I am working with the cumulative emergence of flies over time (taken at irregular intervals) over many summers (though first I am just trying to make one year work). The cumulative emergence follows a sigmoid pattern and I want to create a maximum likelihood estimation of a 3-parameter Weibull cumulative distribution function. The three-parameter models I've been trying to use in the fitdistrplus package keep giving me an error. I think this must have something to do with how my data is structured, but I cannot figure it out. Obviously I want it to read each point as an x (degree days) and a y (emergence) value, but it seems to be unable to read two columns. The main error I'm getting says "Non-numeric argument to mathematical function" or (with slightly different code) "data must be a numeric vector of length greater than 1". Below is my code including added columns in the df_dd_em dataframe for cumulative emergence and percent emergence in case that is useful.
degree_days <- c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94)
emergence <- c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0)
cum_em <- cumsum(emergence)
df_dd_em <- data.frame (degree_days, emergence, cum_em)
df_dd_em$percent <- ave(df_dd_em$emergence, FUN = function(df_dd_em) 100*(df_dd_em)/46)
df_dd_em$cum_per <- ave(df_dd_em$cum_em, FUN = function(df_dd_em) 100*(df_dd_em)/46)
x <- pweibull(df_dd_em[c(1,3)],shape=5)
dframe2.mle <- fitdist(x, "weibull",method='mle')
Here's my best guess at what you're after:
Set up data:
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd <- transform(dd,cum_em=cumsum(emergence))
We're actually going to fit to an "interval-censored" distribution (i.e. probability of emergence between successive degree day observations: this version assumes that the first observation refers to observations before the first degree-day observation, you could change it to refer to observations after the last observation).
library(bbmle)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun <- function(scale,shape,x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull(c(-Inf,x), ## or (c(x,Inf))
shape=shape,scale=scale)),1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
library(bbmle)
I should probably have used something more systematic like the method of moments (i.e. matching the mean and variance of a Weibull distribution with the mean and variance of the data), but I just hacked around a bit to find plausible starting values:
## preliminary look (method of moments would be better)
scvec <- 10^(seq(0,4,length=101))
plot(scvec,sapply(scvec,NLLfun,shape=1))
It's important to use parscale to let R know that the parameters are on very different scales:
startvals <- list(scale=1000,shape=1)
m1 <- mle2(NLLfun,start=startvals,
control=list(parscale=unlist(startvals)))
Now try with a three-parameter Weibull (as originally requested) -- requires only a slight modification of what we already have:
library(FAdist)
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
Looks like the three-parameter fit is much better:
library(emdbook)
AICtab(m1,m2)
## dAIC df
## m2 0.0 3
## m1 21.7 2
And here's the graphical summary:
with(dd,plot(cum_em~degree_days,cex=3))
with(as.list(coef(m1)),curve(sum(dd$emergence)*
pweibull(x,shape=shape,scale=scale),col=2,
add=TRUE))
with(as.list(coef(m2)),curve(sum(dd$emergence)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
(could also do this more elegantly with ggplot2 ...)
These don't seem like spectacularly good fits, but they're sane. (You could in principle do a chi-squared goodness-of-fit test based on the expected number of emergences per interval, and accounting for the fact that you've fitted a three-parameter model, although the values might be a bit low ...)
Confidence intervals on the fit are a bit of a nuisance; your choices are (1) bootstrapping; (2) parametric bootstrapping (resample parameters assuming a multivariate normal distribution of the data); (3) delta method.
Using bbmle::mle2 makes it easy to do things like get profile confidence intervals:
confint(m1)
## 2.5 % 97.5 %
## scale 1576.685652 1777.437283
## shape 4.223867 6.318481
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd$cum_em <- cumsum(dd$emergence)
dd$percent <- ave(dd$emergence, FUN = function(dd) 100*(dd)/46)
dd$cum_per <- ave(dd$cum_em, FUN = function(dd) 100*(dd)/46)
dd <- transform(dd)
#start 3 parameter model
library(FAdist)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$percent) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
summary(m2)
#graphical summary
windows(5,5)
with(dd,plot(cum_per~degree_days,cex=3))
with(as.list(coef(m2)),curve(sum(dd$percent)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
I want to generate 2 continuous random variables Q1, Q2 (quantitative traits, each are normal) and 2 binary random variables Z1, Z2 (binary traits) with given pairwise correlations between all possible pairs of them.
Say
(Q1,Q2):0.23
(Q1,Z1):0.55
(Q1,Z2):0.45
(Q2,Z1):0.4
(Q2,Z2):0.5
(Z1,Z2):0.47
Please help me generate such data in R.
This is crude but might get you started in the right direction.
library(copula)
options(digits=3)
probs <- c(0.5,0.5)
corrs <- c(0.23,0.55,0.45,0.4,0.5,0.47) ## lower triangle
Simulate correlated values (first two quantitative, last two transformed to binary)
sim <- function(n,probs,corrs) {
tmp <- normalCopula( corrs, dim=4 , "un")
getSigma(tmp) ## test
x <- rCopula(1000, tmp)
x2 <- x
x2[,3:4] <- qbinom(x[,3:4],size=1,prob=rep(probs,each=nrow(x)))
x2
}
Test SSQ distance between observed and target correlations:
objfun <- function(corrs,targetcorrs,probs,n=1000) {
cc <- try(cor(sim(n,probs,corrs)),silent=TRUE)
if (is(cc,"try-error")) return(NA)
sum((cc[lower.tri(cc)]-targetcorrs)^2)
}
See how bad things are when input corrs=target:
cc0 <- cor(sim(1000,probs=probs,corrs=corrs))
cc0[lower.tri(cc0)]
corrs
objfun(corrs,corrs,probs=probs) ## 0.112
Now try to optimize.
opt1 <- optim(fn=objfun,
par=corrs,
targetcorrs=corrs,probs=c(0.5,0.5))
opt1$value ## 0.0208
Stops after 501 iterations with "max iterations exceeded". This will never work really well because we're trying to use a deterministic hill-climbing algorithm on a stochastic objective function ...
cc1 <- cor(sim(1000,probs=c(0.5,0.5),corrs=opt1$par))
cc1[lower.tri(cc1)]
corrs
Maybe try simulated annealing?
opt2 <- optim(fn=objfun,
par=corrs,
targetcorrs=corrs,probs=c(0.5,0.5),
method="SANN")
It doesn't seem to do much better than the previous value. Two possible problems (left as an exercise for the reader are) (1) we have specified a set of correlations that are not feasible with the marginal distributions we have chosen, or (2) the error in the objective function surface is getting in the way -- to do better we would have to average over more replicates (i.e. increase n).
I am trying to determine whether there is a significant difference between two Gamm distributions. One distribution has (shape, scale)=(shapeRef,scaleRef) while the other has (shape, scale)=(shapeTarget,scaleTarget). I try to do analysis of variance with the following code
n=10000
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
glmm1 <- gam(y~x,family=Gamma(link=log))
anova(glmm1)
The resulting p values keep changing and can be anywhere from <0.1 to >0.9.
Am I going about this the wrong way?
Edit: I use the following code instead
f <- gl(2, n)
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
xy <- c(x, y)
anova(glm(xy ~ f, family = Gamma(link = log)),test="F")
But, every time I run it I get a different p-value.
You will indeed get a different p-value every time you run this, if you pick different realizations every time. Just like your data values are random variables, which you'd expect to vary each time you ran an experiment, so is the p-value. If the null hypothesis is true (which was the case in your initial attempts), then the p-values will be uniformly distributed between 0 and 1.
Function to generate simulated data:
simfun <- function(n=100,shapeRef=2,shapeTarget=2,
scaleRef=1,scaleTarget=2) {
f <- gl(2, n)
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
xy <- c(x, y)
data.frame(xy,f)
}
Function to run anova() and extract the p-value:
sumfun <- function(d) {
aa <- anova(glm(xy ~ f, family = Gamma(link = log),data=d),test="F")
aa["f","Pr(>F)"]
}
Try it out, 500 times:
set.seed(101)
r <- replicate(500,sumfun(simfun()))
The p-values are always very small (the difference in scale parameters is easily distinguishable), but they do vary:
par(las=1,bty="l") ## cosmetic
hist(log10(r),col="gray",breaks=50)
let X1,...,X25 be a random sampe from normal distribution with mean=37 and sd=45.
Let xbar be the sample mean. How is xbar distributed? I have to verify it by central limit theorem.
Also compute P(xbar>43.1)
my attempt
for(i in 1:1000){
x=rnorm(25,mean=37,sd=45)
xbar=mean(x)
z=(xbar-37)/(45/sqrt(25))
}
z
But i couldn't find the distribution of xbar.
Change your for loop and use replicate instead
set.seed(1)
X <- replicate(1000, rnorm(25,mean=37,sd=45))
X_bar <- colMeans(X)
hist(X_bar) # this is how the distribution of X_bar looks like
xbar=c()
for(i in 1:1000){
x=rnorm(25,mean=37,sd=45)
xbar=c(xbar,mean(x)) #save every time the value of xbar
}
hist(xbar) #plot the hist of xbar
#compute the probability to b e bigger thant 43.1
prob=which(xbar>43.1)/length(xbar)
Just to expand in this a little bit.
The Central Limit Theorem states the distribution of the mean is asymptotically N[mu, sd/sqrt(n)]. Where mu and sd are the mean and standard deviation of the underlying distribution, and n is the sample size used in calculating the mean. So, in the example below data is a dataset of size 2500 drawn from N[37,45], arbitrarily segmented into 100 groups of 25. means is a dataset of the means of each group. Note that both the data and the means are (aprox.) normally distributed, but the distribution of the means is much tighter (lower sigma). From the CLT we expect sd(mean) ~ sd(data)/sqrt(25), which it is.
data <- data.frame(sample=rep(1:100,each=25),x = rnorm(2500,mean=37,sd=45))
means <- aggregate(data$x,by=list(data$sample),mean)
#plot histoggrams
par(mfrow=c(1,2))
hist(data$x,main="",sub="Histogram of Underlying Data",xlim=c(-150,200))
hist(means$x,main="",sub="Histogram of Means", xlim=c(-150,200))
mtext("Underlying Data ~ N[37,45]",outer=T,line=-3)
c(sd.data=sd(data$x), sd.means=sd(means$x))
sd.data sd.means
43.548570 7.184518
But the real power of the CLT is that it shows that the distribution of the means is asymptotically normal, regardless of the distribution of the underlying data. This is shown here, where the underlying data is sampled from a uniform distribution. Again, sd(mean) ~ sd(data)/sqrt(25).
data <- data.frame(sample=rep(1:100,each=25),x = runif(2500,min=-150, max=200))
means <- aggregate(data$x,by=list(data$sample),mean)
#plot histoggrams
par(mfrow=c(1,2))
hist(data$x,main="",sub="Histogram of Underlying Data",xlim=c(-150,200))
hist(means$x,main="",sub="Histogram of Means", xlim=c(-150,200))
mtext("Underlying Data ~ U[-150,200]",outer=T,line=-3)
c(sd.data=sd(data$x), sd.means=sd(means$x))
sd.data sd.means
99.7800 18.8176
In the following code I use bootstrapping to calculate the C.I. and the p-value under the null hypothesis that two different fertilizers applied to tomato plants have no effect in plants yields (and the alternative being that the "improved" fertilizer is better). The first random sample (x) comes from plants where a standard fertilizer has been used, while an "improved" one has been used in the plants where the second sample (y) comes from.
x <- c(11.4,25.3,29.9,16.5,21.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
library(boot)
diff <- function(x,i) mean(x[i[6:11]]) - mean(x[i[1:5]])
b <- boot(total, diff, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
What I don't like about the code above is that resampling is done as if there was only one sample of 11 values (separating the first 5 as belonging to sample x leaving the rest to sample y).
Could you show me how this code should be modified in order to draw resamples of size 5 with replacement from the first sample and separate resamples of size 6 from the second sample, so that bootstrap resampling would mimic the “separate samples” design that produced the original data?
EDIT2 :
Hack deleted as it was a wrong solution. Instead one has to use the argument strata of the boot function :
total <- c(x,y)
id <- as.factor(c(rep("x",length(x)),rep("y",length(y))))
b <- boot(total, diff, strata=id, R = 10000)
...
Be aware you're not going to get even close to a correct estimate of your p.value :
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
b <- boot(total, diff, strata=id, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
> p.value
[1] 0.5162
How would you explain a p-value of 0.51 for two samples where all values of the second are higher than the highest value of the first?
The above code is fine to get a -biased- estimate of the confidence interval, but the significance testing about the difference should be done by permutation over the complete dataset.
Following John, I think the appropriate way to use bootstrap to test if the sums of these two different populations are significantly different is as follows:
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
b_x <- boot(x, sum, R = 10000)
b_y <- boot(y, sum, R = 10000)
z<-(b_x$t0-b_y$t0)/sqrt(var(b_x$t[,1])+var(b_y$t[,1]))
pnorm(z)
So we can clearly reject the null that they are the same population. I may have missed a degree of freedom adjustment, I am not sure how bootstrapping works in that regard, but such an adjustment will not change your results drastically.
While the actual soil beds could be considered a stratified variable in some instances this is not one of them. You only have the one manipulation, between the groups of plants. Therefore, your null hypothesis is that they really do come from the exact same population. Treating the items as if they're from a single set of 11 samples is the correct way to bootstrap in this case.
If you have two plots, and in each plot tried the different fertilizers over different seasons in a counterbalanced fashion then the plots would be statified samples and you'd want to treat them as such. But that isn't the case here.