similar to: Missing shapes in legend with scale_shape_manual

I believe the missing shapes are because you had set alpha=0 for the last geom point. I expect there are better ways, but one way to handle it would be to avoid the filtering, adding columns with med and exercise status, like the following: # setup with data provided Date <- c('2023-10-17', '2023-10-16', '2023-10-15', '2023-10-14',

Hello, I'm trying to plot a graph of blood glucose versus date. I also record conditions, such as missing the previous night's medications, and missing exercise on the previous day. My data looks like: > b2[68:74,] # A tibble: 7 ? 5 Date Time bg missed_meds no_exercise <date> <time> <dbl> <lgl> <lgl> 1 2023-10-17 08:50

Hi, I was trying to replicate one of the graphs given on the ggplot2 website. I have given a sample code below. I would like to combine the legends, since each color is uniquely mapped to a shape. ### library(ggplot2) leaves <- letters[1:8] mat <- matrix(sample(1:1000,32),nrow=16,ncol=2) colnames(mat) <- paste('t',1:2,sep='') df <-

Hi, I am new to R so forgive me if the following query is somewhat simple.  I have a small tab-separated file with n records of glucose values, one record per day, seven measurements per day. It looks like this: date    sober    no    vm    nm    va    na    vs 20091229    NA    6.8    NA    2.7    11.7    2.7    6.2 I'd like to make a graph on which the glucose day curves are plotted

Dear R-experts, I guess I have a problem with my fast function (fast tau estimator) here below. Indeed, zero errors look highly suspicious. I guess there is a bug in my R code. How could I correct my R code ? # install.packages( "robustbase" ) # install.packages( "MASS" ) # install.packages( "quantreg" ) # install.packages( "RobPer" ) #

Greetings R users, I am trying to renumber my groups within the file shown below. The groups are currently set as 8,9,10,etc. I would like to renumber this as 1,2,3,etc. I have searched the help files and only come across using the rownames to renumber the values but I need to match values. Any assistance is always appreciated, Regards, Charles structure(list(Group = structure(c(1L, 1L, 1L,

Dear all, I have the following output generated by linear regression. Since there is only one regression intercept and one slope for one set of data, what is the meaning of std. error for intercept and that of slope? Thanks in advance. Sincerely, Minghua > data(thuesen) > attach(thuesen) > lm(short.velocity~blood.glucose) Call: lm(formula = short.velocity ~ blood.glucose)

Hi all, I would like to test the homoegeneity of variances between several linear model for some analysis of covariance. It seems that the Bartlett's test is a good test to use but I am having problem using with linear model and I cannot find any examples on the internet. There are some examples for comparisons of variances but not linear models. If I take the hellung data set, which is

Hello, If you want to bootstrap a statistic, I suggest you use base package boot. You would need the data in a data.frame, see how you could do it. library(boot) bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median(y - ypred)^2 } dat <-

I forgot, you should also set.seed() before calling boot() to make the results reproducible. Rui Barradas On 5/22/2018 10:00 AM, Rui Barradas wrote: > Hello, > > If you want to bootstrap a statistic, I suggest you use base package boot. > You would need the data in a data.frame, see how you could do it. > > > library(boot) > > bootMedianSE <- function(data,

Hi everybody! I want to find a closer neighbourins observation. This is my code: ########################## library(klaR) library(ipred) library(mlbench) data(PimaIndiansDiabetes2) dane=na.omit(PimaIndiansDiabetes2)[,c(2,5,9)] dane[,2]=log(dane[,2]) dane[,1:2]=scale(dane[,1:2]) zbior.uczacy=sample(1:nrow(dane),nrow(dane)/2,F)

Hi! I'm a bit lost while performing multiple comparisons on a lme model of that type: lmeglu=lme(glucose~Ath*tim,random=~1|Vol,na.action=na.omit,data=data) multc = glht(lmeglu, linfct = mcp(Ath = "Tukey", tim = "Tukey")) This works fine for identifying the effect of each factor. However, when I look for their interactions, l only obtain error messages. For example this

I understand that the hat matrix is a function of the predictor variable alone. So, in the following example why do the values on the diagonal of the hat matrix change when I go from an unweighted fit to a weighted fit? Is the function hatvalues giving me something other than what I think it is? library(ISwR) data(thuesen) attach(thuesen) fit <- lm(short.velocity ~ blood.glucose)

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Dear R-experts, I am trying to bootstrap (and average) the median squared error evaluation metric for a robust regression. I can't get it. What is going wrong ? Here is the reproducible example. ############################# install.packages( "quantreg" ) library(quantreg) crp <-c(12,14,13,24,25,34,45,56,25,34,47,44,35,24,53,44,55,46,36,67) bmi

Hi, I'm trying to figure out how anova works in R by translating the examples in Sokal And Rohlf's (1995 3rd edition) Biometry. I've hit a snag with planned comparisons, their box 9.4 and section 9.6. It's a basic anova design: treatment <- factor(rep(c("control", "glucose", "fructose", "gluc+fruct",

Hi I want to create a nice picture about my result of k -the nearest neighbours algorithm. Here is my easy code: ################################# library(klaR) library(ipred) library(mlbench) data(PimaIndiansDiabetes2) dane=na.omit(PimaIndiansDiabetes2)[,c(2,5,9)] dane[,2]=log(dane[,2]) dane[,1:2]=scale(dane[,1:2]) zbior.uczacy=sample(1:nrow(dane),nrow(dane)/2,F)

Hello, Right! I copied from the OP's question without thinking about it. Corrected would be bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median((y - ypred)^2) } Sorry, rui Barradas On 5/22/2018 11:32 AM, Daniel Nordlund wrote: > On 5/22/2018

Hi All, I have about one hundred patients and all the patients had their glucose measured on three different days. The days are all the same for all he patients. So I have three measurement for each patient . I want to know whether the day when the glucose was measured has an effect on the measurements. I was thinking to use a single factor analysis of variance but I am not sure how to do it in

Hi there, >library(faraway) >pima pregnant glucose diastolic triceps insulin bmi diabetes age test 1 6 148 72 35 0 33.6 0.627 50 1 2 1 85 66 29 0 26.6 0.351 31 0 >data(pima) >pima pregnant glucose diastolic triceps insulin bmi diabetes age test 1 6 148 72 35 0 33.6