I'm trying to use repeat loop to generate 100 data set of Poisson Distribution with sample size n=100 and I would like to arrange the result in by row and column but it is just show me repeating to show me the last set of data while not all the data set. At the same time I would also trying to figure out the way to get the mean, variance and MSE of the 100 data set.
set.seed(124)
a <- 1
repeat{
b = rpois(100, lambda = 3)
Storage100 <- matrix(data=b,nrow=100,ncol=1)
a = a+1
print(b)
if (a>100){break
}
}
Storage100
I'm expecting that my 100 data set can be show like first set of data in first column, second set of data in second column.....
Use replicate with simplify as TRUE to get matrix of dimension 100 X 100 where each column represents the distribution.
set.seed(124)
m1 <- replicate(100, matrix(data=rpois(100, lambda = 3),ncol = 1), simplify = TRUE)
To get the mean for each column we can use colMeans (thanks to #jay.sf)
colMeans(m1)
Related
I'm trying to assess the feasibility of an instrumental variable in my project with a variable I havent seen before. The variable essentially is an interaction between the mean and standard deviation of a sample drawn from a gaussian, and im trying to see what this distribution might look like. Below is what im trying to do, any help is much appreciated.
Generate a set of 1000 individuals with a variable x following the gaussian distribution, draw 50 random samples of 5 individuals from this distribution with replacement, calculate the means and standard deviation of x for each sample, create an interaction variable named y which is calculated by multiplying the mean and standard deviation of x for each sample, plot the distribution of y.
Beginners version
There might be more efficient ways to code this, but this is easy to follow, I guess:
stat_pop <- rnorm(1000, mean = 0, sd = 1)
N = 50
# As Ben suggested, we create a data.frame filled with NA values
samples <- data.frame(mean = rep(NA, N), sd = rep(NA, N))
# Now we use a loop to populate the data.frame
for(i in 1:N){
# draw 5 samples from population (without replacement)
# I assume you want to replace for each turn of taking 5
# If you want to replace between drawing each of the 5,
# I think it should be obvious how to adapt the following code
smpl <- sample(stat_pop, size = 5, replace = FALSE)
# the data.frame currently has two columns. In each row i, we put mean and sd
samples[i, ] <- c(mean(smpl), sd(smpl))
}
# $ is used to get a certain column of the data.frame by the column name.
# Here, we create a new column y based on the existing two columns.
samples$y <- samples$mean * samples$sd
# plot a histogram
hist(samples$y)
Most functions here use positional arguments, i.e., you are not required to name every parameter. E.g., rnorm(1000, mean = 0, sd = 1) is the same as rnorm(1000, 0, 1) and even the same as rnorm(1000), since 0 and 1 are the default values.
Somewhat more efficient version
In R, loops are very inefficient and, thus, ought to be avoided. In case of your question, it does not make any noticeable difference. However, for large data sets, performance should be kept in mind. The following might be a bit harder to follow:
stat_pop <- rnorm(1000, mean = 0, sd = 1)
N = 50
n = 5
# again, I set replace = FALSE here; if you meant to replace each individual
# (so the same individual can be drawn more than once in each "draw 5"),
# set replace = TRUE
# replicate repeats the "draw 5" action N times
smpls <- replicate(N, sample(stat_pop, n, replace = FALSE))
# we transform the output and turn it into a data.frame to make it
# more convenient to work with
samples <- data.frame(t(smpls))
samples$mean <- rowMeans(samples)
samples$sd <- apply(samples[, c(1:n)], 1, sd)
samples$y <- samples$mean * samples$sd
hist(samples$y)
General note
Usually, you should do some research on the problem before posting here. Then, you either find out how it works by yourself, or you can provide an example of what you tried. To this end, you can simply google each of the steps you outlined (e.g., google "generate random standard distribution R" in order to find out about the function rnorm().
Run ?rnorm to get help on the function in RStudio.
For a paper I'm writing I have subsetted a larger dataset into 3 groups, because I thought the strength of correlations between 2 variables in those groups would differ (they did). I want to see if subsetting my data into random groupings would also significantly affect the strength of correlations (i.e., whether what I'm seeing is just an effect of subsetting, or if those groupings are actually significant).
To this end, I am trying to generate n new data frames by randomly sampling 150 rows from an existing dataset, and then want to calculate correlation coefficients for two variables in those n new data frames, saving the correlation coefficient and significance in a new file.
But, HOW?
I can do it manually, e.g., with dplyr, something like
newdata <- sample_n(Random_sample_data, 150)
output <- cor.test(newdata$x, newdata$y, method="kendall")
I'd obviously like to not type this out 1000 or 100000 times, and have been trying things with loops and lapply (see below) but they've not worked (undoubtedly due to something really obvious that I'm missing!).
Here I have tried to assign each row to a different group, with 10 groups in total, and then to do correlations between x and y by those groups:
Random_sample_data<-select(Range_corrected, x, y)
cat <- sample(1:10, 1229, replace=TRUE)
Random_sample_cats<-cbind(Random_sample_data,cat)
correlation <- function(c) {
c <- cor.test(x,y, method="kendall")
return(c)
}
b<- daply(Random_sample_cats, .(cat), correlation)
Error message:
Error in cor.test(x, y, method = "kendall") :
object 'x' not found
Once you have the code for what you want to do once, you can put it in replicate to do it n times. Here's a reproducible example on built-in data
result = replicate(n = 10, expr = {
newdata <- sample_n(mtcars, 10)
output <- cor.test(newdata$wt, newdata$qsec, method="kendall")
})
replicate will save the result of the last line of what you did (output <- ...) for each replication. It will attempt to simplify the result, in this case cor.test returns a list of length 8, so replicate will simplify the results to a matrix with 8 rows and 10 columns (1 column per replication).
You may want to clean up the results a little bit so that, e.g., you only save the p-value. Here, we store only the p-value, so the result is a vector with one p-value per replication, not a matrix:
result = replicate(n = 10, expr = {
newdata <- sample_n(mtcars, 10)
cor.test(newdata$wt, newdata$qsec, method="kendall")$p.value
})
I am creating a function that allows me to multiply my data by random proportions, sums them thus creating a mixture of my data multiplied by this Proportion.
For example, if I have 4 data sets, I create a Proportion of 4 random numbers that sum 100 and multiply each data set by each Proportion and sum the result.
Besides that, I want that my function iterates through my dataset and also through my proportions as to permutate through all possible combinations of Proportion dataset multiplication
A sample data set can be seen:
library(LCF)
data(stdmix)
My function currently stands at this Point:
library(combinat)
props <- function(corr.spec.standards = specdat, size, nprop){
if (size < 2) stop("number must be greater than 1")
## create progress bar
try(pb <- txtProgressBar(min = 1, max = nprop, style = 3), silent = TRUE)
## initial loop for proportions
for (i in 1:nprop) {
prop <- sample.int(100, size = size)
prop <- (prop/sum(prop))
permut <- permn(prop)
## permutation loop
for (i in permut[[i]]) {
mapply(`*`,permut, rep(specdat[i]$data$corr.spec$cor.absorption,each=length(permut)))
}
}
My Problem is that specdat is a nested list, which in this example is a list of 8 and that the only members to be multiplied by the Permutation are specdat[i]data$corr.spec%cor.absorption
Thus my Question is: How to loop through a (very) nested list only on a specific member of the list?
I have got a text file containing 200 models all compared to eachother and a molecular distance for each 2 models compared. It looks like this:
1 2 1.2323
1 3 6.4862
1 4 4.4789
1 5 3.6476
.
.
All the way down to 200, where the first number is the first model, the second number is the second model, and the third number the corresponding molecular distance when these two models are compared.
I can think of a way to import this into R and create a nice 200x200 matrix to perform some clustering analyses on. I am still new to Stack and R but thanks in advance!
Since you don't have the distance between model1 and itself, you would need to insert that yourself, using the answer from this question:
(you can ignore the wrong numbering of the models compared to your input data, it doesn't serve a purpose, really)
# Create some dummy data that has the same shape as your data:
df <- expand.grid(model1 = 1:120, model2 = 2:120)
df$distance <- runif(n = 119*120, min = 1, max = 10)
head(df)
# model1 model2 distance
# 1 2 7.958746
# 2 2 1.083700
# 3 2 9.211113
# 4 2 5.544380
# 5 2 5.498215
# 6 2 1.520450
inds <- seq(0, 200*119, by = 200)
val <- c(df$distance, rep(0, length(inds)))
inds <- c(seq_along(df$distance), inds + 0.5)
val <- val[order(inds)]
Once that's in place, you can use matrix() with the ncol and nrow to "reshape" your vector of distance in the appropriate way:
matrix(val, ncol = 200, nrow = 200)
Edit:
When your data only contains the distance for one direction, so only between e.g. model1 - model5 and not model5 - model1 , you will have to fill the values in the upper triangular part of a matrix, like they do here. Forget about the data I generated in the first part of this answer. Also, forget about adding the ones to your distance column.
dist_mat <- diag(200)
dist_mat[upper.tri(dist_mat)] <- your_data$distance
To copy the upper-triangular entries to below the diagonal, use:
dist_mat[lower.tri(dist_mat)] <- t(dist_mat)[lower.tri(dist_mat)]
As I do not know from your question what format is your file in, I will assume the most general file format, i.e., CSV.
Then you should look at the reading files, read.csv, or fread.
Example code:
dt <- read.csv(file, sep = "", header = TRUE)
I suggest using data.table package. Then:
setDT(dt)
dt[, id := paste0(as.character(col1), "-", as.character(col2))]
This creates a new variable out of the first and the second model and serves as a unique id.
What I do is then removing this id and scale the numerical input.
After scaling, run clustering algorithms.
Merge the result with the id to analyse your results.
Is that what you are looking for?
I want to draw normal random numbers in an array of order ((100*8)*5000) with a specific Mean (M) and Standard Deviation (S) but I want them to be only within the range M±3S, so that I don't have any outliers in my array exceeding those limits.
Any Suggestion? I want to write a program in R based on this array for some simulation studies. I am using following R Code to generate my Data Set:
for(i in 1:5000){
for(j in 1:8){
Dat[,j,i]=rnorm(100,mean=muu[j],sd=sigma[j])
}
}
Now, We want to get rid of those values which are higher than muu±3sigma in the above data. Definitely, We have to replace discarded values with fresh values so that the dimension of the Dat array keep intact.
First Solution
Here is a start but I bet there is a more elegant solution.
First generate a sample next step is to subset it to your desired values. Of course you have to adjust values to your desire.
set.seed(123)
rs <- rnorm(10000, mean = 10, sd = 3)
rs1 <- rs[ rs >= -19 & rs <= 19 ]
Second (better) solution
I think my first solutions didn't work so well. I have just written some code that might be perfect for your purposes. Here are the steps.
create an array of NAs with the required dimensions
fill it with random numbers
create a logical vector where TRUEs are for the desired conditions
subset the data based on that vector and replace the values where TRUE is TRUE (pardon my words game) with the mean used to generate samples
data <- array(NA, dim = c(100, 8, 5000))
for(i in 1:5000){
data[ , , i] <- rnorm(800, 3, 1)
}
bound <- 3 + c(-1, 1)*3*1
pr <- data <= bound[1] | data >= bound[2]
data[pr] <- 3