This package acts as a wrapper to the penalized R package to add the following functionality to that package:
It also provides a function for simulation of collinear high-dimensional data with survival or binary response.
This package was developed in support of the study by @waldron_optimized_2011.
This paper contains greater detail
on proper application of the methods provided here. Please cite this
paper when using the pensim package in your research, as well as the
penalized package (@goeman_l1_2010).
pensim provides example data from a microarray experiment investigating survival of cancer patients with lung adenocarcinomas (@beer_gene-expression_2002). Load this data and do an initial pre-filter of genes with low IQR:
library(pensim) data(beer.exprs) data(beer.survival) ##select just 100 genes to speed computation, just for the sake of example: set.seed(1) beer.exprs.sample <- beer.exprs[sample(1:nrow(beer.exprs), 100),] # gene.quant <- apply(beer.exprs.sample, 1, quantile, probs = 0.75) dat.filt <- beer.exprs.sample[gene.quant > log2(100),] gene.iqr <- apply(dat.filt, 1, IQR) dat.filt <- as.matrix(dat.filt[gene.iqr > 0.5,]) dat.filt <- t(dat.filt) dat.filt <- data.frame(dat.filt) # library(survival) surv.obj <- Surv(beer.survival$os, beer.survival$status)
Note that the expression data are in "wide" format, with one column per predictor (gene). It is recommended to put covariate data in a dataframe object, rather than a matrix.
Unbiased estimation of prediction accuracy involves two levels of cross-validation: an outer level for estimating prediction accuracy, and an inner level for model tuning. This process is simplified by the opt.nested.crossval function.
It is recommended first to establish the arguments for the penalized
regression by testing on the penalized package functions optL1
(for LASSO), optL2
(for Ridge) or cvl
(for Elastic Net). Here we use
LASSO. Setting maxlambda1=5
is not a generally recommended procedure,
but is useful in this toy example to avoid converging on the null
model.
library(penalized) testfit <- optL1( response = surv.obj, penalized = dat.filt, fold = 5, maxlambda1 = 5, positive = FALSE, standardize = TRUE, trace = FALSE )
Now pass these arguments to opt.nested.crossval()
for cross-validated
calculation and assessment of risk scores, with the additional
arguments:
outerfold
and nprocessors
: number of folds for the outer level of cross-validation, and the number of processors to use for the outer level of cross-validation (see ?opt.nested.crossval
)opt1D
for LASSO or Ridge, opt2D
for Elastic Net) - see ?opt.splitval
.nsim
defines the number of times to repeat tuning (see ?opt1D
. opt2D
has different required arguments.)set.seed(1) preds <- opt.nested.crossval( outerfold = 5, nprocessors = 1, #opt.nested.crossval arguments optFUN = "opt1D", scaling = FALSE, #opt.splitval arguments setpen = "L1", nsim = 1, #opt1D arguments response = surv.obj, #rest are penalized::optl1 arguments penalized = dat.filt, fold = 5, positive = FALSE, standardize = TRUE, trace = FALSE )
Ideally nsim would be 50, and outerfold and fold would be 10, but the
values below speed computation 200x compared to these recommended
values. Note that here we are using the standardize=TRUE
argument of
optL1
rather than the scaling=TRUE
argument of opt.splitval. These
two approaches to scaling are roughly equivalent, but the scaling
approaches are not the same (scaling=TRUE
does z-score,
standardize=TRUE
scales to unit central L2 norm), and results will not
be identical. Also, using standardize=TRUE
scales variables but
provides coeffients for the original scale, whereas using scaling=TRUE
scales variables in the training set then applies the same scales to
the test set.
Cox fit on the continuous risk predictions:
coxfit.continuous <- coxph(surv.obj~preds) summary(coxfit.continuous)
Dichotomize the cross-validated risk predictions at the median, for visualization:
preds.dichot <- preds > median(preds)
Plot the ROC curve:
nobs <- length(preds) cutoff <- 12 if (requireNamespace("survivalROC", quietly = TRUE)) { preds.roc <- survivalROC::survivalROC( Stime = beer.survival$os, status = beer.survival$status, marker = preds, predict.time = cutoff, span = 0.01 * nobs ^ (-0.20) ) plot( preds.roc$FP, preds.roc$TP, type = "l", xlim = c(0, 1), ylim = c(0, 1), xlab = paste("FP", "\n", "AUC = ", round(preds.roc$AUC, 3)), lty = 2, ylab = "TP", main = "LASSO predictions\n ROC curve at 12 months" ) abline(0, 1) }
Finally, we can get coefficients for the model fit on all the data,
for future use. Note that nsim should ideally be greater than 1, to
train the model using multiple foldings for cross-validation. The
output of opt1D
or opt2D
will be a matrix with one row per simulation.
The default behavior in opt.nested.crossval()
is to take the
simulation with highest cross-validated partial log likelihood (CVL),
which is the recommended way to select a model from the multiple
simulations.
beer.coefs <- opt1D( setpen = "L1", nsim = 1, response = surv.obj, penalized = dat.filt, fold = 5, maxlambda1 = 5, positive = FALSE, standardize = TRUE, trace = FALSE )
We can also include unpenalized covariates, if desired. Note that
when keeping only one variable for a penalized or unpenalized
covariate, indexing a dataframe like [1]
instead of doing [, 1]
preserves the variable name. With [, 1]
the variable name gets
converted to "".
beer.coefs.unpen <- opt1D( setpen = "L1", nsim = 1, response = surv.obj, penalized = dat.filt[-1], # This is equivalent to dat.filt[,-1] unpenalized = dat.filt[1], fold = 5, maxlambda1 = 5, positive = FALSE, standardize = TRUE, trace = FALSE )
Note the non-zero first coefficient this time, due to it being unpenalized:
beer.coefs[1, 1:5] #example output with no unpenalized covariates beer.coefs.unpen[1, 1:5] #example output with first covariate unpenalized
The pensim also provides a convenient means to simulation high-dimensional expression data with (potentially censored) survival outcome or binary outcome which is dependent on specified covariates.
First, generate the data. Here we simulate 20 variables. The first
15 (group "a") are uncorrelated, and have no association with
outcome. The final five (group "b") have covariance of 0.8 to each
other variable in that group. The response variable is associated
with the first variable group "b" (firstonly=TRUE
) with a
coefficient of 2.
Binary outcomes for $n_s= 50$ samples are simulated as a Bernoulli distribution with probability for patient s:
\begin{equation} p_{s} =\frac{1}{1 + exp(-\beta X_{s})} \end{equation}
with $\beta_{s,16} = 0.5$ and all other $\beta_{s,i}$ equal to zero.
The code for this simulation is as follows:
set.seed(9) x <- create.data( nvars = c(15, 5), cors = c(0, 0.8), associations = c(0, 2), firstonly = c(TRUE, TRUE), nsamples = 50, response = "binary", logisticintercept = 0.5 )
Take a look at the simulated data:
summary(x) x$summary
A simple logistic model fails at variable selection in this case:
simplemodel <- glm(outcome ~ ., data = x$data, family = binomial) summary(simplemodel)
But LASSO does a better job, selecting several of the collinear variables in the "b" group of variables which are associated with outcome:
lassofit <- opt1D( nsim = 3, nprocessors = 1, setpen = "L1", penalized = x$data[1:20], response = x$data[, "outcome"], trace = FALSE, fold = 10 ) print(lassofit)
And visualize the data as a heatmap:
dat <- t(as.matrix(x$data[,-match("outcome", colnames(x$data))])) heatmap(dat, ColSideColors = ifelse(x$data$outcome == 0, "black", "white"))
We simulate these data in the same way, but with
response="timetoevent"
. Here censoring is uniform random between
times 2 and 10, generating approximately 34\% censoring:
set.seed(1) x <- create.data( nvars = c(15, 5), cors = c(0, 0.8), associations = c(0, 0.5), firstonly = c(TRUE, TRUE), nsamples = 50, censoring = c(2, 10), response = "timetoevent" )
How many events are censored?
sum(x$data$cens == 0) / nrow(x$data)
Kaplan-Meier plot of this simulated cohort:
library(survival) surv.obj <- Surv(x$data$time, x$data$cens) plot(survfit(surv.obj ~ 1), ylab = "Survival probability", xlab = "time")
sessionInfo()
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