similar to: error: could not find function "sd"

I need some help for a curious question of a friend of mine. She usually does some experiments (3-5 repeats for each exp) and then she calculates mean and standard deviation. In microsoft excel she writes something like the following sample mean sd a 1.25 0.35 b 2.65 0.65 c 3.45 0.50 She can do a vertical barplot graph just giving mean value and specifying the

I cannot file suggestions on bugzilla, so writing here. As far as I can tell, the manual help page for ``sd`` ?sd does not explicitly mention that the formula for the standard deviation is the so-called "Bessel-corrected" formula (divide by n-1 rather than n). I suggest it should be stated near the top. I would also suggest (feature request!) that either - a population standard

In "An Introduction to R" Version 3.3.1, in "4.2 The function tapply() and ragged arrays", after stderr <- function(x) sqrt(var(x)/length(x)) , there is a note in brackets: Writing functions will be considered later in [Writing your own functions], and in this case was unnecessary as R also has a builtin function sd(). The part "in this case was unnecessary as R also

hi could you help me to solve this issue Question: Using command rweibull(100,8,15), simulate n = 100 realizations from Weibull(8; 15) distribution. Using the simulated sample, compute the sample mean, variance and standard deviation of these observations. I am trying like this sim<-rweibull(100,8,15) # simulated sample SM<-mean(sim) # simulated sample mean var(sim) # variance

While you are editing that, you might change its name from 'stderr' to standardError (or standard_error, etc.) so as not to conflict with base::stderr(). Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Sep 13, 2016 at 8:55 AM, Martin Maechler <maechler at stat.math.ethz.ch > wrote: > >>>>> Suharto Anggono Suharto Anggono via R-devel <r-devel at

Hi, I want to do a meta-analysis with count data for treatement/control cases. Mi problem is that I need to use zero values (an informative zero value) for the mean and standard deviation for one of the treatement, but R has a problem: "Studies with zero values for sd.e or sd.c get no weight in meta-analysis". I can agroup the case by Family (byvar=Family). ¿Can you help me? Thanks!

dear Researchers, i am looking for a function to plot a barplot for each mean value and the related standard deviation, and i can close my week. This is an example of my data set. really Thanks in advance for any help or suggestions Gianni My.mean <- data.frame(Mean=c(0.4108926,0.3949009,0.4520346, 0.4091665,0.4664066,0.3048296,0.4297226,0.4056383,

Hello, I am getting a strange error while computing standard deviation. If the do the following I get the answer NA NA. Any help??? Thank you so much. a<- cbind() a<- cbind(a, 5) a<- cbind(a, 7) print(sd(a)) If I compute the mean it works fine though i.e., print(mean(a)) works just fine.

Hi! I want to do a meta-analysis with count data for treatement/control cases. Mi problem is that I need to use zero values (an informative value) for the mean and standard deviation for one of the treatement, but R has a problem: "Studies with zero values for sd.e or sd.c get no weight in meta-analysis". I can agroup the case by Family (byvar=Family). ¿anybody help me? Thanks! >

Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with

Dear list, I am using R 2.1.1 on a Fedora 3 Linux, 32 bit PC. If I compute the aggregated mean and the standard deviation I get standard deviation values for factors where the mean was not computed. It seems to me that this is somehow related to the NA values. But I don't quite understand what is going wrong? Could it be related to the data import already? Some of the imported data got the

hi there, I need to subtract each value in the dataframe from the mean of the column and divide by the standard deviation of the column. > dim(a) [1] 22011 52 data2 <- sapply(seq(from = 2, by = 1, length = 50), function(e){ rbind((a[[e]] - mean(a[,e]))/sd(a[,e])) }) Does this look right???? Any help would be appreciated. Thanks RT -- View this message in context:

Hi! This is probably completely off base, but your ymin and y max setup lines are different. One uses sqrt(y), while the second uses sqrt(length(y)). Could that play a part, please? Thank you Erin Hodgess, PhD mailto: erinm.hodgess at gmail.com On Sun, Aug 11, 2024 at 10:10?AM SIBYLLE ST?CKLI via R-help < r-help at r-project.org> wrote: > Dear community > > > > Using

Dear community Using after_stat() I was able to visualise ggplot with standard deviations instead of a confidence interval as seen in the R help. p1<-ggplot(data = MS1, aes(x= Jahr, y= QI_A,color=Bio, linetype=Bio)) + geom_smooth(aes(fill=Bio, ymax=after_stat(y+se*sqrt(length(y))), ymin=after_stat(y-se*sqrt(y))) , method = "lm" , formula = y ~ x +

Hi all, Can anyone tell me why I am getting different results in calculating SD of 2 numbers ? > (1.25-0.95)/2 [1] 0.15 > sd(c(1.25, 0.95)) [1] 0.2121320 # why it is different from 0.15? Regards, Send instant messages to your online friends http://uk.messenger.yahoo.com [[alternative HTML version deleted]]

Dear all, I have a dataset of 3 categorical factors, the mean and the standard deviation of each value. I want to use these values to plot a boxplot, grouped by each of the 3 categorical factors (24 boxplots in total). I don't have a clue on how to do the boxplot from mean and SD data already calculated. Thanks in advance [[alternative HTML version deleted]]

Any help with this problem would be greatly appreciated: I need to produce a custom plot i haven't come across in R. Basically, I want to show means, 1st standard deviation and 5th and 95th percentiles visually, using something resembling a boxplot. Is it possible to completely customize a boxplot so that it shows means as the bar (instead of, not as well as medians), standard deviations at

Dear R-helpers, I am developing a Mixed-Effects model for a study of immunoassays using 'lme4' library. The study design is as follows: 10 samples were run using 7 different immunoassays, 3 times each, in duplicates. Data attached. I have developed the following model: c.lme <- lmer(Result~SPL + (SPL|Assay/Run) -1, data=data) This model has excellent predictions - the Rsquared of

How do I trim a matrix to exclude columns that have no standard deviation? Thanks, Chris [[alternative HTML version deleted]]

Dear Prof. Ripley I noted a change in the behaviour of "cov", which is very reasonable: ## R version 2.7.0 Under development (unstable) (2007-11-30 r43565) > cov(as.numeric(NA), as.numeric(NA), use="complete.obs") Error in cov(as.numeric(NA), as.numeric(NA), use = "complete.obs") : no complete element pairs whereas earlier behavior was, for example: ## R