comb <- combn(colnames(data),v) colnames(results) <- apply(comb,2,function(rows) paste0(rows, collapse = “”)) insted of paste use paste0 @henrik and @chargaff solved how paste function working in R? [duplicate]
There are a few things that could be wrong. Primarily, you do need to modify the DESCRIPTION file to include LinkingTo: Rcpp, RcppArmadillo and ensure that #include <RcppArmadillo.h> is present in each .cpp file in the /src directory. You will also need to include two Makevars files. Makevars.win and Makevars with: PKG_LIBS = $(LAPACK_LIBS) $(BLAS_LIBS) … Read more
http://www.r-bloggers.com/creating-guis-in-r-with-gwidgets/ The gWidgets framework is a way of creating graphical user interfaces in a toolkit independent way. That means that you can choose between tcl/tk, Gtk, Java or Qt underneath the bonnet. There’s also a web-version based upon RApache and ExtJS. About web integration features, take a look at the RApache project for documentation and … Read more
With a data.frame named dat that looked like: rank name country category sales profits assets marketvalue 21 21 DaimlerChrysler Germany Consumer_dur 157.13 5.12 195.58 47.43 Try (untested d/t the numerous spaces in the text preventing read.table from making sense of it): aggregate(dat[ , c(“sales”, “profits”, “assets”, “marketvalue”)], # cols to aggregate dat[“country”], # group column … Read more
Where did you come across the expression? It’s very likely that this is just to get you thinking about evaluations and booleans in R. Essentially, work inside of the parentheses and work outward until there is nothing left to evaluate. Let’s work through it step by step: ((111 >= 111) | !(TRUE)) & ((4 + … Read more
You could use paste to create the full name and filter to subset on that new variable library(dplyr) filter(d,paste(NAME,SURNAME) %in% c(“Giovanni Bianchi”,”Luca Rossi”)) NAME SURNAME COLOR 1 Giovanni Bianchi Red 2 Giovanni Bianchi Blue 3 Luca Rossi Blue 4 Luca Rossi Red Data d <- read.table(text=” NAME SURNAME COLOR Giovanni Rossi Red Giovanni Bianchi Red … Read more
You may do: # data head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa the function foo_divide <- function(x, y){ foo <- … Read more
We can try library(dplyr) df1%>% filter(LocationName != “NULL” & LocationID!=0) solved How to remove rows with NULL / Zero (0) in R
Using stringr and tidyr: data.frame(column_1 = str_extract(v, “(?<=\\s\\()[^\\s]*?(?=\\),)”), column_2 = replace_na(str_extract(v, “(?<=transcript\\svariant\\s)\\d”), 1)) 4 solved i want to matche some substring in my vector with regular expressions with R language [closed]
If subset is truly a subset of df and no additional rows have been omitted or added to df, filter using row names will work. xy <- data.frame(sub = rep(letters[1:3], 9), val = runif(9)) xy.sub <- xy[xy$sub %in% “b”, ] xy[!rownames(xy) %in% rownames(xy.sub), ] To match multiple columns, you can do xy[!(xy$val %in% xy.sub$val & … Read more
I used solution provided by NelsonGon, but maked it a bit shorter. Besides i always try to avoid of using third party libraries if it’s possible. So the code working for me was: ggplot(df_test,aes(x=factor(genotype),fill=factor(type)))+geom_bar(position=”fill”)+xlab(“Genotype”)+scale_y_continuous(labels=scales::percent_format()) solved Stacked barplot with percentage in R ggplot2 for categorical variables from scratch
Noting your actual attempted solution posted in the comment to @ChristopherLouden’s answer, which looks suspiciously like the solution offered by @Jilber to a question from earlier today, I can finally reproduce your problem and offer a solution. For the sake of simplicity, here’s a much smaller data.frame to start our work with. Note that the … Read more
The format argument you give should reflect the format that the data is currently in, not the format that you want to convert it to, so you would have to set format = “%Y-%m-%d”. Read the documentation on strptime again and it should make sense. 1 solved strptime returning NA values [closed]
You can filter the data based on the column, then do the count for task : df <- data.frame( student = c( rep(“A”, 4), rep(“B”, 4), rep(“C”, 4), rep(“D”, 4) ), task = rep( c(“Home”, “Class”, “Assign”, “Poster”), 4 ), res = sample( c(“Completed”, “Pending”, “Not performed”), 16, TRUE ) ) library(dplyr) #> #> Attaching … Read more
Try something in the veins of this: library(wordcloud) words <- c(“foo”, “bar”) freqs <- c(10, 3) wordcloud(words = sprintf(“%s (%s)”, words, freqs), freq = freqs) ?sprintf and ?paste might be helpful. 0 solved How to display frequency in wordcloud