I tried to generate boson sampling data using R package BosonSampling. Although it takes long time to generate samples for larger values of n and m, I tried smaller values but I didn't get any output from code. I don't know what is the problem.
The documentation is available in the link:
https://cran.r-project.org/web/packages/BosonSampling/index.html
the code from documentation:
library('BosonSampling')
library('Rcpp')
set.seed(7)
n <- 10
m <- 20
A <- randomUnitary(m)[,1:n]
valueList <- bosonSampler(A, sampleSize=10, perm = FALSE)$values
valueList
Related
I first generated random data from a Gamma distribution using the following code
data <- rgamma(9, shape=32, scale=1/4)
I proceeded to generate a single sample of 9 observations from the population.
sample(data, 9)
I'm trying to run a for loop in R so that I can repeatedly generate samples of 9 observations and save the mean of each sample into a new vector. I want to do this 500,000 times. After the for loop I then want to create a null distribution based on the distribution created from the for loop. I am also wanting to sample with replacement. (I am also very new to R, so any suggestions or help is greatly appreciated).
Here is the code I have tried for the for loop:
v <- 500000
Storage <- numeric(9)
for (i in v) {
Storage[i] <- mean(i)
}
The easiest way to do this is like this...
means <- replicate(500000, mean(rgamma(9, shape=32, scale=1/4)))
This will generate 9 gamma variates, take the mean, and repeat the process 500,000 times, storing the result in the vector means. Definitely no need for a for loop!
Use replicate to create the vectors, then compute the means with the fast colMeans.
set.seed(2023)
data <- rgamma(9, shape=32, scale=1/4)
v <- 500000L
Storage <- replicate(v, sample(data, 9, TRUE))
mean_Storage <- colMeans(Storage)
hist(mean_Storage, freq = FALSE)
Created on 2023-02-03 with reprex v2.0.2
Or maybe you want to sample from a Gamma distribution.
set.seed(2023)
v <- 500000L
Storage <- replicate(v, rgamma(9, shape=32, scale=1/4))
mean_Storage <- colMeans(Storage)
hist(mean_Storage, freq = FALSE)
Created on 2023-02-03 with reprex v2.0.2
So I have been working through a population ecology exercise using the popbio package in R-Studio that focuses on using Leslie Matrix's. I have successfully created a Leslie matrix with the proper dimensions using the Fecundity (mx) and Annual Survival values (sx) that I have calculated with my life table. I then am trying to use the pop.projection function in the popbio package to multiply my Leslie matrix (les.mat) by a starting population vector (N0) followed by the number of time intervals (4 years). It is my understanding that you should be able to take a Leslie matrix and multiply by a population vector to calculate a population size after a set number of time intervals. Have I done something wrong here, when I try to run my pop.projection line of code I get the following error message in R:
"> projA <- pop.projection(les.mat,N0,10)
Error in A %*% n : non-conformable arguments"
Could the problem be an issue with my pop.projection function? I am thinking it may be an issue with by N0 argument (population vector), when I look at my N0 values it seems like it has been saved in R as a "Numeric Type", should I be converting it into its own matrix, or as it's own vector somehow to get my pop.projection line of code to run? Any advice would be greatly appreciated, the short code I have been using will be linked below!
Sx <- c(0.8,0.8,0.7969,0.6078,0.3226,0)
mx <- c(0,0,0.6,1.09,0.2,0)
Fx <- mx # fecundity values
S <- Sx # dropping the first value
F <- Fx
les.mat <- matrix(rep(0,36),nrow=6)
les.mat[1,] <- F
les.mat
for(i in 1:5){
les.mat[(i+1),i] <- S[i]
}
les.mat
N0 <- c(100,80,64,51,31,10,0)
projA <- pop.projection(les.mat,N0,10)
The function uses matrix multiplication on the first and second arguments so they must match. The les.mat matrix is 6x6, but N0 is length 7. Try
projA <- pop.projection(les.mat, N0[-7], 10) # Delete last value
or
projA <- pop.projection(les.mat, N0[-1], 10) # Delete first value
I am trying to compare forecast reconciliation methods from the hts package on previously existing forecasts. The forecast.gts function is not available to me since there is no computationally tractable way to create a user defined function that returns the values in a forecast object. Because of this, I am using the combinef() function in the package to redistribute the forecasts. I have been able to work of the proper weights to get the wls and nseries methods, and the ols version is the default. I was able to get the "bottom up" method using:
# Creates sample forecasts, taken from `combinef()` example
library(hts)
h <- 12
ally <- aggts(htseg1)
allf <- matrix(NA, nrow = h, ncol = ncol(ally))
for(i in 1:ncol(ally))
allf[,i] <- forecast(auto.arima(ally[,i]), h = h, PI = FALSE)$mean
allf <- ts(allf, start = 51)
# create the weight vector
numTS <- ncol(allf) # Get the total number of series
numBaseTS <- sum(tail(htseg1$nodes, 1)[[1]]) # Get the number of bottom level series
# Create weights of 0 for all aggregate ts and 1 for the base level
weightVals <- c(rep(0, numTS - numBaseTS), rep(1, numBaseTS))
y.f <- combinef(allf, htseg1$nodes, weights = weightVals)
I was hoping that something like making the first weight 1 and the rest 0 might give me one of the three top down forecast, but that just results in a bunch of 0s or NaN values depending on how you try to look at it.
combinef(allf, htseg1$nodes, weights = c(1, rep(0, numTS - 1)))
I know the top down methods aren't the hardest thing to compute manually, and I can just write a function to do that, but are there any tools in the hts package that can help with this? I'd like to keep the data format consistent to simplify my analysis. Most specifically, I would like to get the "top down forcasted proportions" or tdfp method.
The functions to reconcile the forecasts using the "top-down" method are currently not exported. Probably I should export them to make the "top-down" results as tractable as combinef() in the next version. The workaround is as follows:
hts:::TdFp(allf, nodes = htseg1$nodes)
Hope it helps.
I am trying to perform KMeans clustering on over a million rows with 4 observations, all numeric. I am using the following code:
kmeansdf<-as.data.frame(rbind(train$V3,train$V5,train$V8,train$length))
km<-kmeans(kmeansdf,2)
As it can be seen, I would like to divide my data into two clusters. The object km is getting populated but I am having trouble plotting the results. Here is the code I am using to plot:
plot(kmeansdf,col=km$cluster)
This piece of code gives me the following error:
Error in plot.new() : figure margins too large
I tried researching online but could not find a solution, I tried working on command line as well but still getting the same error (I am using RStudio at the moment)
Any help to resolve the error would be highly appreciated.
TIA.
When I run your code on a df with 1e6 rows, I don't get the same error, but the system hangs (interrupted after 10 min). It may be that creating a scatterplot matrix with 1e6 points per frame is just too much.
You might consider taking a random sample:
# all this to create a df with two distinct clusters
set.seed(1)
center.1 <- c(2,2,2,2)
center.2 <- c(-2,-2,-2,-2)
n <- 5e5
f <- function(x){return(data.frame(V1=rnorm(n,mean=x[1]),
V2=rnorm(n,mean=x[2]),
V3=rnorm(n,mean=x[3]),
V4=rnorm(n,mean=x[4])))}
df <- do.call("rbind",lapply(list(center.1,center.2),f))
km <- kmeans(df,2) # run kmeans on full dataset
df$cluster <- km$cluster # append cluster column to df
# sample is 10% of population (100,000 rows)
s <- 1e5
df <- df[sample(nrow(df),s),]
plot(df[,1:4],col=df$cluster)
Running the same thing with a 1% sample (50,000 rows) gives this.
I am trying to perform KMeans clustering on over a million rows with 4 observations, all numeric. I am using the following code:
kmeansdf<-as.data.frame(rbind(train$V3,train$V5,train$V8,train$length))
km<-kmeans(kmeansdf,2)
As it can be seen, I would like to divide my data into two clusters. The object km is getting populated but I am having trouble plotting the results. Here is the code I am using to plot:
plot(kmeansdf,col=km$cluster)
This piece of code gives me the following error:
Error in plot.new() : figure margins too large
I tried researching online but could not find a solution, I tried working on command line as well but still getting the same error (I am using RStudio at the moment)
Any help to resolve the error would be highly appreciated.
TIA.
When I run your code on a df with 1e6 rows, I don't get the same error, but the system hangs (interrupted after 10 min). It may be that creating a scatterplot matrix with 1e6 points per frame is just too much.
You might consider taking a random sample:
# all this to create a df with two distinct clusters
set.seed(1)
center.1 <- c(2,2,2,2)
center.2 <- c(-2,-2,-2,-2)
n <- 5e5
f <- function(x){return(data.frame(V1=rnorm(n,mean=x[1]),
V2=rnorm(n,mean=x[2]),
V3=rnorm(n,mean=x[3]),
V4=rnorm(n,mean=x[4])))}
df <- do.call("rbind",lapply(list(center.1,center.2),f))
km <- kmeans(df,2) # run kmeans on full dataset
df$cluster <- km$cluster # append cluster column to df
# sample is 10% of population (100,000 rows)
s <- 1e5
df <- df[sample(nrow(df),s),]
plot(df[,1:4],col=df$cluster)
Running the same thing with a 1% sample (50,000 rows) gives this.