I am trying to create a function that examines how variables with different distributions influence OLS results. I have created two DVs (y1 and y2) but would like to expand this to include five or so. I am trying to change my code to include a loop so I do not have to copy and paste this multiple times, but I am not having much luck. Any suggestions would be greatly appreciated.
library(psych)
library(arm)
library(plyr)
library(fBasics)
regsim <- function(iter, n) {
ek1 <- rnorm(n, 0, 1)
ek2 <- rnorm(n, 0, 5)
x <- rnorm(n, 0, .5)
y1 <- .3*x + ek1
y2 <- .3*x + ek2
#y1
lm1 <- lm(y1 ~ x)
bhat1 <- coef (lm1)[2]
sehat1 <- se.coef (lm1) [2]
skewy1 <- skew(y1)
stdevy1 <- stdev(y1)
#y2
lm2 <- lm(y2 ~ x)
bhat2 <- coef (lm2)[2]
sehat2 <- se.coef (lm2) [2]
skewy2 <- skew(y2)
stdevy2 <- stdev(y2)
results <- c(bhat1, sehat1, stdevy1, skewy1,
bhat2, sehat2, stdevy2, skewy2)
names(results) <- c('b1', 'se1', 'sdy1', 'skewy1',
'b2', 'se2', 'sdy2', 'skewy2')
return(results)
}
iter <-1000
n <-500
results <- NULL
sims <-ldply(1:iter, regsim, n)
sims$n <- n
results <- rbind(results, sims)
Another option...
regsim <- function(n=100,num.y=5,sd=c(1:5)){
if(length(sd) != num.y){stop('length of sd must match number of dependent vars')
} else {
ldply(1:num.y,function(x){
e <- rnorm(n,0,sd=sd[x])
x <- rnorm(n,0,5)
y <- 0.3*x + e
out <- lm(y~x)
b1 <- coef(out)[2]
int <- coef(out)[1]
data.frame(b1=b1,int=int)
})
}
}
regsim(num.y=10,sd=c(1:10))
b1 int
1 0.30817303 0.0781049
2 0.38681600 -0.3359067
3 0.24560773 -0.0277561
4 0.08032659 0.1877233
5 0.39873955 -0.6027522
6 0.21729930 0.7384340
7 0.33761456 -0.1053028
8 0.26502006 -0.1851552
9 0.15452261 -1.6334873
10 -0.10496863 -0.3225169
This will allow you to specify the number of dependent variables and the SD for each error term. You can then use replicate to repeat the function for the desired number of replications.
replicate(10,regsim(),simplify = F)
[[1]]
b1 int
1 0.3047779 -0.01984306
2 0.3133198 -0.20458410
3 0.2833979 -0.25307502
4 0.3066878 -0.03235019
5 0.1374949 0.10958616
[[2]]
b1 int
1 0.2902103 -0.12683502
2 0.3499006 0.06691437
3 0.1949797 -0.14371830
4 0.2358269 0.53117467
5 0.2869511 0.16281380
[[3]]
b1 int
1 0.2952211 0.05905549
2 0.2367774 0.02862166
3 0.0896778 -0.08467935
4 0.2352622 -0.20835837
5 0.3149963 0.07042032
[[4]]
b1 int
1 0.2946468 -0.08266406
2 0.3322577 0.17558135
3 0.2200087 -0.25778150
4 0.1822915 0.34962679
5 0.2442479 0.34433656
[[5]]
b1 int
1 0.2882853 0.12677506
2 0.3455534 -0.27885958
3 0.2981193 0.04598347
4 0.3380173 0.05243198
5 0.2148643 -0.09631672
[[6]]
b1 int
1 0.2962269 0.03743759
2 0.2979327 -0.12830803
3 0.3352781 -0.03935422
4 0.2584965 -0.05924351
5 0.2856802 0.03430055
[[7]]
b1 int
1 0.2968077 -0.10300109
2 0.2954560 0.25979902
3 0.3276077 -0.07001758
4 0.1825841 0.13508932
5 0.4302788 -0.13951914
[[8]]
b1 int
1 0.2992147 0.02084806
2 0.2765976 0.07277813
3 0.2469616 0.44580403
4 0.2601966 -0.09849855
5 0.2679183 0.50501652
[[9]]
b1 int
1 0.2963905 0.03308366
2 0.3356783 -0.06080088
3 0.3199835 0.22533444
4 0.3546083 -0.26909478
5 0.3536241 -0.19795094
[[10]]
b1 int
1 0.3100336 -0.05228032
2 0.4076447 -0.18715063
3 0.3436858 -0.37518649
4 0.4569368 -0.09114672
5 0.3255668 -0.18738138
How about this:
n <- 1000
x <- rnorm(n, 0, .5)
fun_reg <- function(n, ek_mu, ek_sd, x){
s <- list() # list to collect results for output
ek <- rnorm(n, ek_mu, ek_sd)
y <- .3*x + ek
m <- lm(y ~ x)
s$bhat <- coef(m)[2]
s$sehat <- arm::se.coef(m)[2]
s$skewy <- psych::skew(y)
s$stdevy <- fBasics::stdev(y)
return(s)
}
purrr::map_dfr(c(1, 5, 10, 20, 50), ~fun_reg(n, 0, ., x))
Edit:
This now has 500 observations each and the regression is repeated with 1000 draws for each value of the standard deviation. A variable ek_sd has been added to the final output, to reflect with which standard deviation the values were arrived at. Note that x is not redrawn for each iteration, but I'm not entirely sure, that that is what you want. If you want x to be redrawn at each iteration, move it inside the function.
n <- 500
x <- rnorm(n, 0, .5)
fun_reg <- function(n, ek_mu, ek_sd, x){
s <- list()
ek <- rnorm(n, ek_mu, ek_sd)
y <- .3*x + ek
m <- lm(y ~ x)
s$ek_sd <- ek_sd
s$bhat <- coef(m)[2]
s$sehat <- arm::se.coef(m)[2]
s$skewy <- psych::skew(y)
s$stdevy <- fBasics::stdev(y)
return(s)
}
intr <- unlist(lapply(c(1, 5, 10, 20, 50), rep, 1000))
purrr::map_dfr(intr, ~fun_reg(n, 0, ., x))
This reduces the package reliance to just psych::skew and an optional ggplot2 call:
library(psych)
regsim <- function(n, eks) {
x <- rnorm(n, 0, .5)
ek <- sapply(eks, function(x) rnorm(n, 0, x))
y <- 0.3 * x + ek
lms <- lm(y ~ x)
data.frame(b_hat = lms[['coefficients']][2,],
int = lms[['coefficients']][1, ],
skew_y = psych::skew(y),
se_hat = unlist(lapply(summary(lms), function(lst) lst[[4]][2,2]), use.names = FALSE),
sd_y = apply(y, 2, sd),
sd_eks = eks
)
}
iter <-1000
n <-500
eks_sd = c(1,5)
# do the simulations and make them into a nice data.frame
sims <- replicate(iter, regsim(n, eks_sd), simplify = FALSE)
results <- do.call(rbind, sims)
#next parts are optional
results$iter_id <- rep(seq_len(iter), each = length(eks_sd))
tibble::as_tibble(results)
# Random graph because everyone loves graphs
library(ggplot2)
ggplot(results, aes(x = iter_id, y = int)) + geom_point() + facet_grid(vars(sd_eks))
The main thing is that lm can take multiple y arguments. That's why we we create a matrix of ek using sapply.
Related
I get this error when I run the MWE code below. Does anyone know how to resolve this? thanks!
Error: Error 10020: Q matrix is not positive semi-definite (PSD). Set NonConvex parameter to 2 to solve model.
MWE:
library(gurobi)
library(Matrix)
model <- list()
#optimization problem:
# max x + y
# s.t.
# -x + y <= 0
# x^2 - y^2 <= 10
# 0 <= x < = 20
# 0 <= y <= 20
model$obj <- c(1,1)
model$A <- matrix(c(-1,1), nrow=1, byrow=T) # for LHS of linear constraint: -x + y <= 0
model$rhs <- c(0) # for RHS of linear constraint: -x + y <= 0
model$ub[1] = 20 # x < = 20
model$ub[2] = 20 # y < = 20
model$sense <- c('<')
# non-convex quadratic constraint: x^2 - y^2 <= 10
qc1 <- list()
qc1$Qc <- spMatrix(2, 2, c(1, 2), c(1, 2), c(1.0, -1.0))
qc1$rhs <- 10
model$quadcon <- list(qc1)
#the QC constraint is a non-convex quadratic constraint, so set NonConvex = 2
model$params <- list(NonConvex=2)
gurobi_write(model,'quadtest.lp', env)
result <- gurobi(model) # THIS IS WHERE I GET THE ERROR ABOVE
print(result$objval)
print(result$x)
NM...i see that I can fix this by not putting the params as part of the model list, and instead running it as an input to the gurobi(,) call as follows:
params <- list(NonConvex=2)
result <- gurobi(model, params)
In an earlier question (R: Logical Conditions Not Being Respected), I learned how to make the following simulation :
Step 1: Keep generating two random numbers "a" and "b" until both "a" and "b" are greater than 12
Step 2: Track how many random numbers had to be generated until it took for Step 1 to be completed
Step 3: Repeat Step 1 and Step 2 100 times
res <- matrix(0, nrow = 0, ncol = 3)
for (j in 1:100){
a <- rnorm(1, 10, 1)
b <- rnorm(1, 10, 1)
i <- 1
while(a < 12 | b < 12) {
a <- rnorm(1, 10, 1)
b <- rnorm(1, 10, 1)
i <- i + 1
}
x <- c(a,b,i)
res <- rbind(res, x)
}
head(res)
[,1] [,2] [,3]
x 12.14232 12.08977 399
x 12.27158 12.01319 1695
x 12.57345 12.42135 302
x 12.07494 12.64841 600
x 12.03210 12.07949 82
x 12.34006 12.00365 782
Question: Now, I am trying to make a slight modification to the above code - Instead of "a" and "b" being produced separately, I want them to be produced "together" (in math terms: "a" and "b" were being produced from two independent univariate normal distributions, now I want them to come from a bivariate normal distribution).
I tried to modify this code myself:
library(MASS)
Sigma = matrix(
c(1,0.5, 0.5, 1), # the data elements
nrow=2, # number of rows
ncol=2, # number of columns
byrow = TRUE) # fill matrix by rows
res <- matrix(0, nrow = 0, ncol = 3)
for (j in 1:100){
e_i = data.frame(mvrnorm(n = 1, c(10,10), Sigma))
e_i$i <- 1
while(e_i$X1 < 12 | e_i$X2 < 12) {
e_i = data.frame(mvrnorm(n = 1, c(10,10), Sigma))
e_i$i <- i + 1
}
x <- c(e_i$X1, e_i$X2 ,i)
res <- rbind(res, x)
}
res = data.frame(res)
But this is producing the following error:
Error in while (e_i$X1 < 12 | e_i$X2 < 12) { : argument is of length
zero
If I understand your code correctly you are trying to see how many samples occur before both values are >=12 and doing that for 100 trials? This is the approach I would take:
library(MASS)
for(i in 1:100){
n <- 1
while(any((x <- mvrnorm(1, mu=c(10,10), Sigma=diag(0.5, nrow=2)+0.5))<12)) n <- n+1
if(i==1) res <- data.frame("a"=x[1], "b"=x[2], n)
else res <- rbind(res, data.frame("a"=x[1], "b"=x[2], n))
}
Here I am assigning the results of a mvrnorm to x within the while() call. In that same call, it evaluates whether either are less than 12 using the any() function. If that evaluates to FALSE, n (the counter) is increased and the process repeated. Once TRUE, the values are appended to your data.frame and it goes back to the start of the for-loop.
Regarding your code, the mvrnorm() function is returning a vector, not a matrix, when n=1 so both values go into a single variable in the data.frame:
data.frame(mvrnorm(n = 1, c(10,10), Sigma))
Returns:
mvrnorm.n...1..c.10..10...Sigma.
1 9.148089
2 10.605546
The matrix() function within your data.frame() calls, along with some tweaks to your use of i, will fix your code:
library(MASS)
Sigma = matrix(
c(1,0.5, 0.5, 1), # the data elements
nrow=2, # number of rows
ncol=2, # number of columns
byrow = TRUE) # fill matrix by rows
res <- matrix(0, nrow = 0, ncol = 3)
for (j in 1:10){
e_i = data.frame(matrix(mvrnorm(n = 1, c(10,10), Sigma), ncol=2))
i <- 1
while(e_i$X1[1] < 12 | e_i$X2[1] < 12) {
e_i = data.frame(matrix(mvrnorm(n = 1, c(10,10), Sigma), ncol=2))
i <- i + 1
}
x <- c(e_i$X1, e_i$X2 ,i)
res <- rbind(res, x)
}
res = data.frame(res)
I have gotten two lists of values in R.
daily_max_car: (List 1)
21 21 22 22 22 22 21
daily_0.8: (List 2)
16 17 17 17 18 17 17
Trying to write a For Loop in R-Studio to generate multiple matrix by using the one of the values from these two lists (One by One).
Here is the code I have been using to generate one matrix!
Lambda <- 21 (From List 1)
Mue <- 4
Rho <- Lambda/Mue
N <- 16 (From List 2)
All of these four parameters will be used in the "calculatewq" Function.
calculatewq <- function(c)
{....Some thing happening }
##Create Matrix
matrix1 <- matrix(0,Lambda,4)
matrix1[,1] <- 1:Lambda
### Create a column of matrix with repeated "N"
rep.row<-function(x,y)
{matrix(rep(x,each=y),nrow=y)}
created_mar_1 <- rep.row(N,Lambda)
car_n<- created_mar_1-matrix1[,1]
created_mar_3 <- rep.row(69*60*24,Lambda)
## Add into Matrix
for (i in 1:Lambda)
{matrix1[i,2] <- calculatewq(i)[2]
matrix1[i,3] <- calculatewq(i)[5]
matrix1[,4] = car_n*created_mar_3}`
Once I change one of the parameters it will generate a new matrix.
Thus, how can I write a for loop to generate multiple matrix while I am putting different value in Lambda and N.
Thank you so much!
Sampson
I removed for loop inside calculatewq function. Please make sure you needed a for loop in it.
myfun <- function(Lambda, N, mu )
{
# browser()
var1 <- seq_len( Lambda )
var2 <- ( rep( N, each = Lambda) ) - var1
var3 <- rep( 69*60*24, each = Lambda )
var4 <- var2 * var3
fun_vals <- do.call( 'rbind',
lapply( var1, function( x ) calculatewq( x, Lambda = Lambda, N = N, mu = mu ) ) )
mat <- matrix( NA, nrow = Lambda, ncol = mu )
mat[, 1] <- var1
mat[, 2] <- fun_vals[, 'Wq']
mat[, 3] <- fun_vals[, 'customer_serviced']
mat[, 4] <- var4
return(mat)
}
calculatewq <- function( x, Lambda, N, mu )
{
# browser()
Rho <- Lambda / mu
p0_inv <- ( Rho^x * (1-(( Rho/x )^( N-x+1)))) / (factorial( x ) * ( 1 - ( Rho / x ) ) )
p0_inv <- p0_inv + ( Rho^x) / factorial( x )
P0 <- 1/p0_inv
Lq <- ( Rho^(x+1)) * (1-((Rho/x)^(N-x+1))-((N-x+1)*(1-(Rho/x))*((Rho/x)^(N-x))))*P0/(factorial(x-1)*(x-Rho)^2)
Wq <- 60*Lq/Lambda
Ls <- Lq + Rho
Ws <- 60*Ls/Lambda
PN <- (Rho^N)*P0/(factorial(x)*x^(N-x))
customer_serviced <- (1 - PN)*100
a <- cbind( Lq, Wq, Ls, Ws, customer_serviced )
return(a)
}
mu <- 4
res <- Map( myfun,
list( 21 ,21, 22, 22 ,22, 22 ,21 ),
list( 16, 17, 17, 17, 18, 17 ,17 ),
mu)
head( res[[1]])
# [,1] [,2] [,3] [,4]
# [1,] 1 42.184874 19.04762 1490400
# [2,] 2 38.241748 38.09526 1391040
# [3,] 3 33.339271 57.13862 1291680
# [4,] 4 26.014138 75.70348 1192320
# [5,] 5 16.339462 89.88989 1092960
# [6,] 6 9.121053 96.32498 993600
daily_max_car <- list(21,21,22,22,22,22,21)
daily_0.8 <- list(16,17,17,17,18,17,17)
myfunction <- function(Lambda, N){
Mue <- 4
Rho <- Lambda/Mue
df <- as.data.frame(matrix(0, ncol = 4, nrow = Lambda))
names(df) <- c("A","B","C","D")
df[,1] <- 1:Lambda
df[,2] <- N
df[,3] <- df[,1] - df[,2]
df[,4] <- 69*60*24
return(df)
}
myfunction(21,16)
result <- mapply(myfunction, daily_max_car, daily_0.8)
Result
Lambda <- 21
Mue <- 4
Rho <- Lambda/Mue
N <- 19
matrix1 <- matrix(0,Lambda,4)
matrix1[,1] <- 1:Lambda
rep.row<-function(x,y)
{
matrix(rep(x,each=y),nrow=y)
}
created_mar_1 <- rep.row(N,Lambda)
car_n<- created_mar_1-matrix1[,1]
created_mar_3 <- rep.row(69*60*24,Lambda)
calculatewq(7)
calculatewq <- function(c)
{
P0inv <- (Rho^c*(1-((Rho/c)^(N-c+1))))/(factorial(c)*(1-(Rho/c)))
for (i in 1:c-1)
{
P0inv = P0inv + (Rho^i)/factorial(i)
}
P0 = 1/P0inv
Lq = (Rho^(c+1))*(1-((Rho/c)^(N-c+1))-((N-c+1)*(1-(Rho/c))*((Rho/c)^(N- c))))*P0/(factorial(c-1)*(c-Rho)^2)
Wq = 60*Lq/Lambda
Ls <- Lq + Rho
Ws <- 60*Ls/Lambda
PN <- (Rho^N)*P0/(factorial(c)*c^(N-c))
customer_serviced <- (1 - PN)*100
a <- cbind(Lq,Wq,Ls,Ws,customer_serviced)
return(a)
}
for (i in 1:Lambda)
{
matrix1[i,2] <- calculatewq(i)[2]
matrix1[i,3] <- calculatewq(i)[5]
matrix1[,4] = car_n*created_mar_3
}
I have a question on manipulating data in data.frame.
Essentially I have a large data set - abbreviated version below:
structure(list(nm_mean = c(194213914.326, 194213914.326, 194213914.326,
194213914.326, 194213914.326, 217947112.739), nm_se = c(9984735.05918367,
9984735.05918367, 9984735.05918367, 9984735.05918367, 9984735.05918367,
11010386.0760204), alpha = c(193.197697846336, 214.592588477741,
240.246557258741, 258.116959355425, 282.560024775668, 306.610038660465
), beta = c(61526.2664158025, 57950.9563448233, 56085.1512614369,
52919.4794239927, 51483.4591654126, 50405.8186695088)), .Names = c("nm_mean",
"nm_se", "alpha", "beta"), row.names = c(NA, 6L), class = "data.frame")
I want to use rbeta to generate probabilities using the beta distribution and alpha and beta as the parameters
Similarly I want to use rnorm to generate random numbers using the normal distribution with nm_mean and nm_se as the mean and sd.
I then want to multiply the rbeta values generated by the rnorm values and extract the 50th, 25th and 75th quantile back into the dataframe
So as an example for row 1
x <- rbeta(1000,193.1977,61526.27)
y <- rnorm(1000,194213914,9984735)
z <- x*y
dat$ce <- quantile(z,0.5)
dat$ll <- quantile(z,0.25)
dat$ul <- quantile(z,0.975)
In essence i get a ce, ll and ul for product of the rbeta and rnorm appended back to the database.
Motivated by #HackR's code, what I think is a functional vectorized version:
set.seed(42)
n <- 1000
nrows <- nrow(dat)
rn <- matrix(rnorm(nrows * n, dat$nm_mean, dat$nm_se), ncol = nrows, byrow = TRUE)
rb <- matrix(rbeta(nrows * n, shape1 = dat$alpha, shape2 = dat$beta),
ncol = nrows, byrow = TRUE)
cbind(dat,
structure(t(apply(rn * rb, 2, function(z) quantile(z, c(0.5, 0.25, 0.975)))),
.Dimnames = list(NULL, c("ce", "ll", "ul"))))
# nm_mean nm_se alpha beta ce ll ul
# 1 194213914 9984735 193.1977 61526.27 608455.3 570100.5 710373.6
# 2 194213914 9984735 214.5926 57950.96 715305.0 677754.3 856570.7
# 3 194213914 9984735 240.2466 56085.15 825143.7 778351.2 979361.1
# 4 194213914 9984735 258.1170 52919.48 943261.4 895832.6 1091899.3
# 5 194213914 9984735 282.5600 51483.46 1054514.3 995640.8 1226176.4
# 6 217947113 11010386 306.6100 50405.82 1312325.0 1247030.8 1515630.5
This is vectorized solution based on my conversation with #thelatemail:
n <- 1000
grp <- nrow(dat)
z <- with(dat, rnorm(grp*n, nm_mean, nm_se) * rbeta(grp*n, alpha, beta) )
m <- 1
for(i in 1:nrow(dat)){
dat$ce[i] <- quantile(z[m:(i*1000)],0.5)
dat$ll[i] <- quantile(z[m:(i*1000)],0.25)
dat$ul[i] <- quantile(z[m:(i*1000)],0.975)
m <- m + 1000
}
A less vectorized solution is:
for(i in 1:nrow(dat)){
x <- rbeta(1000, shape1 = dat$alpha[i], shape2 = dat$beta[i])
y <- rnorm(n=1000,dat$nm_mean[i],dat$nm_se[i])
z <- x*y
dat$ce[i] <- quantile(z,0.5)
dat$ll[i] <- quantile(z,0.25)
dat$ul[i] <- quantile(z,0.975)
}
dat
nm_mean nm_se alpha beta ce ll ul
1 194213914 9984735 193.1977 61526.27 607563.9 573229.9 713057.2
2 194213914 9984735 214.5926 57950.96 712268.5 674826.3 836950.8
3 194213914 9984735 240.2466 56085.15 823322.9 777482.8 981156.7
4 194213914 9984735 258.1170 52919.48 937331.2 884945.0 1095876.3
5 194213914 9984735 282.5600 51483.46 1059980.4 1003596.4 1225615.6
6 217947113 11010386 306.6100 50405.82 1316733.1 1250190.1 1515185.0
I wrote the following code, and I need to repeat this for 100 times, and I know I need to user another for loop, but I don't know how to do it. Here is the code:
mean <- c(5,5,10,10,5,5,5)
x <- NULL
u <- NULL
delta1 <- NULL
w1 <- NULL
for (i in 1:7 ) {
x[i] <- rexp(1, rate = mean[i])
u[i] <- (1/1.2)*runif(1, min=0, max=1)
y1 <- min(x,u)
if (y1 == min(x)) {
delta1 <- 1
}
else {
delta1 <- 0
}
if (delta1 == 0)
{
w1 <- NULL
}
else {
if(y1== x[[1]])
{
w1 <- "x1"
}
}
}
output <- cbind(delta1,w1)
output
I want the final output to be 100 rows* 3 columns matrix representing run number, delta1, and w1.
Any thought will be truly appreciated.
Here's what I gather you're trying to achieve from your code:
Given two vectors drawn from different distributions (Exponential and Uniform)
Find out which distribution the smallest number comes from
Repeat this 100 times.
Theres a couple of problems with your code if you want to achieve this, so here's a cleaned up example:
rates <- c(5, 5, 10, 10, 5, 5, 5) # 'mean' is an inbuilt function
# Initialise the output data frame:
output <- data.frame(number=rep(0, 100), delta1=rep(1, 100), w1=rep("x1", 100))
for (i in 1:100) {
# Generating u doesn't require a for loop. Additionally, can bring in
# the (1/1.2) out the front.
u <- runif(7, min=0, max=5/6)
# Generating x doesn't need a loop either. It's better to use apply functions
# when you can!
x <- sapply(rates, function(x) { rexp(1, rate=x) })
y1 <- min(x, u)
# Now we can store the output
output[i, "number"] <- y1
# Two things here:
# 1) use all.equal instead of == to compare floating point numbers
# 2) We initialised the data frame to assume they always came from x.
# So we only need to overwrite it where it comes from u.
if (isTRUE(all.equal(y1, min(u)))) {
output[i, "delta1"] <- 0
output[i, "w1"] <- NA # Can't use NULL in a character vector.
}
}
output
Here's an alternative, more efficient approach with replicate:
Mean <- c(5, 5, 10, 10, 5, 5, 5)
n <- 100 # number of runs
res <- t(replicate(n, {
x <- rexp(n = length(Mean), rate = Mean)
u <- runif(n = length(Mean), min = 0, max = 1/1.2)
mx <- min(x)
delta1 <- mx <= min(u)
w1 <- delta1 & mx == x[1]
c(delta1, w1)
}))
output <- data.frame(run = seq.int(n), delta1 = as.integer(res[ , 1]),
w1 = c(NA, "x1")[res[ , 2] + 1])
The result:
head(output)
# run delta1 w1
# 1 1 1 <NA>
# 2 2 1 <NA>
# 3 3 1 <NA>
# 4 4 1 x1
# 5 5 1 <NA>
# 6 6 0 <NA>