I have a categorical data column (Urban.f) in a dataset (edu2018) and would like to calculate a division and repeat the process 1000 times.
This is the code I have to get a random sample of size 300. It shows me how many cases of "Urban" and "Rural" there is, along with the rest of columns in the dataset. enter image description here
x1 <- sample_n(edu2018, 300, fac = "urban.f")
summary(x1)
I want to calculate the number that appears in "Urban" by the sum of "Urban" + "Rural".
Then, I want to repeat this step 1000 times. I have tried this code that tries to englobe everything I want, but cannot make it work.
edu2018_df <- edu2018$urban.f
n <- 1000
sample_rural <- replicate(n, {edu2018_df <- sample(edu2018, 300, replace = TRUE)
count(sample_rural = "Urban")/count(sample_rural == "Urban","Rural) * n})
hist(sample_rural)
You could do:
sample_rural <- replicate(1000, sum(sample(edu2018_df, 300, F) == 'Rural')/300)
hist(sample_rural)
Data used
Obviously, we don't have your actual data, so I recreated your vector edu2018_df, which is simply a vector of "Rural" and "Urban"
set.seed(1)
edu2018_df <- sample(c("Rural", "Urban"), 1000, TRUE)
Related
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)
I'm still new to the programming world and looking for some guidance on a model I am building for individual animal growths over time.
The goal for the code I'm working with is to
i) Generate random starting sizes of animals from a given distribution
ii) Give each of these individuals a starting growth rate from a given distribution
iii) Calculate new size of individual after 1 year
iv) Assign a new growth rate from above distribution
v) Calculate the new size of individual after another year.
So far I have the code below, and what I want to do is repeat the last two lines of code x amount of times without I having to physically run the code over and over.
# Generate starting lengths
lengths <- seq(from=4.4, to=5.4, by =0.1)
# Generate starting ks (growth rate)
ks <- seq(from=0.0358, to=0.0437, by =0.0001)
#Create individuals
create.inds <- function(id = NaN, length0=NaN, k1=NaN){
inds <- data.frame(id=id, length0 = length0, k1=k1)
inds
}
# Generate individuals
inds <- create.inds(id=1:n.initial,
length=sample(lengths,100,replace=TRUE),
k1=sample(ks, 100, replace=TRUE))
# Calculate new lengths based on last and 2nd last columns and insert into next column
inds[,ncol(inds)+1] <- 326*(1-exp(-(inds[,ncol(inds)])))+
(inds[,ncol(inds)-1]*exp(-(inds[,ncol(inds)])))
# Calculate new ks and insert into last column
inds[,ncol(inds)+1] <- sample(ks, 100, replace=TRUE)
Any and all assistance would be appreciated, also if you think there is a better way to write this please let me know.
i think what you are asking is a simple loop:
for (i in 1:100) { #replace 100 with the desired times you want this to excecute
inds[,ncol(inds)+1] <- 326*(1-exp(-(inds[,ncol(inds)])))+
(inds[,ncol(inds)-1]*exp(-(inds[,ncol(inds)])))
# Calculate new ks and insert into last column
inds[,ncol(inds)+1] <- sample(ks, 100, replace=TRUE)
}
I want to cluster my data to say 5 clusters, then we need to select 50 individuals with most dissimilar relationship from all the data. That means if cluster one contains 100, two contains 200, three contains 400, four contains 200, and five 100, I have to select 5 from the first cluster + 10 from the second cluster + 20 from the third + 10 from the fourth + 5 from the fifth.
Data example:
mydata<-matrix(nrow=100,ncol=10,rnorm(1000, mean = 0, sd = 1))
What I did till now is clustering the data and rank the individuals within each cluster, then export it to excel and go from there …
That has become became a problem since my data has became really big.
I will appreciate any help or suggestion on how to apply the previous in R
.
I´m not sure if it is exactly what you are searching, but maybe it helps:
mydata<-matrix(nrow=100, ncol=10, rnorm(1000, mean = 0, sd = 1))
rownames(mydata) <- paste0("id", 1:100) # some id for identification
# cluster objects and calculate dissimilarity matrix
cl <- cutree(hclust(
sim <- dist(mydata, diag = TRUE, upper=TRUE)), 5)
# combine results, take sum to aggregate dissimilarity
res <- data.frame(id=rownames(mydata),
cluster=cl, dis_sim=rowSums(as.matrix(sim)))
# order, lowest overall dissimilarity will be first
res <- res[order(res$dis_sim), ]
# split object
reslist <- split(res, f=res$cluster)
## takes first three items with highest overall dissim.
lapply(reslist, tail, n=3)
## returns id´s with highest overall dissimilarity, top 20%
lapply(reslist, function(x, p) tail(x, round(nrow(x)*p)), p=0.2)
regarding you comment, find the code below:
pleas note that the code can be improved in terms of beauty and efficiency.
Further I used a second answer, because otherwise it would be to messy.
# calculation of centroits based on:
# https://stat.ethz.ch/pipermail/r-help/2006-May/105328.html
cl <- hclust(dist(mydata, diag = TRUE, upper=TRUE))
cent <- tapply(mydata,
list(rep(cutree(cl, 5), ncol(mydata)), col(mydata)), mean)
dimnames(cent) <- list(NULL, dimnames(mydata)[[2]])
# add up cluster number and data and split by cluster
newdf <- data.frame(data=mydata, cluster=cutree(cl, k=5))
newdfl <- split(newdf, f=newdf$cluster)
# add centroids and drop cluster info
totaldf <- lapply(1:5,
function(i, li, cen) rbind(cen[i, ], li[[i]][ , -11]),
li=newdfl, cen=cent)
# calculate new distance to centroits and sort them
dist_to_cent <- lapply(totaldf, function(x)
sort(as.matrix(dist(x, diag=TRUE, upper=TRUE))[1, ]))
dist_to_cent
for calculation of centroids out of hclust see R-Mailinglist
I'm just starting to learn R, and my assignment was to create a vector of 10000 values with normal distribution, mean = 0 and sd = 100. Which I did.
x <- rnorm(10000, mean = 0, sd = 100)
But now I'm asked to introduce values between 500 and 700 at 1000 random positions in that vector.
Can anyone help me?
If you mean to replace 1000 elements in the x vector with values between 500 and 700, you first need to generate these 1000 elements:
r <- runif(1000, min=500, max=700)
I am assuming here that random values are uniformly between 500 and 700.
Then you need to select places to put these values in:
idx <- sample(10000, 1000)
Finally, replace the values at these places:
x[ idx ] <- r
Finally, to see the results of your action:
hist(x)
It should look like:
If I have a large dataset in R, how can I take random sample of the data taking into consideration the distribution of the original data, particularly if the data are skewed and only 1% belong to a minor class and I want to take a biased sample of the data?
The sample(x, n, replace = FALSE, prob = NULL) function takes a sample from a vector x of size n. This sample can be with or without replacement, and the probabilities of selecting each element to the sample can be either the same for each element, or a vector informed by the user.
If you want to take a sample of same probabilities for each element with 50 cases, all you have to do is
n <- 50
smpl <- df[sample(nrow(df), 50),]
However, if you want to give different probabilities of being selected for the elements, let's say, elements that sex is M has probability 0.25, while those whose sex is F has prob 0.75, you should do
n <- 50
prb <- ifelse(sex=="M",0.25,0.75)
smpl <- df[sample(nrow(df), 50, prob = prb),]