search for: 0.170

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

Tom Cohen <tom.cohen78@yahoo.se> skrev: Thanks Prof Brian for your suggestion. I should know that for right-skewed data, one should generate the samples from a lognormal. My problem is that x and y are two instruments that were thought to be measured the same thing but somehow show a wide confidence interval of the difference between the two intruments.This may be true that these

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

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

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.

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

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

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, 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)

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

Dear R users, I'm trying to do betareg on my dataset. Dependent variable is not normally distributed and is proportion (of condom use (0,1)). But I'm having problems: gyl<-betareg(cond ~ alcoh + drug, data=results) Error in optim(par = start, fn = loglikfun, gr = gradfun, method = method, : initial value in 'vmmin' is not finite Why is R returning me error in optim()? What

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 folks, after upgrading a Centos container (LXC) from 7.4 to 7.5 I got this: # systemctl status Failed to get D-Bus connection: Connection refused # busctl Failed to connect to bus: Connection refused # ps -ef | grep db[u]s dbus 55 1 0 Jul02 ? 00:00:00 /usr/bin/dbus-daemon --system --address=systemd: --nofork --nopidfile --systemd-activation # lsof -p 55 COMMAND PID

Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2<-as.matrix(as.numeric(mat1>0.25)) but that returns a 1:100000 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre

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!

Dear all, I have a quasipoisson glm for which I need confidence bands in a graphic: gm6 <- glm(num_leaves ~ b_dist_min_new, family = quasipoisson, data = beva) summary(gm6) library('VIM') b_dist_min_new <- as.numeric(prepare(beva$dist_min, scaling="classical", transformation="logarithm")). My first steps for the solution are following: range(b_dist_min_new)