I have:
x = rnorm(100)
# Partie b
z = rbinom(100,1,0.60)
# Partie c
y = 1.4 + 0.7*x - 0.5*z
# Partie d
x1 = abs(x)
y1 = abs(y)
Don<-cbind(y1,x1,z)
Don1 <- data.frame(Don)
Reg <- glm(y1~x1+z,family=poisson(link="log"),Don1)
# Partie e
#Biais de beta
Reg.cf <- coef(Reg)
biais0 = Reg.cf[1] - 1.4
biais1 = Reg.cf[2] - 0.7
biais2 = Reg.cf[3] + 0.5
And I need to repeat all this 100 times in order to have different coefficient and calculate the bias and then put the mean of each biais in a text file.
I don't know how to implement I taught about repeat{if()break;} But how do I do that? I tried the loop for but it didn't work out.
I'd be inclined to do it this way.
get.bias <- function(i) { # the argument i is not used
x <- rnorm(100)
z <- rbinom(100,1,0.60)
y <- 1.4 + 0.7*x - 0.5*z
df <- data.frame(y1=abs(y), x1=abs(x), z)
coef(glm(y1~x1+z,family=poisson(link="log"),df)) - c(1.4,0.7,-0.5)
}
set.seed(1) # for reproducible example; you may want to comment out this line
result <- t(sapply(1:100,get.bias))
head(result)
# (Intercept) x1 z
# [1,] -1.129329 -0.4992925 0.076027012
# [2,] -1.205608 -0.5642966 0.215998775
# [3,] -1.089448 -0.5834090 0.081211412
# [4,] -1.206076 -0.4629789 0.004513795
# [5,] -1.203938 -0.6980701 0.201001466
# [6,] -1.366077 -0.5640367 0.452784690
colMeans(result)
# (Intercept) x1 z
# -1.1686845 -0.5787492 0.1242588
sapply(list,fun) "applies" the list element-wise to the function; e.g. it calls the function once for each element in the list, and assembles the results into a matrix. So here get.bias(...) will be called 100 times and the results returned each time will be assembled into a matrix. This matrix has one column for each result, but we want the results in rows with one column for each parameter, so we transpose with t(...).
Related
how can I use rnorm_multiI() to simulate 100 observations from 3 sets of zscores (a, b and c) which all correlated with each other at 0.25?
Here's a way using MASS::mvrnorm, don't know rnorm_multi(). First, set up parameters, then use them in mvrnorm.
## parameters
s <- c(.5, 1, 2) ## define sds of z1-3
r <- .25 ## corr.
## define covariance matrix
Sigma <- matrix(c(
s[1]^2, r/2, r,
r/2, s[2]^2, 2*r,
r, 2*r, s[3]^2
), ncol=3, nrow=3)
## simulation
n <- 100
set.seed(42)
library(MASS)
M <- mvrnorm(n=n, numeric(3), Sigma, empirical=TRUE) |> `colnames<-`(letters[1:3])
Result
head(M)
# a b c
# [1,] -0.50980732 0.8981857 4.078068
# [2,] -0.58623896 0.2514551 -1.370179
# [3,] 0.71829541 0.2693594 -0.985852
# [4,] -0.19684349 2.2982353 -2.238448
# [5,] 0.08263476 -0.5594775 2.335499
# [6,] -0.35133993 -0.4720599 1.282973
matrixStats::colSds(M)
# [1] 0.5 1.0 2.0
colMeans(M)
# a b c
# 8.326673e-18 1.748601e-17 -2.872702e-17
## i.e. zero
cor(M)
# a b c
# a 1.00 0.25 0.25
# b 0.25 1.00 0.25
# c 0.25 0.25 1.00
Note: The empirical=TRUE flag forces the result to have the exact Sigma given, i.e. sample mean. You might want to set it to FALSE to simulate sampling from a population with given Sigma. Apart from that, it is also helpful to check if the parameters were specified correctly.
library(faux)
set.seed(96)
rnorm_multi(vars = 3, r = 0.25, varnames = c("a", "b", "c"))
You can get the same correlation between every pair of variables by specifying a single value. More on it in the vignette.
Create a simulated dataset of 100 observations, where x is a random normal variable with mean 0 and standard deviation 1, and y = 0.1 + 2 * X + e, where epsilon is also a random normal error with mean 0 and sd 1.
set.seed(1)
# simulate a data set of 100 observations
x <- rnorm(100)
y.1 <- 0.1 + 2*x + rnorm(100)
Now extract the first 5 observations.
y1.FirstFive <- (y.1[1:5]) # extract first 5 observations from y
x.FirstFive <- (x[1:5]) # extract first 5 observations from x
y1.FirstFive # extracted 5 observations from y1
[1] -1.7732743 0.5094025 -2.4821789 3.4485904 0.1044309
x.FirstFive # extracted 5 observations from x
[1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078
Assuming the mean and sd of the sample that you calculated from the first five observations would not change, what is the minimum total number of additional observations you would need to be able to conclude that the true mean of the population is different from 0 at the p = 0.01 confidence level?
alpha <- 0.01
mu <- 0
for (i in 5:2000) {
# Recalculate the standard error and CI
stand_err <- Sd_y1 / sqrt(i)
ci <- sample_mean_y1 + c(qt(alpha/2, i-1), qt(1-alpha/2, i-1))*stand_err
if (ci[2] < mu)
break # condition met, exit loop
}
i
[1] 2000
Here, I wrote a loop that iteratively increases n from the initial n=5 to n=2000, uses pt to find the p value (given a fixed y-bar and sd), and stops when p < 0.01. However I keep getting the wrong output. Such that, the output is always the number of the maximum range that I give (here, it is 2000) instead of giving me the specific minimum n sample in order to reject the null that mu_y = 0 at the p=0.01 level. Any suggestions as to how to fix the code?
additional info: the sd of y1.FirstFive = 2.3 and mean of y1.FirstFive = -0.04
Assuming:
Sd_y1 = sd(y1.FirstFive)
sample_mean_y1 = mean(y1.FirstFive)
sample_mean_y1
[1] -0.03860587
As pointed out by #jblood94, you need to go for larger sample size.
You don't need a for loop for this, most of your functions are vectorized, so something like this:
n = 5:30000
stand_err = Sd_y1 / sqrt(n)
ub = sample_mean_y1 + qt(1-alpha/2, n-1)*stand_err
n[min(which(ub<0))]
[1] 23889
It's because n > 2000.
set.seed(1)
x <- rnorm(100)
y.1 <- 0.1 + 2*x + rnorm(100)
Sd_y1 <- sd(y.1[1:5])
sample_mean_y1 <- mean(y.1[1:5])
alpha <- 0.01
sgn <- 2*(sample_mean_y1 > 0) - 1
f <- function(n) qt(alpha/2, n - 1)*Sd_y1 + sgn*sample_mean_y1*sqrt(n)
upper <- 2
while (f(upper) < 0) upper <- upper*2
(n <- ceiling(uniroot(f, lower = upper/2, upper = upper, tol = 0.5)$root))
#> [1] 23889
I have defined a function to calculate the relationship between height (h) and diameter (dbh) of trees based on equations extracted from 2 publications. My goal is to use the relationship established in paper 1 (Xiangtao) to predict the values of variables in an equation in paper 2 (Marechaux and Chave). I would like to test to see over what diameter range [x:y] the generated nls() curve of paper 2 fits paper 1. Currently, I keep getting an error (I believe in plot())
Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ
if I use anything except x=1 for [x:y] i.e. dbh.min:dbh.max
My function is as follows:
# Plant.Functional.Type constants...
Dsb1 <- 2.09
Dsb2 <- 0.54
Db1 <- 0.93
Db2 <- 0.84
BDb1 <- 2.66
BDb2 <- 0.48
Eb1 <- 1.41
Eb2 <- 0.65
# # # # # # # # # # # # # # # # # # # # # # # # # # #
Generate.curve <- function(b1, b2, dbh.min, dbh.max){
# calculate Xiangtao's allometry...
tmp_h <- c(dbh.min:dbh.max)
for (dbh in dbh.min:dbh.max)
{
h = b1*dbh^(b2)
tmp_h[dbh] = h
}
# plot to check curve
plot(dbh.min:dbh.max, tmp_h)
# define secondary function for Marechaux and Chave allometry
h_fxn <- function(hlim,dbh,ah){
h = hlim * (dbh / (dbh + ah))
return(h)
}
# use nonlinear least squares model to solve for ah and hlim
# set model inputs
start.ah <- 1
start.hlim <- 5
tmp_v <- cbind(dbh.min:dbh.max,tmp_h)
tmp.fit <- nls(tmp_h ~ h_fxn(hlim,dbh.min:dbh.max,ah), start = list(hlim = start.hlim,
ah = start.ah), algorithm = "port", upper = list(hlim = 75, ah = 99))
# seems to be no way of extracting ah and hlim from tmp.fit via subset
# extract manually and then check fit with
# lines(dbh.min:dbh.max, hlim * (dbh.min:dbh.max/(dbh.min:dbh.max + ah)))
# for equation h = hlim * (dbh / (dbh + ah)) from Marechaux and Chave
return(tmp.fit)
}
# # # # # # # # # # # # # # # # # # # # # # # # # # #
This works great for
Generate.curve(Dsb1,Dsb2,1,100)
lines(1:100, 36.75 * (1:100/(1:100 + 52.51)))
But I would like to be able to examine the curve fit in ranges such as [80:100] as well.
I have been trying to figure out why Generate.curve(Dsb1,Dsb2,80,100) returns an error for about 3 days now. Thanks for any help.
Your problem lies in this section:
tmp_h <- c(dbh.min:dbh.max)
for (dbh in dbh.min:dbh.max)
{
h = b1*dbh^(b2)
tmp_h[dbh] = h
}
Think about what happens when you set dbh.min to 80 and dbh.max to 100:
tmp_h <- 80:100
for (dbh in 80:100)
{
h = b1*dbh^(b2)
tmp_h[dbh] = h
}
What happens on the first cycle of the loop? Well, tmp_h is length 20, but on the first cycle, dbh is 80, and you are assigning a number to tmp_h[dbh], which is tmp_h[80]. By the time the loop has finished, tmp_h will have the correct values stored, but they will be in the indices 80:100. So tmp_h will have the numbers 80:100 stored in the first 21 indices, then a bunch of NAs then the correct numbers in the last 21 indices.
So change it to:
tmp_h <- c(dbh.min:dbh.max)
for (dbh in dbh.min:dbh.max)
{
h = b1*dbh^(b2)
tmp_h[dbh - dbh.min + 1] = h
}
and it will work.
However, you don't actually need a loop at all here, since R uses vectorized operations, so this whole section can be replaced with:
tmp_h <- b1 * (dbh.min:dbh.max)^(b2)
and then when you do
Generate.curve(Dsb1,Dsb2,80,100)
lines(80:100, 36.75 * (80:100/(80:100 + 52.51)))
you get this:
I'm looking to simulate an age variable (constrained range 18-35) that is correlated 0.1 with an existing binary variable called use. Most of the examples I've come across demonstrate how to simulate both variables simultaneously.
# setup
set.seed(493)
n <- 134
dat <- data.frame(partID=seq(1, n, 1),
trt=c(rep(0, n/2),
rep(1, n/2)))
# set proportion
a <- .8
b <- .2
dat$use <- c(rbinom(n/2, 1, b),
rbinom(n/2, 1, a))
Not sure if this is the best way to approach this, but you might get close using the answer from here: https://stats.stackexchange.com/questions/15011/generate-a-random-variable-with-a-defined-correlation-to-an-existing-variable
For example (using the code from the link):
x1 <- dat$use # fixed given data
rho <- 0.1 # desired correlation = cos(angle)
theta <- acos(rho) # corresponding angle
x2 <- rnorm(n, 2, 0.5) # new random data
X <- cbind(x1, x2) # matrix
Xctr <- scale(X, center=TRUE, scale=FALSE) # centered columns (mean 0)
Id <- diag(n) # identity matrix
Q <- qr.Q(qr(Xctr[ , 1, drop=FALSE])) # QR-decomposition, just matrix Q
P <- tcrossprod(Q) # = Q Q' # projection onto space defined by x1
x2o <- (Id-P) %*% Xctr[ , 2] # x2ctr made orthogonal to x1ctr
Xc2 <- cbind(Xctr[ , 1], x2o) # bind to matrix
Y <- Xc2 %*% diag(1/sqrt(colSums(Xc2^2))) # scale columns to length 1
x <- Y[ , 2] + (1 / tan(theta)) * Y[ , 1] # final new vector
dat$age <- (1 + x) * 25
cor(dat$use, dat$age)
# 0.1
summary(dat$age)
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 20.17 23.53 25.00 25.00 26.59 30.50
I have correlated one set number with .9, .5, .0
A derives from rnorm(30,-0.5,1)
B derives from rnorm(30,.5,2)
and want to make A & B correlated with .9, .5, .0.
You are describing a multivariate normal distribution, which can be computed with the mvrnorm function:
library(MASS)
meanA <- -0.5
meanB <- 0.5
sdA <- 1
sdB <- 2
correlation <- 0.9
set.seed(144)
vals <- mvrnorm(10000, c(meanA, meanB), matrix(c(sdA^2, correlation*sdA*sdB,
correlation*sdA*sdB, sdB^2), nrow=2))
mean(vals[,1])
# [1] -0.4883265
mean(vals[,2])
# [1] 0.5201586
sd(vals[,1])
# [1] 0.9994628
sd(vals[,2])
# [1] 1.992816
cor(vals[,1], vals[,2])
# [1] 0.8999285
As an alternative, please consider the following. Let the random variables X ~ N(0,1) and Y ~ N(0,1) independently. Then the random variables X and rho X + sqrt(1 - rho^2) Y are both distributed N(0,1), but are now correlated with correlation rho. So possible R code could be
# Define the parameters
meanA <- -0.5
meanB <- 0.5
sdA <- 1
sdB <- 2
correlation <- 0.9
n <- 10000 # You want 30
# Generate from independent standard normals
x <- rnorm(n, 0, 1)
y <- rnorm(n, 0, 1)
# Transform
x2 <- x # could be avoided
y2 <- correlation*x + sqrt(1 - correlation^2)*y
# Fix up means and standard deviations
x3 <- meanA + sdA*x2
y3 <- meanB + sdB*y2
# Check summary statistics
mean(x3)
# [1] -0.4981958
mean(y3)
# [1] 0.4999068
sd(x3)
# [1] 1.014299
sd(y3)
# [1] 2.022377
cor(x3, y3)
# [1] 0.9002529
I created the correlate package to be able to create a correlation between any type of variable (regardless of distribution) given a certain amount of toleration. It does so by permutations.
install.packages('correlate')
library('correlate')
A <- rnorm(30, -0.5, 1)
B <- rnorm(30, .5, 2)
C <- correlate(cbind(A,B), 0.9)
# 0.9012749
D <- correlate(cbind(A,B), 0.5)
# 0.5018054
E <- correlate(cbind(A,B), 0.0)
# -0.00407327
You can pretty much decide the whole matrix if you want (for multiple variables), by giving a matrix as second argument.
Ironically, you can also use it to create a multivariate normal.....