similar to: geom_smooth

?s 05:17 de 12/08/2023, Thomas Subia via R-help escreveu: > Colleagues, > > Here is my reproducible code for a graph using geom_smooth > set.seed(55) > scatter_data <- tibble(x_var = runif(100, min = 0, max = 25) > ?????????????????????? ,y_var = log2(x_var) + rnorm(100)) > > library(ggplot2) > library(cowplot) > > ggplot(scatter_data,aes(x=x_var,y=y_var))+

G'day Thomas, On Sat, 12 Aug 2023 04:17:42 +0000 (UTC) Thomas Subia via R-help <r-help at r-project.org> wrote: > Here is my reproducible code for a graph using geom_smooth The call "library(tidyverse)" was missing. :) > I'd like to add a black boundary around the shaded area. I suspect > this can be done with geom_ribbon but I cannot figure this out. Some >

Dear R-users Please excuse me if this topic has been covered before, but I was unable to find anything relevant by searching I am currently doing a comparison of two biological variables that have a highly significant linear relationship. I know that the p-value of linear regression is not so interesting in itself, but this particular value does raise a question. How does R calculate

Hi Everyone, I just don't know how to extract the information I want from the summary of a linear regression model fitting. For example, I fit the following simple linear regression model: results = lm(y_var ~ x_var) summary(results) gives me: Call: lm(formula = y_var ~ x_var) Residuals: Min 1Q Median 3Q Max -5.9859 -1.5849 0.4574 2.0163 4.6015 Coefficients:

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! 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

Default level = 0.95. Does this mean +/- 0.025 from estimate? [[alternative HTML version deleted]]

Rhelpers: Having a problem solving this. I have an xyplot call that looks like this: print(xyplot(temp_species_EAM_Pred_Pop$x+temp_species_NULL_Pred_Pop$x~temp_species_EAM_Pred_Pop$Action,main=current_species, xlab="Action",ylab="Predicted Pop", xlim=c(xmin,xmax),ylim=c(ymin,ymax),

I think remember this error from trying to write an Excel file that already existed. If this file already exists, try to delete it and see, if this solves the issue. Besides that you're writing that you are "Using write.xlsx to extract data from an Excel file", write.xlsx() is to write an Excel file, not to read from it. Should be read.xlsx() then iirc. ? ? Gesendet:?Freitag, 15.

Hallo I am not familiar with any of packages you use (except of MASS and ggplot2) and the code is too complicated without any hint where the error could come from and what is the message you get. I wonder if anybody would like to go through your whole code. 1. data seems to be read correctly ICUData <- read.csv(file = "ICUData.csv", stringsAsFactors = TRUE) ICUData.neuro <-

Colleagues, Many of my colleagues come to me for a recommendation for statistical software. Since I am an R user, that's my typical answer. Some colleagues of mine refuse to use it because of its steep learning curve and lack of a GUI. They wanted a statistical software that's free and that had a GUI. I recently learned about JASP. See https://jasp-stats.org/ for more details This may be

plotly is _not_ associated with posit. I think you are unlikely to find expertise with plotly in their forums. You might find help at stackoverflow.com. On July 21, 2023 1:40:49 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote: >As you apparently haven't received any responses yet, I'll try to >suggest something useful. However, I have absolutely zero experience >with

Colleagues Here is my reproducible code plot_ly( ? domain = list(x = c(0, 1), y = c(0, 1)), ? value = 2874, ? title = list(text = "Generic"), ? type = "indicator", ? mode = "gauge+number+delta", ? delta = list(reference = 4800), ? gauge = list( ??? axis =list(range = list(NULL, 5000)), ??? steps = list( ??? list(range = c(0, 4800), color = "white"), ???

> On Jul 20, 2016, at 10:01 AM, Tom Subia <tgs77m at gmail.com> wrote: > > Default level = 0.95. > Does this mean +/- 0.025 from estimate? > > [[alternative HTML version deleted]] I would have guessed that it meant something along the lines of localized (or one might say "loess-ized") mean +/- 2* similarly localized standard error of the estimate. To find out

What do you mean "Rounded"? What do you expect, what do you get instead? ? > On Jul 22, 2023, at 10:40, Thomas Subia via R-help <r-help at r-project.org> wrote: > > Colleagues, > Thanks for the update. > My colleagues at work have run this script but the resulting graph output for value is rounded. How can one turn this annoying feature off? > I've

As you apparently haven't received any responses yet, I'll try to suggest something useful. However, I have absolutely zero experience with plotly, so this is just from general principles and reading the plot_ly Help file, which says for the "..." arguments: "Arguments (i.e., attributes) passed along to the trace type. See schema() for a list of acceptable attributes for a

ChatGPT-4: ------ Query: ---------------------------------- Thank you. The `value` in the code example is 2874. The plot shows a large number at the center and a smaller value computed as a delta relative to the `reference` of 4800. But the large value is given as 2870, and the smaller value is given as -1930 i.e. both values are rounded. Can I control the precision of these two numbers?

Thanks for writing this great piece of code. x = rnorm(100) boxplot(x) # you shouldn't see any outliers here although sometimes yow will # lets add some outliers intentionally x = c(21, 20, 25, x) # now 10, 15 and 20 are outliers myboxplot <- boxplot(x) # now you should see your three outliers myboxplot$out # it will print the values of the outliers How does one amend

Colleagues, Here is my reproducible code. library(plotly) t <- list( ? family = "Arial", ? size = 12, ? color = "black", ? face="bold") t3 <- list( ? family = "Arial", ? size = 12, ? color = "black", ? face="bold") t5 <- list( ? family = "Arial", ? size = 12, ? color = "black", ? face="bold")

Colleagues, Thanks for the update. My colleagues at work have run this script but the resulting graph output for value is rounded. How can one turn this annoying feature off? I've googled this but to no avail. [[alternative HTML version deleted]]