This should solve your problem: mod <- survreg((Surv(as.numeric(Time), event=Status)) ~ Prison+Dose+Clinic, data = meth, dist = “lognormal”) as you reported to me that: names(meth) # [1] “ID” “Clinic” “Status” “Time” “Prison” “Dose” Note, it is Time, not time; Also, it is Status, not status. In fact, all variables names start with a capital letter! 5 … Read more
It looks like you’re talking about a data frame, not a matrix. You can use names(Fires) to get a vector of column names, and use that to go from your variables to a column reference: X <- 1 Y <- 1 column.name <- paste0( X, Y ) Fires[ , column.name ] EDIT: in fact, you … Read more
To plot two plots (or more) in one figure, you can use the facet_grid function in ggplot2 (http://docs.ggplot2.org/current/facet_grid.html for more info). You can choose on which trait the plots will be facetted by using a formula. On the left hand side you can mention rows (graphs under each other) and on the left hand side … Read more
If all your compound terms are separated only by blanks, you can use gsub: > x = c(“hello World”, “good Morning”, “good Night”) > y = gsub(pattern = ” “, replacement = “”, x = x) > print(y) [1] “helloWorld” “goodMorning” “goodNight” You can always add more patterns to pattern argument. Read more about regular … Read more
We can try library(data.table) setDT(data)[, list(avg = mean(Score)), by = AD] 0 solved How to average score from different elements [duplicate]
I don’t word with data tables, but this is a solution that would work for data frames, and should hopefully generalize. The strategy is to use the fact that you can fill one vector with another vector, without ever having to use a loop. # make the example data sets D1 <- as.data.frame(matrix(data=(1:(20*181)), nrow=20, ncol=181)) … Read more
As Brian Hannays says: Change the definition of Alive.prob to 1-Dead.prob. A good programming habit is to try to avoid redundant (and then possibly conflicting) definitions… function(Dead.prob=.010,Dead.cost=40000,Alive.prob=1-Dead.prob,Alive.cost=50000,high=100,p.payoff=1) { return((Dead.prob * Dead.cost) + (Alive.prob * Alive.cost)) } solved Calculate simple function with R [closed]
what kind pf Problem do you have while installing “qicharts”? Do you get an error? I am asking because it is working for me. Nevertheless you might also want to have a lock at the “qcc”-Package. Here you can find something like a small introduction and here is the cran help to the package. But … Read more
Is this what you want? : test <- read.table(“test.txt”, header = T, fill = T) for(i in 1:nrow(test)){ for(j in 1:ncol(test)) { print(test[i,j]) } } 2 solved How to access data in R from read.table
Here is one. It uses the interesting property found here: https://math.stackexchange.com/a/157716 check.solutions <- function(solutions) { board <- matrix(1:81, 9, 9) row.idx <- row(board) col.idx <- col(board) squ.idx <- (row.idx – 1L) %/% 3L + 1L + 3L * (col.idx – 1L) %/% 3L grp.mat <- t(cbind(sapply(1:9, `==`, row.idx), sapply(1:9, `==`, col.idx), sapply(1:9, `==`, squ.idx))) flat.sol … Read more
I’ve copied the data from your post above so you can skip the variable assignment library(“tidyverse”) df <- read.table(file = “clipboard”, header = T) %>% as_tibble() You need to modify your data structure slightly before you pass it to ggplot. Get each of your test names into a single variable with tidyr::gather. Then pipe to … Read more
I think what you’re looking for is a sum of an infinite series. We basically need to model every scenario – that is we need to sum the expected value of getting a 5 on the 1st, 2nd, 3rd, 4th, etc rolls. To do this we can use a simple sum (not infinite, but a … Read more
You should give a reproducible example. Here some data: set.seed(1234) dat <- data.frame(Smoker=sample(c(‘Smoke’,’No_Smoke’),20,rep=TRUE), CANCER=sample(c(‘Cancer’,’NO_Cancer’),20,rep=TRUE)) Then using table you can get your contingency table: table(dat$Smoker,dat$CANCER) Cancer NO_Cancer No_Smoke 7 4 Smoke 5 4 For more information see ?table Description table uses the cross-classifying factors to build a contingency table of the counts at each combination of … Read more
We can use filter from dplyr library(dplyr) df %>% filter(ID == “WTN” & Time == “A”) 4 solved select a row based on different column value (same row) [duplicate]
You need a barplot. Height = c(4,17,12,6) Bins = c(“0-19”, “20-39”, “40-59”, “60-99”) barplot(Height, names.arg=Bins) barplot has many parameters that you might consider adjusting to make this prettier. 1 solved Create histogram from summarized data R [closed]