Summary statistics for imputed data from Zelig & Amelia - r

I'm using Amelia to impute the missing values.
While I'm able to use Zelig and Amelia to do some calculations...
How do I use these packages to find the pooled means and standard deviations of the newly imputed data?
library(Amelia)
library(Zelig)
n= 100
x1= rnorm(n,0,1) #random normal distribution
x2= .4*x1+rnorm(n,0,sqrt(1-.4)^2) #x2 is correlated with x1, r=.4
x1= ifelse(rbinom(n,1,.2)==1,NA,x1) #randomly creating missing values
d= data.frame(cbind(x1,x2))
m=5 #set 5 imputed data frames
d.imp=amelia(d,m=m) #imputed data
summary(d.imp) #provides summary of imputation process


I couldn't figure out how to format the code in a comment so here it is.
foo <- function(x, fcn) apply(x, 2, fcn)
lapply(d.imp$imputations, foo, fcn = mean)
lapply(d.imp$imputations, foo, fcn = sd)
d.imp$imputations gives a list of all the imputed data sets. You can work with that list however you are comfortable with to get out the means and sds by column and then pool as you see fit. Same with correlations.
lapply(d.imp$imputations, cor)
Edit: After some discussion in the comments I see that what you are looking for is how to combine results using Rubin's rules for, for example, the mean of imputed data sets generated by Amelia. I think you should clarify in the title and body of your post that what you are looking for is how to combine results over imputations to get appropriate standard errors with Rubin's rules after imputing with package Amelia. This was not clear from the title or original description. "Pooling" can mean different things, particularly w.r.t. variances.
The mi.meld function is looking for a q matrix of estimates from each imputation, an se matrix of the corresponding se estimates, and a logical byrow argument. See ?mi.meld for an example. In your case, you want the sample means and se_hat(sample means) for each of your imputed data sets in the q and se matrices to pass to mi_meld, respectively.
q <- t(sapply(d.imp$imputations, foo, fcn = mean))
se <- t(sapply(d.imp$imputations, foo, fcn = sd)) / sqrt(100)
output <- mi.meld(q = q, se = se, byrow = TRUE)
should get you what you're looking for. For other statistics than the mean, you will need to get an SE either analytically, if available, or by, say, bootstrapping, if not.

Related

How to conduct parametric bootstrapping in R?

I am working with the orings data set in the faraway package in R. I have written the following grouped binomial model:
orings_model <- glm(cbind(damage, 6-damage) ~ temp, family = binomial, data = orings)
summary(orings_model)
I then constructed the Chi-Square test statistic and calculated the p-value:
pchisq(orings_model$null.deviance, orings_model$df.null,lower=FALSE)
First, I would like to generate data under the null distribution for this test statistic using rbinom with the average proportion of damaged o-rings (i.e., the variable "damage"). Second, I would like to recompute the above test statistic with this new data. I am not sure how to do this.
And second, I want to the process above 1000 times, saving the test statistic
each time. I am also not sure how to do this. My inclination is to use a for loop, but I am not sure how to set it up. Any help would be really appreciated!

It is not completely clear what you're looking to do here, but we can at least show some quick principles of how we can achieve this, and then hopefully you can get to your goal.
1) Simulating the null model
It is not entirely clear that you would like to simulate the null model here. It seems more like you're interested in simulating the actual model fit. Note that the null model is the model with form cbind(damage, 6-damage) ~ 1, and the null deviance and df are from this model. Either way, we can simulate data from the model using the simulate function in base R.
sims <- simulate(orings_model, 1000)
If you want to go the manual way estimate the mean vector of your model and use this for the probabilities in your call to rbinom
nsim <- 1000 * nrow(orings)
probs <- predict(orings_model, type = 'response')
sims_man <- matrix(rbinom(nsim, 6, probs),
ncol = 1000)
# Check they are equal:
# rowMeans(sims_man) - probs
In the first version we get a data.frame with 1000 columns each with a n times 2 matrix (damage vs not damage). In the latter we just summon the damage outcome.
2) Perform the bootstrapping
You could do this manually with the data above.
# Data from simulate
statfun <- function(x){
data <- orings_model$data
data$damage <- if(length(dim(x)) > 1)
x[, 1]
else
x
newmod <- update(orings_model, data = data)
pchisq(newmod$null.deviance, newmod$df.null, lower=FALSE)
}
sapply(sims, statfun)
# data from manual method
apply(sims_man, 2, statfun)
or alternatively one could take a bit of time with the boot function, allowing for a standardized way to perform the bootstrap:
library(boot)
# See help("boot")
ran_gen <- function(data, mle){
data$damage <- simulate(orings_model)[[1]][,1]
data
}
boot_metric <- function(data, w){
model <- glm(cbind(damage = damage, not_damage = 6 - damage) ~ temp,
family = binomial, data = data)
pchisq(model$null.deviance,
model$df.null,
lower=FALSE)
}
boots <- boot(orings, boot_metric,
R = 1000,
sim = 'parametric',
ran.gen = ran_gen,
mle = pchisq(orings_model$null.deviance,
orings_model$df.null,
lower=FALSE))
At which point we have the statistic in boots$t and the null statistic in boots$t0, so a simple statistic can be estimated using sum(boots$t > boots$t0) / boots$R (R being the number of replication).

How to use lapply with get.confusion_matrix() in R?

I am performing a PLS-DA analysis in R using the mixOmics package. I have one binary Y variable (presence or absence of wetland) and 21 continuous predictor variables (X) with values ranging from 1 to 100.
I have made the model with the data_training dataset and want to predict new outcomes with the data_validation dataset. These datasets have exactly the same structure.
My code looks like:
library(mixOmics)
model.plsda<-plsda(X,Y, ncomp = 10)
myPredictions <- predict(model.plsda, newdata = data_validation[,-1], dist = "max.dist")
I want to predict the outcome based on 10, 9, 8, ... to 2 principal components. By using the get.confusion_matrix function, I want to estimate the error rate for every number of principal components.
prediction <- myPredictions$class$max.dist[,10] #prediction based on 10 components
confusion.mat = get.confusion_matrix(truth = data_validatie[,1], predicted = prediction)
get.BER(confusion.mat)
I can do this seperately for 10 times, but I want do that a little faster. Therefore I was thinking of making a list with the results of prediction for every number of components...
library(BBmisc)
prediction_test <- myPredictions$class$max.dist
predictions_components <- convertColsToList(prediction_test, name.list = T, name.vector = T, factors.as.char = T)
...and then using lapply with the get.confusion_matrix and get.BER function. But then I don't know how to do that. I have searched on the internet, but I can't find a solution that works. How can I do this?
Many thanks for your help!

Without reproducible there is no way to test this but you need to convert the code you want to run each time into a function. Something like this:
confmat <- function(x) {
prediction <- myPredictions$class$max.dist[,x] #prediction based on 10 components
confusion.mat = get.confusion_matrix(truth = data_validatie[,1], predicted = prediction)
get.BER(confusion.mat)
}
Now lapply:
results <- lapply(10:2, confmat)
That will return a list with the get.BER results for each number of PCs so results[[1]] will be the results for 10 PCs. You will not get values for prediction or confusionmat unless they are included in the results returned by get.BER. If you want all of that, you need to replace the last line to the function with return(list(prediction, confusionmat, get.BER(confusion.mat)). This will produce a list of the lists so that results[[1]][[1]] will be the results of prediction for 10 PCs and results[[1]][[2]] and results[[1]][[3]] will be confusionmat and get.BER(confusion.mat) respectively.

Standardization and inclusion of intercept in sparse lasso GLM

I found some problems while practicing the sparse group lasso method using the cvSGL function forom the SGL package.
My questions are as follows:
Looking at the code for SGL:::center_scale, it doesn't seem to consider the sample size of the data.
SGL:::center_scale
#Output
function (X, standardize) {
means <- apply(X, 2, mean)
X <- t(t(X) - means)
X.transform <- list(X.means = means)
if (standardize == TRUE) {
var <- apply(X, 2, function(x) (sqrt(sum(x^2))))
X <- t(t(X)/var)
X.transform$X.scale <- var
}
else {
X.transform$X.scale <- 1
}
return(list(x = X, X.transform = X.transform))
}
Therefore, the sample standard deviation of the predicted variable is measured somewhat larger.
Is my understanding correct that this may cause the coefficients to be excessively large?
whether the model can be estimated by SGL package with intercept term (or constant term)
The SGL package does not seem to provide a function for estimating by including a intercept term.
In cvFit[["fit"]], I can see only the beta of the predict variables according to the lambda's except for the constant term. The value of cvFit[["fit"]][["intercept"]] is the mean of the y variable.
It can be estimated by adding 1 to first column of predict variable X, but in this case, it is expected to cause problems in centering and standardizing predict variables.
In addition, the SPG package seems to add a penalty to all predict variables. Even if the estimation is performed by adding 1 to the first column of the explanatory variable X as described above, the constant term may be estimated as 0.

Different results when performing PCA in R with princomp() and principal ()

I tried to use princomp() and principal() to do PCA in R with data set USArressts. However, I got two different results for loadings/rotaion and scores.
First, I centered and normalised the original data frame so it is easier to compare the outputs.
library(psych)
trans_func <- function(x){
x <- (x-mean(x))/sd(x)
return(x)
}
A <- USArrests
USArrests <- apply(USArrests, 2, trans_func)
princompPCA <- princomp(USArrests, cor = TRUE)
principalPCA <- principal(USArrests, nfactors=4 , scores=TRUE, rotate = "none",scale=TRUE)
Then I got the results for the loadings and scores using the following commands:
princompPCA$loadings
principalPCA$loadings
Could you please help me to explain why there is a difference? and how can we interprete these results?

At the very end of the help document of ?principal:
"The eigen vectors are rescaled by the sqrt of the eigen values to produce the component loadings more typical in factor analysis."
So principal returns the scaled loadings. In fact, principal produces a factor model estimated by the principal component method.

In 4 years, I would like to provide a more accurate answer to this question. I use iris data as an example.
data = iris[, 1:4]
First, do PCA by the eigen-decomposition
eigen_res = eigen(cov(data))
l = eigen_res$values
q = eigen_res$vectors
Then the eigenvector corresponding to the largest eigenvalue is the factor loadings
q[,1]
We can treat this as a reference or the correct answer. Now we check the results by different r functions.
First, by function 'princomp'
res1 = princomp(data)
res1$loadings[,1]
# compare with
q[,1]
No problem, this function actually just return the same results as 'eigen'. Now move to 'principal'
library(psych)
res2 = principal(data, nfactors=4, rotate="none")
# the loadings of the first PC is
res2$loadings[,1]
# compare it with the results by eigendecomposition
sqrt(l[1])*q[,1] # re-scale the eigen vector by sqrt of eigen value
You may find they are still different. The problem is the 'principal' function does eigendecomposition on the correlation matrix by default. Note: PCA is not invariant with rescaling the variables. If you modify the code as
res2 = principal(data, nfactors=4, rotate="none", cor="cov")
# the loadings of the first PC is
res2$loadings[,1]
# compare it with the results by eigendecomposition
sqrt(l[1])*q[,1] # re-scale the eigen vector by sqrt of eigen value
Now, you will get the same results as 'eigen' and 'princomp'.
Summarize:
If you want to do PCA, you'd better apply 'princomp' function.
PCA is a special case of the Factor model or a simplified version of the factor model. It is just equivalent to eigendecomposition.
We can apply PCA to get an approximation of a factor model. It doesn't care about the specific factors, i.e. epsilons in a factor model. So, if you change the number of factors in your model, you will get the same estimations of the loadings. It is different from the maximum likelihood estimation.
If you are estimating a factor model, you'd better use 'principal' function, since it provides more functions, like rotation, calculating the scores by different methods, and so on.
Rescale the loadings of a PCA model doesn't affect the results too much. Since you still project the data onto the same optimal direction, i.e. maximize the variation in the resulting PC.

ev <- eigen(R) # R is a correlation matrix of DATA
ev$vectors %*% diag(ev$values) %*% t(ev$vectors)
pc <- princomp(scale(DATA, center = F, scale = T),cor=TRUE)
p <-principal(DATA, rotate="none")
#eigen values
ev$values^0.5
pc$sdev
p$values^0.5
#eigen vectors - loadings
ev$vectors
pc$loadings
p$weights %*% diag(p$values^0.5)
pc$loading %*% diag(pc$sdev)
p$loadings
#weights
ee <- diag(0,2)
for (j in 1:2) {
for (i in 1:2) {
ee[i,j] <- ev$vectors[i,j]/p$values[j]^0.5
}
};ee
#scores
s <- as.matrix(scale(DATA, center = T, scale = T)) %*% ev$vectors
scale(s)
p$scores
scale(pc$scores)

Errors running Maximum Likelihood Estimation on a three parameter Weibull cdf

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))

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