I am trying not to use a for loop to assign values to the elements of a list.
Here, I create an empty list, gives it a length of 20 and name each of the 20 elements.
mylist <- list()
length(mylist) <- 20
names(mylist) <- paste0("element", 1:20, sep = "")
I want each element of mylist to contain samples drawn from a pool of randomly generated numbers denoted as x:
x <- runif(100, 0, 1)
I tried the following codes, which do not get to the desired result:
mylist[[]] <- sample(x = x, size = 20, replace = TRUE) # Gives an error
mylist[[1:length(mylist)]] <- sample(x = x, size = 20, replace = TRUE) # Does not give the desired result
mylist[1:length(mylist)] <- sample(x = x, size = 20, replace = TRUE) # Gives the same undesired result as the previous line of code
mylist[] <- sample(x = x, size = 20, replace = TRUE) # Gives the same undesired result as the previous line of code
P.S. As explained above, the desired result is a list of 20 elements, which individually contains 20 numeric values. I can do it using a for loop, but I would like to become a better R user and use vectorized operations as much as possible.
Thank you for your help.
Maybe replicate is what you're looking for.
mylist <- replicate(20, sample(x = x, size = 20, replace = TRUE), simplify=FALSE)
names(mylist) <- paste0("element", 1:20, sep = "")
Note that there is no need to first create a list, replicate will do it for you.
Since you're using replace=TRUE you could also generate all 400 at once and then split them up. If you were doing this many times, this probably would be faster than replicate. For only 20 times, the speed difference won't matter hardly at all and tje code using replicate is perhaps easier to read and understand and so might be preferred for that reason.
foo <- sample(x = x, size = 20*20, replace = TRUE)
mylist <- split(foo, rep(1:20, each=20))
Alternatively, you could split them by converting to a data frame first. Not sure which would be faster.
mylist <- as.list(as.data.frame(matrix(foo, ncol=20)))
Related
I am trying to generate a for loop that will repeat a sequence of the following:
sample(x = 1:14, size = 10, replace = TRUE, prob = c(1/4,1/4,1/4,1/4)
I want it to repeat 5000 times. So far, I include the above as the body of the loop and added
for (i in seq_along[1:5000]){
at the beginning but I am getting an error message saying
Error in seq_along[1:10000] : object of type 'builtin' is not subsettable
We need replicate
out <- replicate(5000, sample(x = 1:14, size = 10, replace = TRUE, prob = c(1/4,1/4,1/4,1/4)), simplify = FALSE)
There are a few issues here.
#MartinGal noted the syntax issues with seq_along and the missing ). Note that you can use seq(n) or 1:n in defining the number of loops.
You are not storing the sampled vectors anywhere, so the for loop will run the code but you won't capture the output.
You have x = 1:14 but you only have 4 prob values, which suggests you intended x = 1:4 (either that or you are 10 prob values short).
Here's one way to address these issues using a for loop.
n <- 5
s <- 10
xmax <- 4
p <- 1/4
out <- matrix(nrow = n, ncol = s, byrow = TRUE)
set.seed(1L)
for (i in seq(n)) {
out[i, ] <- sample(x = seq(xmax), size = s, replace = TRUE, prob = rep(p, xmax))
}
As andrew reece notes in his comment, it looks like you want x = 1:4 Depending what you want to do with your result you could generate all of the realizations at one time since you are sampling with replacement and then store the result in a matrix with 5000 rows of 10 realizations per row. So:
x <- sample(1:4, size = 5000 * 10, replace = TRUE, prob = c(1/4,1/4,1/4,1/4))
result <- matrix(x, nrow = 5000)
I want to make a little summary table for my colleagues in R-Markdown using qwraps::summary_table. The data.frame contains information of different exposures. All the variables are coded as binary.
library(qwraps2)
library(dplyr)
pop <- rbinom(n = 1000, size = 1, prob = runif(n = 10, min = 0, max = 1))
exp <- rbinom(n = 1000, size = 1, prob = .5)
ID <- c(1:500)
therapy <- factor(sample(x = pop, size = 500, replace = TRUE), labels = c("Control", "Intervention"))
exp_1 <- sample(x = exp, size = 500, replace = TRUE)
exp_2 <- sample(x = exp, size = 500, replace = TRUE)
exp_3 <- sample(x = exp, size = 500, replace = TRUE)
exp_4 <- sample(x = exp, size = 500, replace = TRUE)
df <- data.frame(ID, exp_1, exp_2, exp_3, exp_4, therapy)
head(df)
In the next step, I create a simple summary table as follows. In the table I want to have the groups (control vs. intervention) as columns and the exposures as rows:
my_summary <-
list(list("Exposure 1" = ~ n_perc(exp_1 %in% 1),
"Exposure 2" = ~ n_perc(exp_2 %in% 1),
"Exposure 3" = ~ n_perc(exp_3 %in% 1),
"Exposure 4" = ~ n_perc(exp_4 %in% 1))
)
my_table <- summary_table(group_by(df, therapy), my_summary)
my_table
In the next step I wanted to add a further column containing p-values for the group differences between control and intervention group, e. g. with fisher.test. I read in ?qwraps::summary_table that cbind is a suitable method for class qwraps2_summary_table, but to be honest, I'm struggling with it. I tried different ways but failed, unfortunately.
Is there a convenient way to add individual columns via qwraps::summary_table especially p-values according to the grouped columns?
Thanks for your help!
Best,
Florian
[SOLVED]
Meanwhile, after a lot of research on this topic, I found a convenient and easy way to add a p.values column. Maybe it is not the smartest solution, but worked, at least for me.
First I calculated the p.values with a function, which extracts the p.values from the returned output of fisher.test and stored them in an object, in my case a simple numeric vector:
# write function to extract fishers.test
fisher.pvalue <- function(x) {
value <- fisher.test(x)$p.value
return(value)
}
# fisher test/generate pvalues
p.vals <- round(sapply(list(
table(df$exp_1, df$therapy),
table(df$exp_2, df$therapy),
table(df$exp_3, df$therapy),
table(df$exp_4, df$therapy)), fisher.pvalue), digits = 2)
In the following step I simply added an empty table column called P-Values and added the p.vals to the column cells.
overall_table <- cbind(my_table, "P-Value" = "") # create empty column
overall_table[9:12] <- p.vals # add vals to empty column
# overall_table <- cbind(my_table, "P-Value" = p.vals) works the same way in one line of code
overall_table
In my case, I simply looked for the corresponding cell indices in overall_table (for P-Values = 9:12) and filled them using base syntax. In the vignette of qwraps2 (https://cran.r-project.org/web/packages/qwraps2/vignettes/summary-statistics.html), the author used regular expressions to identify the right cells (see section 3.2).
If there are other methods to add individual columns to qwraps2::summary_table I would appreciate to see how it is possible.
Best,
Florian
I have a data frame which looks like this
value <- c(1:1000)
group <- c(1:5)
df <- data.frame(value,group)
And I want to use this function on my data frame
myfun <- function(){
wz1 <- df[sample(nrow(df), size = 300, replace = FALSE),]
wz2 <- df[sample(nrow(df), size = 10, replace = FALSE),]
wz3 <- df[sample(nrow(df), size = 100, replace = FALSE),]
wz4 <- df[sample(nrow(df), size = 40, replace = FALSE),]
wz5 <- df[sample(nrow(df), size = 50, replace = FALSE),]
wza <- rbind(wz1,wz2, wz3, wz4, wz5)
wza_sum <- aggregate(wza, by = list(group_ID=wza$group), FUN = sum)
return(list(wza = wza,wza_sum = wza_sum))
}
Right now I am returning one list which includes wza and wza_sum.
Is there a way to return two separate list in which one contains wza and the other list contains wza_sum?
The aggregate() function needs to be in myfun() because I want to replicate myfun() 100 times using
dfx <- replicate(100,myfun(),simplify = FALSE,)
A function should take one input (or set of inputs), and return only one output (or a set of outputs). Consider the simple example of
myfunction <- function(x) {
x
x ** 2
}
Unless you are calling return() early (which you usually don't), the last object is returned. In fact, if you try to return two objects, e.g. return(1,2) you are met with
Error in return(1, 2) : multi-argument returns are not permitted
That is why the solution proposed by #StupidWolf in the comments is the most appropriate one, where you use return(list(wza = list(wza),wza_sum = list(wza_sum))). You then have to perform the necessary post-processing of splitting the lists if appropriate.
How does one draw a sample within a sapply function without replacement? Consider the following MWE below. What I am trying to achieve is for a number in idDRAW to receive a letter from chrSMPL (given the sample size of chrSMPL). Whether a number from idDRAW receives a letter is determined by the respective probabilities, risk factors and categories. This is calculated in the sapply function and stored in tmp.
The issue is sample replacement, leading to a number being named with a letter more than once. How can one avoid replacement whilst still using the sapply function? I have tried to adjust the code from this question (Alternative for sample) to suit my needs, but no luck. Thanks in advance.
set.seed(3)
chr<- LETTERS[1:8]
chrSMPL<- sample(chr, size = 30, replace = TRUE)
idDRAW<- sort(sample(1:100, size = 70, replace = FALSE))
p_mat<- matrix(runif(16, min = 0, max = 0.15), ncol = 2); rownames(p_mat) <- chr ## probability matrix
r_mat <- matrix(rep(c(0.8, 1.2), each = length(chr)), ncol = 2); rownames(r_mat) <- chr ## risk factor matrix
r_cat<- sample(1:2, 70, replace = TRUE) ## risk categories
# find number from `idDRAW` to be named a letter:
Out<- sapply(chrSMPL, function(x){
tmp<- p_mat[x, 1] * r_mat[x, r_cat]
sample(idDRAW, 1, prob = tmp)
})
> sort(Out)[1:3]
G B B
5 5 5
I managed with an alternative solution using a for loop as seen below. If anyone can offer suggestions on how the desired result can be achieved without using a for loop it would be greatly appreciated.
set.seed(3)
Out <- c()
for(i in 1:length(chrSMPL)){
tmp <- p_mat[chrSMPL[i], 1] * r_mat[chrSMPL[i], r_cat]
Out <- c(Out, sample(idDRAW, 1, prob = tmp))
rm <- which(idDRAW == Out[i])
idDRAW <- idDRAW[-rm]
r_cat <- r_cat[-rm]
}
names(Out) <- chrSMPL
sort(Out)[1:3]
I would like to read a large .csv into R. It'd handy to split it into various objects and treat them separately. I managed to do this with a while loop, assigning each tenth to an object:
# The dataset is larger, numbers are fictitious
n <- 0
while(n < 10000){
a <- paste('a_', n, sep = '')
assign(a, read.csv('df.csv',
header = F, stringsAsFactors = F, nrows = 1000, skip = 0 + n)))
# There will be some additional processing here (omitted)
n <- n + 1000
}
Is there a more R-like way of doing this? I immediately thought of lapply. According to my understanding each object would be the element of a list that I would then have to unlist.
I gave a shot to the following but it didn't work and my list only has one element:
A <- lapply('df.csv', read.csv,
header = F, stringsAsFactors = F, nrows = 1000, skip = seq(0, 10000, 1000))
What am I missing? How do I proceed from here? How do I then unlist A and specify each element of the list as a separate data.frame?
If you apply lapply to a single element you'll have only one element as an output.
You probably want to do this:
a <- paste0('a_', 1:1000) # all your 'a's
A <- lapply(a,function(x){
read.csv('df.csv', header = F, stringsAsFactors = F, nrows = 1000, skip = 0 + n)
})
for each element of a, called x because it's the name I chose as my function parameter, I execute your command. A will be a list of the results.
Edit: As #Val mentions in comments, assign seems not needed here, so I removed it, you'll end up with a list of data.frames coming from your csvs if all works fine.