ????-0.086 3: ? ? ? ????????chr1 800001 ?850000 ????chr1-0003 ? ?????0.006 4: ? ? ? ????????chr1 850001 ?900000 ????chr1-0004 ?

Hi, folks, Here are the codes: ############## y=1:10 x=c(1:9,1) lin=lm(log(y)~x) ### log(y) is following Normal distribution x=5:14 prediction=predict(lin,newdata=x) ##prediction=predict(lin) ############### 1. The codes do not work, and give the error message: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one. But if I use the code after the pound sign, it

Hi, I am experimenting with using glht() from multcomp package together with coxph(), and glad to find that glht() can work on coph object, for example: > (fit<-coxph(Surv(stop, status>0)~treatment,bladder1)) coxph(formula = Surv(stop, status > 0) ~ treatment, data = bladder1) coef exp(coef) se(coef) z p treatmentpyridoxine -0.063 0.939 0.161

Hi all, I am trying to run a glm with mixed effects. My response variable is number of seedlings emerging; my fixed effects are the tree species and distance from the tree (in two classes - near and far).; my random effect is the individual tree itself (here called Plot). The command I've used is: mod <- glmer(number ~ Species + distance + offset(area) + (1|Plot), family = poisson)

Hello. Has anyone any idea how a function would look like of a model based bootstrap, when the underlying time series follows an ARIMA(1,1,1)-process? A pure AR-process is no problem, but what is, if the time series need to be differentiated of order one or above and the additional MA-part? Sample code for a series, which follows a pure AR-process: #Series y of 192 observations, which follows

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first

I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). month X2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 1 1.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 2 2.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380

Hi, I had a logit regression, but don't really know how to handle the "Warning message: non-integer #successes in a binomial glm! in: eval(expr, envir, enclos)" problem. I had the same logit regression without weights and it worked out without the warning, but I figured it makes more sense to add the weights. The weights sum up to one. Could anyone give me some hint? Thanks a lot!