I am trying to perform a regression analysis but controlling for genetic relatedness among individuals (genetic relatedness matrix). See the code for an example including identical twins (“1” and siblings “0.5”). N=10
I know I should use a mixed model. However, I am not able to include the genetic relatedness matrix into the model. I have seen that these two packages are often used for this (“Kindship2” and “pedigreemm”.
Here is the code but I am unable to fit the model.
require("pedigreemm")
require("lme4")
library(kinship2)
library(car)
FAMID <- c(1,1,2,2,3,3,4,4,5,5)
UNIQUEID <- 1:10
Pheno <- c(2,4,5,5,8,10,15,20,0,0)
PRS <- c(0.1,0.5, 1,1, 2,2 ,3,3, -0.5, -0.4)
data <- as.data.frame(cbind(FAMID,UNIQUEID,Pheno,PRS))
RELMAT <- matrix(c(1,1,0.04,0.03,0.03,0.03,0.03,0.03,0.03,0.03,
1,1,0.03,0.03,0.03,0.03,0.03,0.03,0.03,0.03,
0.04,0.03,1,1,0.03,0.05,0.03,0.03,0.03,0.03,
0.03,0.03,1,1,0.03,0.03,0.03,0.03,0.03,0.03,
0.03,0.03,0.03,0.03,1,0.5,0.03,0.03,0.03,0.03,
0.03,0.03,0.05,0.03,0.5,1,0.03,0.03,0.03,0.03,
0.03,0.03,0.03,0.03,0.03,0.03,1,0.5,0.03,0.03,
0.03,0.03,0.03,0.03,0.03,0.03,0.5,1,0.03,0.03,
0.03,0.03,0.03,0.03,0.03,0.03,0.03,0.03,1,0.5,
0.03,0.03,0.03,0.03,0.03,0.03,0.03,0.03,0.5,1), nrow = 10, ncol = 10, byrow = T)
This is the model that I want to fit:
m1 = lmer(Pheno ~ PRS + (1 | RELMAT), data = data)
Thank you so much in advance.
Related
covariates <- c("age", "sex", "ph.karno", "ph.ecog", "wt.loss")
univ_formulas <- sapply(covariates,
function(x) as.formula(paste('Surv(time, status)~', x)))
univ_models <- lapply( univ_formulas, function(x){coxph(x, data = lung)})
# Extract data
univ_results <- lapply(univ_models,
function(x){
x <- summary(x)
p.value<-signif(x$wald["pvalue"], digits=2)
wald.test<-signif(x$wald["test"], digits=2)
beta<-signif(x$coef[1], digits=2);#coeficient beta
HR <-signif(x$coef[2], digits=2);#exp(beta)
HR.confint.lower <- signif(x$conf.int[,"lower .95"], 2)
HR.confint.upper <- signif(x$conf.int[,"upper .95"],2)
HR <- paste0(HR, " (",
HR.confint.lower, "-", HR.confint.upper, ")")
res<-c(beta, HR, wald.test, p.value)
names(res)<-c("beta", "HR (95% CI for HR)", "wald.test",
"p.value")
return(res)
#return(exp(cbind(coef(x),confint(x))))
})
res <- t(as.data.frame(univ_results, check.names = FALSE))
as.data.frame(res)
Normally I use this code for univariate cox regression analysis but I have multiple genes >20000 that I want to run as independent variables in a univariate cox regression analysis and I am not sure how I can run this code without typing the individual covariates (gene names) out. All my column names for the genes begin with "ENSG..".
Is there a way to do univariate cox regression on so many genes in an efficient way please? Thanks in advance.
There are several ways to make the list of variable names without typing them out. Probably one of the easiest is to use names() to get all of the variable names in the data, then remove time and status from that list (as well as any other variables you don't want to include). For example, for the veteran dataset:
covariates <- names(survival::veteran)
covariates # Look at which names were detected
#> [1] "trt" "celltype" "time" "status" "karno" "diagtime" "age"
#> [8] "prior"
covariates <- covariates[-which(covariates %in% c("time", "status"))]
covariates # Confirm time and status have been removed
#> [1] "trt" "celltype" "karno" "diagtime" "age" "prior"
Created on 2022-08-30 by the reprex package (v2.0.1)
You could also programmatically create a list of names. For example:
covariates <- sapply(1:10, FUN = function(x) paste0("ENSG.", x))
covariates
#> [1] "ENSG.1" "ENSG.2" "ENSG.3" "ENSG.4" "ENSG.5" "ENSG.6" "ENSG.7"
#> [8] "ENSG.8" "ENSG.9" "ENSG.10"
This approach might be better if the naming is easy to program. If the gene names are irregular it might be more difficult.
As far as efficiency, I don't think much can be done to improve the overall runtime. The bulk of the runtime is doing the actually coxph() calculations. There are other questions/answers on the site about optimizing R code. If you want to pursue optimization I'd suggest looking through those, and then making a new question with a reproducible example if you need more help.
I would like to extract an adjacency matrix of a giant component of a graph using R.
For example, I can create Erdos-Renyi g(n,p)
n = 100
p = 1.5/n
g = erdos.renyi.game(n, p)
coords = layout.fruchterman.reingold(g)
plot(g, layout=coords, vertex.size = 3, vertex.label=NA)
# Get the components of an undirected graph
cl = clusters(g)
# How many components?
cl$no
# How big are these (the first row is size, the second is the number of components of that size)?
table(cl$csize)
cl$membership
# Get the giant component
nodes = which(cl$membership == which.max(cl$csize))
# Color in red the nodes in the giant component and in sky blue the rest
V(g)$color = "SkyBlue2"
V(g)[nodes]$color = "red"
plot(g, layout=coords, vertex.size = 3, vertex.label=NA)
here, I only want to extract the adjacency matrix of those red nodes.
enter image description here
It's easy to get the giant component as a new graph like below and then get the adjacency matrix.
g <- erdos.renyi.game(100, .015, directed = TRUE)
# if you have directed graph, decide if you want
# strongly or weakly connected components
co <- components(g, mode = 'STRONG')
gi <- induced.subgraph(g, which(co$membership == which.max(co$csize)))
# if you want here you can decide if you want values only
# in the upper or lower triangle or both
ad <- get.adjacency(gi)
But you might want to keep the vertex IDs of the original graph. In this case just subset the adjacency matrix:
g <- erdos.renyi.game(100, .015)
co <- components(g)
gi_vids <- which(co$membership == which.max(co$csize))
gi_ad <- get.adjacency(g)[gi_vids, gi_vids]
# you can even add the names of the nodes
# as row and column names.
# generating dummy node names:
V(g)$name <- sapply(
seq(vcount(g)),
function(i){
paste(letters[ceiling(runif(5) * 26)], collapse = '')
}
)
rownames(gi_ad) <- V(g)$name[gi_vids]
colnames(gi_ad) <- V(g)$name[gi_vids]
What are the options for visualising the results of a benchmark experiment of regression learners? For instance, generateCalibrationData doesn't accept a benchmark result object derived from a set of regr. learners. I would like something similar to the calibration plots available for classification.
In response to #LarsKotthoff's comment, I (the OP) have edited my original post to provide greater detail as to what functionality I am seeking.
Edit:
I'm looking for actual vs predicted calibration type plots such as simple scatterplots or something like the plots that exist under Classifier Calibration. If I'm not mistaken, the following would make sense for regression problems (and seems to be what is done for Classifier Calibration):
decide on a number of buckets to discretize the predictions on the x-axis, say 10 equal length bins (obviously you could continue with the breaks and groups interface to generateCalibrationData that currently exists)
for each of those bins 10, calculate the mean "predicted" and plot (say via a dot) on the x-axis (possibly with some measure of variability) and join the dots across the 10 bins
for each of those bins 10, calculate the mean "actual" and plot on the y-axis (possibly with some measure of variability) and join the dots
provide some representation of volume in each bucket (as you've done for Classifier Calibration via "rag/rug" plots)
The basic premise behind my question is what kind of visualisation can be provided to help interpret an rsq, mae etc performance measure. There are many configurations of actual vs predicted that can lead to the same rsq, mae etc.
Once some plot exists, switching aggregation on/off would allow individual resampling results to be examined.
I would hope that the combination:
cal <- generateCalibrationData(bmr)
plotCalibration(cal)
would be available for regression tasks, at present it doesn't seem to be (reproducible example below):
# Practice Data
library("mlr")
library(mlbench)
data(BostonHousing)
dim(BostonHousing)
head(BostonHousing)
# Define Nested Cross-Validation Strategy
cv.inner <- makeResampleDesc("CV", iters = 2)
cv.outer <- makeResampleDesc("CV", iters = 6)
# Define Performance Measures
perf.measures <- list(rsq, mae)
# Create Task
bh.task <- makeRegrTask(id = "bh",
data = BostonHousing,
target = "medv")
# Create Tuned KSVM Learner
ksvm <- makeLearner("regr.ksvm")
ksvm.ps <- makeParamSet(makeDiscreteParam("C", values = 2^(-2:2)),
makeDiscreteParam("sigma", values = 2^(-2:2)))
ksvm.ctrl <- makeTuneControlGrid()
ksvm.lrn = makeTuneWrapper(ksvm,
resampling = cv.inner,
measures = perf.measures,
par.set = ksvm.ps,
control = ksvm.ctrl,
show.info = FALSE)
# Create Tuned Random Forest Learner
rf <- makeLearner("regr.randomForest",
fix.factors.prediction = TRUE)
rf.ps <- makeParamSet(makeDiscreteParam("mtry", values = c(2, 3, 5)))
rf.ctrl <- makeTuneControlGrid()
rf.lrn = makeTuneWrapper(rf,
resampling = cv.inner,
measures = perf.measures,
par.set = rf.ps,
control = rf.ctrl,
show.info = FALSE)
# Run Cross-Validation Experiments
bh.lrns = list(ksvm.lrn, rf.lrn)
bh.bmr <- benchmark(learners = bh.lrns,
tasks = bh.task,
resampling = cv.outer,
measures = perf.measures,
show.info = FALSE)
# Calibration Charts
bh.cal <- generateCalibrationData(bh.bmr)
plotCalibration(bh.cal)
which yields:
> bh.cal <- generateCalibrationData(bh.bmr)
Error in checkPrediction(x, task.type = "classif", predict.type = "prob") :
Prediction must be one of 'classif', but is: 'regr'
> sessionInfo()
R version 3.2.3 (2015-12-10)
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] mlbench_2.1-1 ROCR_1.0-7 gplots_3.0.1 mlr_2.9
[5] stringi_1.1.1 ParamHelpers_1.10 ggplot2_2.1.0 BBmisc_1.10
loaded via a namespace (and not attached):
[1] digest_0.6.9 htmltools_0.3.5 R6_2.2.0 splines_3.2.3
[5] scales_0.4.0 assertthat_0.1 grid_3.2.3 stringr_1.0.0
[9] bitops_1.0-6 checkmate_1.8.2 gdata_2.17.0 survival_2.38-3
[13] munsell_0.4.3 tibble_1.2 randomForest_4.6-12 httpuv_1.3.3
[17] parallelMap_1.3 mime_0.5 DBI_0.5-1 labeling_0.3
[21] chron_2.3-47 shiny_1.0.0 KernSmooth_2.23-15 plyr_1.8.4
[25] data.table_1.9.6 magrittr_1.5 reshape2_1.4.1 kernlab_0.9-25
[29] ggvis_0.4.3 caTools_1.17.1 gtable_0.2.0 colorspace_1.2-6
[33] tools_3.2.3 parallel_3.2.3 dplyr_0.5.0 xtable_1.8-2
[37] gtools_3.5.0 backports_1.0.4 Rcpp_0.12.4
I want to plot a learning curve in my application.
A sample curve image is shown below.
Learning curve is a plot between the following Variance,
X-Axis: Number of samples (Training set size).
Y-axis: Error(RSS/J(theta)/cost function )
It helps in observing whether our model is having the high bias or high variance problem.
Is there any package in R which can help in getting this plot?
You can make such a plot using the excellent Caret package. The section on Customizing the tuning process will be very helpful.
Also, you can check out the well written blog posts on R-Bloggers by Joseph Rickert. They are titled "Why Big Data? Learning Curves" and "Learning from Learning Curves".
UPDATE
I just did a post on this question Plot learning curves with caret package and R. I think my answer will be more useful to you. For convenience sake, I have reproduced the same answer here on plotting a learning curve with R. However, I used the popular caret package to train my model and get the RMSE error for the training and test set.
# set seed for reproducibility
set.seed(7)
# randomize mtcars
mtcars <- mtcars[sample(nrow(mtcars)),]
# split iris data into training and test sets
mtcarsIndex <- createDataPartition(mtcars$mpg, p = .625, list = F)
mtcarsTrain <- mtcars[mtcarsIndex,]
mtcarsTest <- mtcars[-mtcarsIndex,]
# create empty data frame
learnCurve <- data.frame(m = integer(21),
trainRMSE = integer(21),
cvRMSE = integer(21))
# test data response feature
testY <- mtcarsTest$mpg
# Run algorithms using 10-fold cross validation with 3 repeats
trainControl <- trainControl(method="repeatedcv", number=10, repeats=3)
metric <- "RMSE"
# loop over training examples
for (i in 3:21) {
learnCurve$m[i] <- i
# train learning algorithm with size i
fit.lm <- train(mpg~., data=mtcarsTrain[1:i,], method="lm", metric=metric,
preProc=c("center", "scale"), trControl=trainControl)
learnCurve$trainRMSE[i] <- fit.lm$results$RMSE
# use trained parameters to predict on test data
prediction <- predict(fit.lm, newdata = mtcarsTest[,-1])
rmse <- postResample(prediction, testY)
learnCurve$cvRMSE[i] <- rmse[1]
}
pdf("LinearRegressionLearningCurve.pdf", width = 7, height = 7, pointsize=12)
# plot learning curves of training set size vs. error measure
# for training set and test set
plot(log(learnCurve$trainRMSE),type = "o",col = "red", xlab = "Training set size",
ylab = "Error (RMSE)", main = "Linear Model Learning Curve")
lines(log(learnCurve$cvRMSE), type = "o", col = "blue")
legend('topright', c("Train error", "Test error"), lty = c(1,1), lwd = c(2.5, 2.5),
col = c("red", "blue"))
dev.off()
The output plot is as shown below:
I am trying to call rq() of the package quantreg within a function. Herebelow is a simplified explanation of my problem.
If I follow the recommendations found at
http://developer.r-project.org/model-fitting-functions.txt, I have a design matrix after the line
x <- model.matrix(mt, mf, contrasts)
with the first column full of 1's to create an intercept.
Now, when I call rq(), I am obliged to use something like
fit <- rq (y ~ x [,2], tau = 0.5, ...)
My problem happens if there is more than 1 explanatory variable. I don't know how to find an automatic way to write:
x [,2] + x [,3] + x [,4] + ...
Here is the complete simplified code:
ao_qr <- function (formula, data, method = "br",...) {
cl <- match.call ()
## keep only the arguments which should go into the model
## frame
mf <- match.call (expand.dots = FALSE)
m <- match (c ("formula", "data"), names (mf), 0)
mf <- mf[c (1, m)]
mf$drop.unused.levels <- TRUE
mf[[1]] <- as.name ("model.frame")
mf <- eval.parent (mf)
if (method == "model.frame") return (mf)
## allow model.frame to update the terms object before
## saving it
mt <- attr (mf, "terms")
y <- model.response (mf, "numeric")
x <- model.matrix (mt, mf, contrasts)
## proceed with the quantile regression
fit <- rq (y ~ x[,2], tau = 0.5, ...)
print (summary (fit, se = "boot", R = 100))
}
I call the function with:
ao_qr(pain ~ treatment + extra, data = data.subset)
And here is how to get the data:
require (lqmm)
data(labor)
data <- labor
data.subset <- subset (data, time == 90)
data.subset$extra <- rnorm (65)
In this case, with this code, my linear predictor only includes "treatment". If I want "extra", I have to manually add x[,3] in the linear predictor of rq() in the code. This is not automatic and will not work on other datasets with unknown number of variables.
Does anyone know how to tackle this ?
Any help would be greatly appreciated !!!
I found a simple solution:
x[,2:ncol(x)]