If your original is: h1 <- hist(t1,breaks=15) plot(h1,xlim=c(0,200),col=”red”) then does this do it? h1 <- hist(t1,breaks=15) plot(h1,xlim=c(-10,190),col=”red”) Not having your t1 data, I can’t tell. But is that what you are trying to do? solved Manipulating a histogram plot in R
The following will convert all non-zero numeric values to 1: df.richness %>% mutate_if(is.numeric, ~1 * (. != 0)) while df.richness %>% mutate_if(is.numeric, ~1 * (. > 0)) will do that with those greater than zero. solved Replace values greater than zero with 1 in r
It would help if you provided more information. Here is an example: df <- data.frame(x=1:26, y=as.factor(LETTERS)) paste(df$x, df$y) [1] “1 A” “2 B” “3 C” “4 D” “5 E”… paste(df$x, df$y, sep=””) [1] “1A” “2B” “3C” “4D” “5E”… It doesn’t matter what class the elements are, the engine will convert them to character class. If … Read more
library(ggplot2) ggplot(dat, aes(x = Experiment, y = Mean, colour = Method, linetype = as.factor(Values), shape = as.factor(Values))) + geom_line() + geom_point() where dat is the name of your data frame. 3 solved comparing two protocol for different values in R [closed]
Here you go. I acknowledge you specified a loop in your question, but in R I avoid loops wherever possible. This is better. This uses plyr::join_all to join all your data frames by Item and LC, then dplyr::mutate to do the calculations. Note you can put multiple mutations in one mutate() function: library(plyr) library(dplyr) library(tidyr) … Read more
Assuming that your table is called df and the columns of your table are named X1, X2, X3 try the following: library(dplyr) library(tidyr) df %>% spread(X2, X3) 1 solved Conditional columns to row [duplicate]
First, you must to define exactly your time-space. If you want a daily model you must to use “date” as your variable time. For this case, I think waht you need is a time series forecasting model. Linear Regression is an option but there are more sofisticated and useful models for this case. I leave … Read more
This is a very basic use of subset. data |> subset(ex != 0 | is.na(ex)) Output: #> ex #> 1 2 #> 2 2 #> 3 NA #> 4 NA #> 6 1 #> 8 NA #> 9 2 #> 10 3 PS – create reproducible example data <- structure(list(ex = c(2, 2, NA, NA, … Read more
df <- X – Y plyr::ldply(1:nrow(df), function(x) data.frame( min=min(df[x,], na.rm=T), max=max(df[x,], na.rm=T))) 1 solved How to write a function in R to calculate the min/max difference between two dataframes?
I would suggest using the function expand.grid which will automatically generate all the combinations. Also in your code unique(planningItems[“Subject”]), it will return a data frame which is actually not a good idea for this case. A vector would be better. Here is my code: uniquesubject = unique(dfSubClass$Subject) uniqueclass = unique(dfSubClass$Class) newDF=expand.grid(uniquesubject,uniqueclass) If using for loops, … Read more
See ?likert (from HH). Use the box.ratio argument: likert( Q, col=PCWP, main=”projects”, xlab=”Percentage”, ylab=”Question”, box.ratio=0.5, par.settings= list( layout.widths=list(axis.right=1, axis.key.padding=0, ylab.right=1, right.padding=1) ) ) 4 solved Decrease vertical thicknes of bars in Likert Chart R
This is not really a merge so much as a concatenation, since you don’t have to worry about matching up by values in common columns: dfA = read.table(text=” a b c 1 1 11 21 2 2 12 22 3 3 13 23 4 4 14 24 5 5 15 25″) dfB = read.table(text=” x … Read more
Many options here. Unfortunately you don’t show any effort to resolve the problem or even to make it clear and reproducible. For example, using ifelse : dat <- read.table(text=””aaa” 28 “aac” 8 “aag” 20″) dat$V1 <- with(dat,ifelse(V1 %in% c(‘aaa’,’aag’),’K’,’N’)) dat V1 V2 1 K 28 2 N 8 3 K 20 EDIT If I want … Read more
You needed to store objects from qnorm() and qcauchy(). Try: p <- seq(0, 1, .01) q1 <- qnorm(p, mean = 0, sd = 1) q2 <- qcauchy(p, location = 0, scale = 1) plot(q1, q2, main = “Normal-Cauchy”, xlab = “Normal”, ylab = “Cauchy”) 2 solved Error in seq.int(from, to, length.out = n) : ‘from’ … Read more
First put all files in a single folder. filenames <- list.files(pattern = “.csv”) all <- lapply(filenames, function(name) { readr:: read_csv(name) }) Based on your description the list should have a length of 300, each item containing a dataframe. If you want to bind the rows of all dataframes you can use dplyr’s bind_rows(). Below is … Read more