How can I do difference in means (ttest) for a multivariate using R and WinBUGS14
I have a multivariate outcome y and the categorical variable x. I am able to get the means of the MCMC sampled values from the multivariate using the code below, but how can I test for the difference in means by variable x?
Here is the R code
library(R2WinBUGS)
library(MASS) # need to mvrnorm
library(MCMCpack) # need for rwish
# Generate synthetic data
N <- 500
#we use this to simulate the data
S <- matrix(c(1,.2,.2,5),nrow=2)
#Produces one or more samples from the specified multivariate normal distribution.
#produces 2 variables with the given distribution
y <- mvrnorm(n=N,mu=c(1,3),Sigma=S)
x <- rbinom(500, 1, 0.5)
# Set up for WinBUGS
#set up of the mu0 values
mu0 <- as.vector(c(0,0))
#covariance matrices
# the precisions
S2 <- matrix(c(1,0,0,1),nrow=2)/1000 #precision for unkown mu
# precison matrix to be passes to the wishart distribution for the tau
S3 <- matrix(c(1,0,0,1),nrow=2)/10000
#the data for the winbug code
data <- list("y","N","S2","S3","mu0")
inits <- function(){
list( mu=mvrnorm(1,mu0,matrix(c(10,0,0,10),nrow=2) ),
tau <- rwish(3,matrix(c(.02,0,0,.04),nrow=2)) )
}
# Run WinBUGS
bug_file <- paste0(getwd(), "/codes/mult_normal.bug")
multi_norm.sim <- bugs(data,inits,model.file=bug_file,
parameters=c("mu","tau"),n.chains = 2,n.iter=4010,n.burnin=10,n.thin=1,
bugs.directory="../WinBUGS14/",codaPkg=F)
print(multi_norm.sim,digits=3)
and this is the WinBUGS14 code called mult_normal.bug
model{
for(i in 1:N)
{
y[i,1:2] ~ dmnorm(mu[],tau[,])
}
mu[1:2] ~ dmnorm(mu0[],S2[,])
#parameters of a wishart
tau[1:2,1:2] ~ dwish(S3[,],3)
}
2 Steps:
Load a function to run the t.test using sample statistics instead of doing it directly.
t.test2 <- function(m1,m2,s1,s2,n1,n2,m0=0,equal.variance=FALSE)
{
if( equal.variance==FALSE )
{
se <- sqrt( (s1^2/n1) + (s2^2/n2) )
# welch-satterthwaite df
df <- ( (s1^2/n1 + s2^2/n2)^2 )/( (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) )
} else
{
# pooled standard deviation, scaled by the sample sizes
se <- sqrt( (1/n1 + 1/n2) * ((n1-1)*s1^2 + (n2-1)*s2^2)/(n1+n2-2) )
df <- n1+n2-2
}
t <- (m1-m2-m0)/se
dat <- c(m1-m2, se, t, 2*pt(-abs(t),df))
names(dat) <- c("Difference of means", "Std Error", "t", "p-value")
return(dat)
}
Parse out the mean and standard deviation of the things we want to test against x, then pass them to the function.
mu1 <- as.data.frame(multi_norm.sim$mean)$mu[1]
sdmu1 <- multi_norm.sim$sd$mu[1]
t.test2( mean(x), as.numeric(mu1), s1 = sd(x), s2 = sdmu1, 500, 500)
Difference of means Std Error t p-value
-4.950656e-01 2.246905e-02 -2.203323e+01 5.862968e-76
When I copied the results from my screen to SO it was hard to make the labels of the results properly spaced apart, my apologies.
Related
I have run a multiple imputation (m=45, 10 iterations) using the MICE package, and want to calculate the cronbach's alpha for a number of ordinal scales in the data. Is there a function in r that could assist me in calculating the alpha coefficient across the imputed datasets in a manner that would satisfy Rubin's rules for pooling estimates?
We may exploit pool.scalar from the mice package, which performs pooling of univariate estimates according to Rubin's rules.
Since you have not provided a reproducible example yourself, I will provide one.
set.seed(123)
# sample survey responses
df <- data.frame(
x1 = c(1,2,2,3,2,2,3,3,2,3,
1,2,2,3,2,2,3,3,2,3,
1,2,2,3,2,2,3,3,2,3),
x2 = c(1,1,1,2,3,3,2,3,3,3,
1,1,1,2,3,3,2,3,3,3,
1,2,2,3,2,2,3,3,2,3),
x3 = c(1,1,2,1,2,3,3,3,2,3,
1,1,2,1,2,3,3,3,2,3,
1,2,2,3,2,2,3,3,2,3)
)
# function to column-wise generate missing values (MCAR)
create_missings <- function(data, prob) {
x <- replicate(ncol(data),rbinom(nrow(data), 1, prob))
for(k in 1:ncol(data)) {
data[, k] <- ifelse(x[, k] == 1, NA, data[,k])
}
data
}
df <- create_missings(df, prob = 0.2)
# multiple imputation ----------------------------------
library(mice)
imp <- mice(df, m = 10, maxit = 20)
# extract the completed data in long format
implong <- complete(imp, 'long')
We need a function to compute cronbach's alpha and obtain an estimate of the standard error of alpha, which can be used in a call to pool.scalar() later on. Since there is no available formula with which we can analytically estimate the standard error of alpha, we also need to deploy a bootstrapping procedure to estimate this standard error.
The function cronbach_fun() takes the following arguments:
list_compl_data: a character string specifying the list of completed data from a mids object.
boot: a logical indicating whether a non-parametrical bootstrap should be conducted.
B: an integer specifying the number of bootstrap samples to be taken.
ci: a logical indicating whether a confidence interval around alpha should be estimated.
cronbach_fun <- function(list_compl_data, boot = TRUE, B = 1e4, ci = FALSE) {
n <- nrow(list_compl_data); p <- ncol(list_compl_data)
total_variance <- var(rowSums(list_compl_data))
item_variance <- sum(apply(list_compl_data, 2, sd)^2)
alpha <- (p/(p - 1)) * (1 - (item_variance/total_variance))
out <- list(alpha = alpha)
boot_alpha <- numeric(B)
if (boot) {
for (i in seq_len(B)) {
boot_dat <- list_compl_data[sample(seq_len(n), replace = TRUE), ]
total_variance <- var(rowSums(boot_dat))
item_variance <- sum(apply(boot_dat, 2, sd)^2)
boot_alpha[i] <- (p/(p - 1)) * (1 - (item_variance/total_variance))
}
out$var <- var(boot_alpha)
}
if (ci){
out$ci <- quantile(boot_alpha, c(.025,.975))
}
return(out)
}
Now that we have our function to do the 'heavy lifting', we can run it on all m completed data sets, after which we can obtain Q and U (which are required for the pooling of the estimates). Consult ?pool.scalar for more information.
m <- length(unique(implong$.imp))
boot_alpha <- rep(list(NA), m)
for (i in seq_len(m)) {
set.seed(i) # fix random number generator
sub <- implong[implong$.imp == i, -c(1,2)]
boot_alpha[[i]] <- cronbach_fun(sub)
}
# obtain Q and U (see ?pool.scalar)
Q <- sapply(boot_alpha, function(x) x$alpha)
U <- sapply(boot_alpha, function(x) x$var)
# pooled estimates
pool_estimates <- function(x) {
out <- c(
alpha = x$qbar,
lwr = x$qbar - qt(0.975, x$df) * sqrt(x$t),
upr = x$qbar + qt(0.975, x$df) * sqrt(x$t)
)
return(out)
}
Output
# Pooled estimate of alpha (95% CI)
> pool_estimates(pool.scalar(Q, U))
alpha lwr upr
0.7809977 0.5776041 0.9843913
I have a problem when using replicate to repeat the function.
I tried to use the bootstrap to fit
a quadratic model using concentration as the predictor and Total_lignin as the response and going to report an estimate of the maximum with a corresponding standard error.
My idea is to create a function called bootFun that essentially did everything within one iteration of a for loop. bootFun took in only the data set the predictor, and the response to use (both variable names in quotes).
However, the SD is 0, not correct. I do not know where is the wrong place. Could you please help me with it?
# Load the libraries
library(dplyr)
library(tidyverse)
# Read the .csv and only use M.giganteus and S.ravennae.
dat <- read_csv('concentration.csv') %>%
filter(variety == 'M.giganteus' | variety == 'S.ravennae') %>%
arrange(variety)
# Check the data
head(dat)
# sample size
n <- nrow(dat)
# A function to do one iteration
bootFun <- function(dat, pred, resp){
# Draw the sample size from the dataset
sample <- sample_n(dat, n, replace = TRUE)
# A quadratic model fit
formula <- paste0('resp', '~', 'pred', '+', 'I(pred^2)')
fit <- lm(formula, data = sample)
# Derive the max of the value of concentration
max <- -fit$coefficients[2]/(2*fit$coefficients[3])
return(max)
}
max <- bootFun(dat = dat, pred = 'concentration', resp = 'Total_lignin' )
# Iterated times
N <- 5000
# Use 'replicate' function to do a loop
maxs <- replicate(N, max)
# An estimate of the max of predictor and corresponding SE
mean(maxs)
sd(maxs)
Base package boot, function boot, can ease the job of calling the bootstrap function repeatedly. The first argument must be the data set, the second argument is an indices argument, that the user does not set and other arguments can also be passed toit. In this case those other arguments are the predictor and the response names.
library(boot)
bootFun <- function(dat, indices, pred, resp){
# Draw the sample size from the dataset
dat.sample <- dat[indices, ]
# A quadratic model fit
formula <- paste0(resp, '~', pred, '+', 'I(', pred, '^2)')
formula <- as.formula(formula)
fit <- lm(formula, data = dat.sample)
# Derive the max of the value of concentration
max <- -fit$coefficients[2]/(2*fit$coefficients[3])
return(max)
}
N <- 5000
set.seed(1234) # Make the bootstrap results reproducible
results <- boot(dat, bootFun, R = N, pred = 'concentration', resp = 'Total_lignin')
results
#
#ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
#Call:
#boot(data = dat, statistic = bootFun, R = N, pred = "concentration",
# resp = "Total_lignin")
#
#
#Bootstrap Statistics :
# original bias std. error
#t1* -0.4629808 -0.0004433889 0.03014259
#
results$t0 # this is the statistic, not bootstrapped
#concentration
# -0.4629808
mean(results$t) # bootstrap value
#[1] -0.4633233
Note that to fit a polynomial, function poly is much simpler than to explicitly write down the polynomial terms one by one.
formula <- paste0(resp, '~ poly(', pred, ',2, raw = TRUE)')
Check the distribution of the bootstrapped statistic.
op <- par(mfrow = c(1, 2))
hist(results$t)
qqnorm(results$t)
qqline(results$t)
par(op)
Test data
set.seed(2020) # Make the results reproducible
x <- cumsum(rnorm(100))
y <- x + x^2 + rnorm(100)
dat <- data.frame(concentration = x, Total_lignin = y)
I have an array of outputs from hundreds of segmented linear models (made using the segmented package in R). I want to be able to use these outputs on new data, using the predict function. To be clear, I do not have the segmented linear model objects in my workspace; I just saved and reimported the relevant outputs (e.g. the coefficients and breakpoints). For this reason I can't simply use the predict.segmented function from the segmented package.
Below is a toy example based on this link that seems promising, but does not match the output of the predict.segmented function.
library(segmented)
set.seed(12)
xx <- 1:100
zz <- runif(100)
yy <- 2 + 1.5*pmax(xx-35,0) - 1.5*pmax(xx-70,0) +
15*pmax(zz-0.5,0) + rnorm(100,0,2)
dati <- data.frame(x=xx,y=yy,z=zz)
out.lm<-lm(y~x,data=dati)
o<-## S3 method for class 'lm':
segmented(out.lm,seg.Z=~x,psi=list(x=c(30,60)),
control=seg.control(display=FALSE))
# Note that coefficients with U in the name are differences in slopes, not slopes.
# Compare:
slope(o)
coef(o)[2] + coef(o)[3]
coef(o)[2] + coef(o)[3] + coef(o)[4]
# prediction
pred <- data.frame(x = 1:100)
pred$dummy1 <- pmax(pred$x - o$psi[1,2], 0)
pred$dummy2 <- pmax(pred$x - o$psi[2,2], 0)
pred$dummy3 <- I(pred$x > o$psi[1,2]) * (coef(o)[2] + coef(o)[3])
pred$dummy4 <- I(pred$x > o$psi[2,2]) * (coef(o)[2] + coef(o)[3] + coef(o)[4])
names(pred)[-1]<- names(model.frame(o))[-c(1,2)]
# compute the prediction, using standard predict function
# computing confidence intervals further
# suppose that the breakpoints are fixed
pred <- data.frame(pred, predict(o, newdata= pred,
interval="confidence"))
# Try prediction using the predict.segment version to compare
test <- predict.segmented(o)
plot(pred$fit, test, ylim = c(0, 100))
abline(0,1, col = "red")
# At least one segment not being predicted correctly?
Can I use the base r predict() function (not the segmented.predict() function) with the coefficients and break points saved from segmented linear models?
UPDATE
I figured out that the code above has issues (don't use it). Through some reverse engineering of the segmented.predict() function, I produced the design matrix and use that to predict values instead of directly using the predict() function. I do not consider this a full answer of the original question yet because predict() can also produce confidence intervals for the prediction, and I have not yet implemented that--question still open for someone to add confidence intervals.
library(segmented)
## Define function for making matrix of dummy variables (this is based on code from predict.segmented())
dummy.matrix <- function(x.values, x_names, psi.est = TRUE, nameU, nameV, diffSlope, est.psi) {
# This function creates a model matrix with dummy variables for a segmented lm with two breakpoints.
# Inputs:
# x.values: the x values of the segmented lm
# x_names: the name of the column of x values
# psi.est: this is legacy from the predict.segmented function, leave it set to 'TRUE'
# obj: the segmented lm object
# nameU: names (class character) of 3rd and 4th coef, which are "U1.x" "U2.x" for lm with two breaks. Example: names(c(obj$coef[3], obj$coef[4]))
# nameV: names (class character) of 5th and 6th coef, which are "psi1.x" "psi2.x" for lm with two breaks. Example: names(c(obj$coef[5], obj$coef[6]))
# diffSlope: the coefficients (class numeric) with the slope differences; called U1.x and U2.x for lm with two breaks. Example: c(o$coef[3], o$coef[4])
# est.psi: the estimated break points (class numeric); these are the estimated breakpoints from segmented.lm. Example: c(obj$psi[1,2], obj$psi[2,2])
#
n <- length(x.values)
k <- length(est.psi)
PSI <- matrix(rep(est.psi, rep(n, k)), ncol = k)
newZ <- matrix(x.values, nrow = n, ncol = k, byrow = FALSE)
dummy1 <- pmax(newZ - PSI, 0)
if (psi.est) {
V <- ifelse(newZ > PSI, -1, 0)
dummy2 <- if (k == 1)
V * diffSlope
else V %*% diag(diffSlope)
newd <- cbind(x.values, dummy1, dummy2)
colnames(newd) <- c(x_names, nameU, nameV)
} else {
newd <- cbind(x.values, dummy1)
colnames(newd) <- c(x_names, nameU)
}
# if (!x_names %in% names(coef(obj.seg)))
# newd <- newd[, -1, drop = FALSE]
return(newd)
}
## Test dummy matrix function----------------------------------------------
set.seed(12)
xx<-1:100
zz<-runif(100)
yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2)
dati<-data.frame(x=xx,y=yy,z=zz)
out.lm<-lm(y~x,data=dati)
#1 segmented variable, 2 breakpoints: you have to specify starting values (vector) for psi:
o<-segmented(out.lm,seg.Z=~x,psi=c(30,60),
control=seg.control(display=FALSE))
slope(o)
plot.segmented(o)
summary(o)
# Test dummy matrix fn with the same dataset
newdata <- dati
nameU1 <- c("U1.x", "U2.x")
nameV1 <- c("psi1.x", "psi2.x")
diffSlope1 <- c(o$coef[3], o$coef[4])
est.psi1 <- c(o$psi[1,2], o$psi[2,2])
test <- dummy.matrix(x.values = newdata$x, x_names = "x", psi.est = TRUE,
nameU = nameU1, nameV = nameV1, diffSlope = diffSlope1, est.psi = est.psi1)
# Predict response variable using matrix multiplication
col1 <- matrix(1, nrow = dim(test)[1])
test <- cbind(col1, test) # Now test is the same as model.matrix(o)
predY <- coef(o) %*% t(test)
plot(predY[1,])
lines(predict.segmented(o), col = "blue") # good, predict.segmented gives same answer
I have written a small function that simulates data from a normal distribution, how it is usual in linear models. My question is how to get a model with a pvalue of sim[, 1] == 0.05. I want to show that if I add a random variable even it is normal distributed around zero with small variance N(0,0.0023) , that pvalue of sim[,1] changes. The code below shows the true model.
set.seed(37) # seed for reproducability
simulation <- function(b_0, b_1,n,min_x_1 ,max_x_1,sd_e){
mat <- NA
x_1 <- runif(n = n, min = min_x_1, max =max_x_1)
error <- rnorm(mean = 0,sd = sd_e, n = n )
y <- b_0 + b_1*x_1 + error
mat <- matrix(cbind(x_1,y), ncol = 2)
return(mat)
#plot(mat[,1],mat[,2])
}
sim <- simulation(10,-2,10000,-10,70,0.003)
summary(lm(sim[,2] ~ sim[,1] ))
I am trying to create a linear mixed model (lmm) that allows for a spatial correlation between points (have lat/long for each point). I would like the spatial correlation to be based upon the great circular distance between points.
The package ramps includes a correlation structure that computes the ‘haversine’ distance – although I am having trouble implementing it. I have previously used other correlation structures (corGaus, corExp) and not had any difficulties. I am assuming the corRGaus with the 'haversine' metric can be implemented in the same way.
I am able to successfully create an lmm with spatial correlation calculated on a planar distance using the lme function.
I am also able to create a linear model (not mixed) with spatial correlation calculated using great circular distance although there are errors with the correlation structure using the gls command.
When trying to the use the gls command for a linear model with the great circular distance I have the following errors:
x = runif(20, 1,50)
y = runif(20, 1,50)
gls(x ~ y, cor = corRGaus(form = ~ x + y))
Generalized least squares fit by REML
Model: x ~ y
Data: NULL
Log-restricted-likelihood: -78.44925
Coefficients:
(Intercept) y
24.762656602 0.007822469
Correlation Structure: corRGaus
Formula: ~x + y
Parameter estimate(s):
Error in attr(object, "fixed") && unconstrained :
invalid 'x' type in 'x && y'
When I increase the size of the data there are memory allocation errors (still a very small dataset):
x = runif(100, 1, 50)
y = runif(100, 1, 50)
lat = runif(100, -90, 90)
long = runif(100, -180, 180)
gls(x ~ y, cor = corRGaus(form = ~ x + y))
Error in glsEstimate(glsSt, control = glsEstControl) :
'Calloc' could not allocate memory (18446744073709551616 of 8 bytes)
When trying to run a mixed model using the lme command and the corRGaus from the ramps package the following results:
x = runif(100, 1, 50)
y = runif(100, 1, 50)
LC = c(rep(1, 50) , rep(2, 50))
lat = runif(100, -90, 90)
long = runif(100, -180, 180)
lme(x ~ y,random = ~ y|LC, cor = corRGaus(form = ~ long + lat))
Error in `coef<-.corSpatial`(`*tmp*`, value = value[parMap[, i]]) :
NA/NaN/Inf in foreign function call (arg 1)
In addition: Warning messages:
1: In nlminb(c(coef(lmeSt)), function(lmePars) -logLik(lmeSt, lmePars), :
NA/NaN function evaluation
2: In nlminb(c(coef(lmeSt)), function(lmePars) -logLik(lmeSt, lmePars), :
NA/NaN function evaluation
I am unsure about how to proceed with this method. The "haversine" function is what I want to use to complete my models, but I am having trouble implementing them. There are very few questions anywhere about the ramps package, and I have seen very few implementations. Any helps would be greatly appreciated.
I have previously attempted to modify the nlme package and was unable to do so. I posted a question about this, where I was recommended to use the ramps package.
I am using R 3.0.0 on a Windows 8 computer.
OK, here is an option that implements various spatial correlation structures in gls/nlme with haversine distance.
The various corSpatial-type classes already have machinery in place to construct a correlation matrix from spatial covariates, given a distance metric. Unfortunately, dist does not implement haversine distance, and dist is the function called by corSpatial to compute a distance matrix from the spatial covariates.
The distance matrix computations are performed in getCovariate.corSpatial. A modified form of this method will pass the appropriate distance to other methods, and the majority of methods will not need to be modified.
Here, I create a new corStruct class, corHaversine, and modify only getCovariate and one other method (Dim) that determines which correlation function is used. Those methods which do not need modification, are copied from equivalent corSpatial methods. The (new) mimic argument in corHaversine takes the name of the spatial class with the correlation function of interest: by default, it is set to "corSpher".
Caveat: beyond ensuring that this code runs for spherical and Gaussian correlation functions, I haven't really done a lot of checking.
#### corHaversine - spatial correlation with haversine distance
# Calculates the geodesic distance between two points specified by radian latitude/longitude using Haversine formula.
# output in km
haversine <- function(x0, x1, y0, y1) {
a <- sin( (y1 - y0)/2 )^2 + cos(y0) * cos(y1) * sin( (x1 - x0)/2 )^2
v <- 2 * asin( min(1, sqrt(a) ) )
6371 * v
}
# function to compute geodesic haversine distance given two-column matrix of longitude/latitude
# input is assumed in form decimal degrees if radians = F
# note fields::rdist.earth is more efficient
haversineDist <- function(xy, radians = F) {
if (ncol(xy) > 2) stop("Input must have two columns (longitude and latitude)")
if (radians == F) xy <- xy * pi/180
hMat <- matrix(NA, ncol = nrow(xy), nrow = nrow(xy))
for (i in 1:nrow(xy) ) {
for (j in i:nrow(xy) ) {
hMat[j,i] <- haversine(xy[i,1], xy[j,1], xy[i,2], xy[j,2])
}
}
as.dist(hMat)
}
## for most methods, machinery from corSpatial will work without modification
Initialize.corHaversine <- nlme:::Initialize.corSpatial
recalc.corHaversine <- nlme:::recalc.corSpatial
Variogram.corHaversine <- nlme:::Variogram.corSpatial
corFactor.corHaversine <- nlme:::corFactor.corSpatial
corMatrix.corHaversine <- nlme:::corMatrix.corSpatial
coef.corHaversine <- nlme:::coef.corSpatial
"coef<-.corHaversine" <- nlme:::"coef<-.corSpatial"
## Constructor for the corHaversine class
corHaversine <- function(value = numeric(0), form = ~ 1, mimic = "corSpher", nugget = FALSE, fixed = FALSE) {
spClass <- "corHaversine"
attr(value, "formula") <- form
attr(value, "nugget") <- nugget
attr(value, "fixed") <- fixed
attr(value, "function") <- mimic
class(value) <- c(spClass, "corStruct")
value
} # end corHaversine class
environment(corHaversine) <- asNamespace("nlme")
Dim.corHaversine <- function(object, groups, ...) {
if (missing(groups)) return(attr(object, "Dim"))
val <- Dim.corStruct(object, groups)
val[["start"]] <- c(0, cumsum(val[["len"]] * (val[["len"]] - 1)/2)[-val[["M"]]])
## will use third component of Dim list for spClass
names(val)[3] <- "spClass"
val[[3]] <- match(attr(object, "function"), c("corSpher", "corExp", "corGaus", "corLin", "corRatio"), 0)
val
}
environment(Dim.corHaversine) <- asNamespace("nlme")
## getCovariate method for corHaversine class
getCovariate.corHaversine <- function(object, form = formula(object), data) {
if (is.null(covar <- attr(object, "covariate"))) { # if object lacks covariate attribute
if (missing(data)) { # if object lacks data
stop("need data to calculate covariate")
}
covForm <- getCovariateFormula(form)
if (length(all.vars(covForm)) > 0) { # if covariate present
if (attr(terms(covForm), "intercept") == 1) { # if formula includes intercept
covForm <- eval(parse(text = paste("~", deparse(covForm[[2]]),"-1",sep=""))) # remove intercept
}
# can only take covariates with correct names
if (length(all.vars(covForm)) > 2) stop("corHaversine can only take two covariates, 'lon' and 'lat'")
if ( !all(all.vars(covForm) %in% c("lon", "lat")) ) stop("covariates must be named 'lon' and 'lat'")
covar <- as.data.frame(unclass(model.matrix(covForm, model.frame(covForm, data, drop.unused.levels = TRUE) ) ) )
covar <- covar[,order(colnames(covar), decreasing = T)] # order as lon ... lat
}
else {
covar <- NULL
}
if (!is.null(getGroupsFormula(form))) { # if groups in formula extract covar by groups
grps <- getGroups(object, data = data)
if (is.null(covar)) {
covar <- lapply(split(grps, grps), function(x) as.vector(dist(1:length(x) ) ) ) # filler?
}
else {
giveDist <- function(el) {
el <- as.matrix(el)
if (nrow(el) > 1) as.vector(haversineDist(el))
else numeric(0)
}
covar <- lapply(split(covar, grps), giveDist )
}
covar <- covar[sapply(covar, length) > 0] # no 1-obs groups
}
else { # if no groups in formula extract distance
if (is.null(covar)) {
covar <- as.vector(dist(1:nrow(data) ) )
}
else {
covar <- as.vector(haversineDist(as.matrix(covar) ) )
}
}
if (any(unlist(covar) == 0)) { # check that no distances are zero
stop("cannot have zero distances in \"corHaversine\"")
}
}
covar
} # end method getCovariate
environment(getCovariate.corHaversine) <- asNamespace("nlme")
To test that this runs, given range parameter of 1000:
## test that corHaversine runs with spherical correlation (not testing that it WORKS ...)
library(MASS)
set.seed(1001)
sample_data <- data.frame(lon = -121:-22, lat = -50:49)
ran <- 1000 # 'range' parameter for spherical correlation
dist_matrix <- as.matrix(haversineDist(sample_data)) # haversine distance matrix
# set up correlation matrix of response
corr_matrix <- 1-1.5*(dist_matrix/ran)+0.5*(dist_matrix/ran)^3
corr_matrix[dist_matrix > ran] = 0
diag(corr_matrix) <- 1
# set up covariance matrix of response
sigma <- 2 # residual standard deviation
cov_matrix <- (diag(100)*sigma) %*% corr_matrix %*% (diag(100)*sigma) # correlated response
# generate response
sample_data$y <- mvrnorm(1, mu = rep(0, 100), Sigma = cov_matrix)
# fit model
gls_haversine <- gls(y ~ 1, correlation = corHaversine(form=~lon+lat, mimic="corSpher"), data = sample_data)
summary(gls_haversine)
# Correlation Structure: corHaversine
# Formula: ~lon + lat
# Parameter estimate(s):
# range
# 1426.818
#
# Coefficients:
# Value Std.Error t-value p-value
# (Intercept) 0.9397666 0.7471089 1.257871 0.2114
#
# Standardized residuals:
# Min Q1 Med Q3 Max
# -2.1467696 -0.4140958 0.1376988 0.5484481 1.9240042
#
# Residual standard error: 2.735971
# Degrees of freedom: 100 total; 99 residual
Testing that it runs with Gaussian correlation, with range parameter = 100:
## test that corHaversine runs with Gaussian correlation
ran = 100 # parameter for Gaussian correlation
corr_matrix_gauss <- exp(-(dist_matrix/ran)^2)
diag(corr_matrix_gauss) <- 1
# set up covariance matrix of response
cov_matrix_gauss <- (diag(100)*sigma) %*% corr_matrix_gauss %*% (diag(100)*sigma) # correlated response
# generate response
sample_data$y_gauss <- mvrnorm(1, mu = rep(0, 100), Sigma = cov_matrix_gauss)
# fit model
gls_haversine_gauss <- gls(y_gauss ~ 1, correlation = corHaversine(form=~lon+lat, mimic = "corGaus"), data = sample_data)
summary(gls_haversine_gauss)
With lme:
## runs with lme
# set up data with group effects
group_y <- as.vector(sapply(1:5, function(.) mvrnorm(1, mu = rep(0, 100), Sigma = cov_matrix_gauss)))
group_effect <- rep(-2:2, each = 100)
group_y = group_y + group_effect
group_name <- factor(group_effect)
lme_dat <- data.frame(y = group_y, group = group_name, lon = sample_data$lon, lat = sample_data$lat)
# fit model
lme_haversine <- lme(y ~ 1, random = ~ 1|group, correlation = corHaversine(form=~lon+lat, mimic = "corGaus"), data = lme_dat, control=lmeControl(opt = "optim") )
summary(lme_haversine)
# Correlation Structure: corHaversine
# Formula: ~lon + lat | group
# Parameter estimate(s):
# range
# 106.3482
# Fixed effects: y ~ 1
# Value Std.Error DF t-value p-value
# (Intercept) -0.0161861 0.6861328 495 -0.02359033 0.9812
#
# Standardized Within-Group Residuals:
# Min Q1 Med Q3 Max
# -3.0393708 -0.6469423 0.0348155 0.7132133 2.5921573
#
# Number of Observations: 500
# Number of Groups: 5
See if this answer on R-Help is useful: http://markmail.org/search/?q=list%3Aorg.r-project.r-help+winsemius+haversine#query:list%3Aorg.r-project.r-help%20winsemius%20haversine+page:1+mid:ugecbw3jjwphu2pb+state:results
I just checked and and doesn't appear that the ramps or nlme packages have been modified to incorporate those changes suggested by Malcolm Fairbrother, so you will need to do some hacking. I don't want to be considered for the bounty since I am not posting a tested solution and I didn't dream it up either.