# How to compare conStruct model runs

## Model comparison

This document describes how to do model comparison for conStruct analyses. It assumes that you are already familiar with the companion vignette for running conStruct.

Caveat user! Although it may sometimes be necessary to simplify the presentation of the results of several analyses by only showing the output from a single “best” run, it is important to remember several things:

1. First, choice of best K is always relative to the data at hand, and, as the amount of data increases, statistical support for larger K will likely increase. With infinite data, the “best” value of K would probably be the number of samples in the dataset.

2. Although we think that conStruct is less likely to falsely ascribe continuous patterns of genetic variation to discrete population clusters than other existing methods, that does not mean that the discrete groups identified by conStruct are biologically real. See “A tutorial on how not to over-interpret STRUCTURE and ADMIXTURE bar plots” (Lawson, van Dorp, and Falush 2018) for a more in-depth discussion of these issues.

3. Finally, as with all other statistical inference, output should be interpreted with care and a large grain of salt. We strongly recommend that users check whether individual runs seem to have performed well and whether results are consistent across independent runs. We also recommend that users compare output across runs with different values of K to see which samples split out into their own layers in the different analyses.

So, you’ve run two or more conStruct analyses and you want to compare them to see which might be the best model to describe the variation in your data. There are two methods in the conStruct package for doing model comparison:

1. Cross-validation

2. Calculating layer contributions

Below, I describe both options and give examples for how to use their associated functions in the conStruct package and visualize the output they generate.

Note that if you are interested in visually comparing two independent conStruct runs, you can use the function compare.two.runs, the documentation for which can be found with the
command help(compare.two.runs). This function is further described in the companion vignette for visualizing results.

## Cross-validation

Cross-validation is a tool for testing how the results of an analysis will generalize to an independent dataset.

### How it works

In general, the more parameters included in the model, the better the fit to the data. To determine an appropriate level of parameterization for a given dataset, we can use cross-validation. In conStruct, this works by fitting a model to a “training” subset of the data, then testing the fit to the remaining “testing” subset. If the parameter values estimated from the training data parameterize a model that describes the testing data well, the predictive accuracy of the model is good. If the model is overparameterized, it will fit the training data very well, but may not fit the testing better any better than (or even as well as) a less parameter-rich model. By fitting a given model to many training partitions and testing its fit to the accompanying testing partitions, we can get a mean predictive accuracy for each model. We can then compare predictive accuracies across models to determine which model strikes has the best goodness-of-fit without overfitting.

### How to run a cross-validation analysis

To run a cross-validation analysis in conStruct, you can use the x.validation function.

# load the library
library(conStruct)

data(conStruct.data)

# to run a cross-validation analysis
# you have to specify:
#       the numbers of layers you want to compare (K)
#       the allele frequency data (freqs)
#       the geographic distance matrix (geoDist)
#       the sampling coordinates (coords)

my.xvals <- x.validation(train.prop = 0.9,
n.reps = 8,
K = 1:3,
freqs = conStruct.data$allele.frequencies, data.partitions = NULL, geoDist = conStruct.data$geoDist,
nsp.results <- Reduce("cbind",lapply(my.xvals,function(x){unlist(x$nsp)}),init=NULL) The results look like this: Spatial cross-validation results rep1 rep2 rep3 rep4 rep5 rep6 rep7 rep8 K=1 -1.201 -4.579 0.000 -7.315 0.000 0.000 -3.650 0.000 K=2 0.000 -5.730 -5.346 -8.853 -6.125 -11.155 0.000 -4.799 K=3 -1.819 0.000 -1.114 0.000 -3.602 -5.506 -2.909 -9.890 A quick and dirty plot of the output is given below:  # first, get the 95% confidence intervals for the spatial and nonspatial # models over values of K (mean +/- 1.96 the standard error) sp.CIs <- apply(sp.results,1,function(x){mean(x) + c(-1.96,1.96) * sd(x)/length(x)}) nsp.CIs <- apply(nsp.results,1,function(x){mean(x) + c(-1.96,1.96) * sd(x)/length(x)}) # then, plot cross-validation results for K=1:3 with 8 replicates par(mfrow=c(1,2)) plot(rowMeans(sp.results), pch=19,col="blue", ylab="predictive accuracy",xlab="values of K", ylim=range(sp.results,nsp.results), main="cross-validation results") points(rowMeans(nsp.results),col="green",pch=19) # finally, visualize results for the spatial model # separately with its confidence interval bars # # note that you could do the same with the spatial model, # but the confidence intervals don't really show up # because the differences between predictive accuracies # across values of K are so large. plot(rowMeans(sp.results), pch=19,col="blue", ylab="predictive accuracy",xlab="values of K", ylim=range(sp.CIs), main="spatial cross-validation results") segments(x0 = 1:nrow(sp.results), y0 = sp.CIs[1,], x1 = 1:nrow(sp.results), y1 = sp.CIs[2,], col = "blue",lwd=2) ### Interpreting results The model with the highest mean predictive accuracy is the “best” model, but, as noted above, we caution against overinterpretation of these cross-validation results. If a significance test for the “best” number of layers is required, you can use a t-test comparing cross-validation scores across values of K, paired by replicate. E.g., t.test(sp.results[2,],sp.results[1,],paired=TRUE,alternative="greater"). I would interpret the results above as strong evidence that the spatial model is preferred over the nonspatial model over all tested values of K (indicating that isolation by distance is probably a feature of the data). The cross-validation analyses also strongly support the conclusion that a single spatial layer (K = 1) is sufficient to describe the variation in the data. A final caveat of this section is that, with sufficient data, it is possible to get strong statistical support for layers that contribute little to overall patterns of covariance. Therefore, it’s good to interpret cross-validation results alongside calculated layer contributions (discussed further in Layer Contributions. ### Parallelization Because each cross-validation replicate consists of several analyses (one for each specified value of K), and because several cross-validation replicates are required for model comparison, a single call to x.validation can take a long time. To reduce computational burden, we have introduced an option for users to parallelize their analyses across replicates. The simplest way to parallelize is to use the parallel and n.nodes arguments of in the x.validation function, which we illustrate using the same x.validation given above in How it works:  # load the example dataset data(conStruct.data) # to run a cross-validation analysis # you have to specify: # the numbers of layers you want to compare (K) # the allele frequency data (freqs) # the geographic distance matrix (geoDist) # the sampling coordinates (coords) # in addition, here we run our analysis parallelized # across all replicates using 4 nodes my.xvals <- x.validation(train.prop = 0.9, n.reps = 8, K = 1:3, freqs = conStruct.data$allele.frequencies,
data.partitions = NULL,
geoDist = conStruct.data$geoDist, coords = conStruct.data$coords,
prefix = "example",
n.iter = 1e3,
make.figs = TRUE,
save.files = FALSE,
parallel = TRUE,
n.nodes = 4)

The example above should run ~4 times as fast as cross-validation with the same number of replicates not run in parallel. At the end of the cross-validation analysis, the parallel workers generated at the beginning of the run will be terminated.

To facilitate greater flexibility in parallelization, users can also specify their own parallelization scheme before running a cross-validation analysis, in which case they should simply set parallel=TRUE and make sure that n.nodes is equal to the number of nodes they’ve set up. If you’ve set up your own parallelization beforehand (as in the example that follows), x.validation will use that set-up rather than initializing one itself. E.g.,


library(parallel)
library(foreach)
library(doParallel)

cl <- makeCluster(4,type="FORK")
registerDoParallel(cl)

my.xvals <- x.validation(train.prop = 0.9,
n.reps = 8,
K = 1:3,
freqs = conStruct.data$allele.frequencies, data.partitions = NULL, geoDist = conStruct.data$geoDist,
coords = conStruct.data$coords, prefix = "example", n.iter = 1e3, make.figs = TRUE, save.files = FALSE, parallel = TRUE, n.nodes = 4) stopCluster(cl) Note that if you have prespecified a parallelization scheme, you are responsible for ending the parallelization yourself, as shown above with the stopCluster() call. Linux and Mac users may wish use makeCluster(N,type="FORK"), as it does better with memory usage. Windows users should user the default PSOCK cluster (e.g., makeCluster(N,type="PSOCK")). ## Layer contributions Layer contributions offer a second metric users can employ to compare models with different numbers of layers. ### How it works In a conStruct run, users are estimating a parametric covariance matrix to fit their sample allelic covariance. Each layer in the model contributes to that parametric covariance, and those contributions can be calculated and compared. If there is a layer that no samples draw appreciable admixture from, it will contribute almost nothing to overall covariance, and is therefore of little biological importance in the model. By comparing layer contributions across different conStruct analyses run with different values of K, users can identify the point at which layers included in the analysis contribute little to overall covariance, and pick a “best” value of K below that point. ### How to calculate layer contributions Layer contributions are calculated from the output of a standard conStruct analysis using the function calculate.layer.contribution.  # Loop through output files generated by conStruct # runs with K=1 through 5 and calculate the # layer contributions for each layer in each run layer.contributions <- matrix(NA,nrow=5,ncol=5) # load the conStruct.results.Robj and data.block.Robj # files saved at the end of a conStruct run load("K1_sp_conStruct.results.Robj") load("K1_sp_data.block.Robj") # calculate layer contributions layer.contributions[,1] <- c(calculate.layer.contribution(conStruct.results[[1]],data.block),rep(0,4)) tmp <- conStruct.results[[1]]$MAP$admix.proportions for(i in 2:5){ # load the conStruct.results.Robj and data.block.Robj # files saved at the end of a conStruct run load(sprintf("K%s_sp_conStruct.results.Robj",i)) load(sprintf("K%s_sp_data.block.Robj",i)) # match layers up across runs to keep plotting colors consistent # for the same layers in different runs tmp.order <- match.layers.x.runs(tmp,conStruct.results[[1]]$MAP$admix.proportions) # calculate layer contributions layer.contributions[,i] <- c(calculate.layer.contribution(conStruct.results=conStruct.results[[1]], data.block=data.block, layer.order=tmp.order), rep(0,5-i)) tmp <- conStruct.results[[1]]$MAP$admix.proportions[,tmp.order] } Note that, because layers can label switch across runs, the example above uses the match.layers.x.runs function to determine which layers correspond to each other across analyses run with different values of K. ### Visualizing results The table of layer contributions looks like this: Contributions for each layer for runs done with K=1 through 5 K=1 K=2 K=3 K=4 K=5 Layer_1 1 0.68 0.682 0.678 0.684 Layer_2 0 0.32 0.318 0.322 0.315 Layer_3 0 0.00 0.000 0.000 0.000 Layer_4 0 0.00 0.000 0.000 0.000 Layer_5 0 0.00 0.000 0.000 0.000 Layer contributions can be easily plotted across values of K using a stacked barplot: barplot(layer.contributions, col=c("blue", "red", "goldenrod1", "forestgreen", "darkorchid1"), xlab="", ylab="layer contributions", names.arg=paste0("K=",1:5)) In this case, the contributions of layers beyond K = 2 is so small that they don’t even show up on the barplot. ### Interpreting results If a layer in a given model contributes very little to overall covariance, it is unlikely to have much biological significance. If you run conStruct analyses across values of K, and see that, after a certain value of K, no additional clusters contribute much to overall covariance, that may be a good indication that that value of K (or at least, no larger value of K) is best for describing the variation in your data. For example, in the layer contributions plotted above in Visualizing results, additional layers after K = 2 have negligible layer contributions, so we might reasonably conclude that the best value of K for describing our data is no greater than 2. Users can also set some threshold (e.g., 0.01) below which they count a layer’s contribution as negligible, and, by setting this threshold a priori, can use layer contributions as a metric for model selection. ## Cross-validation vs. Layer contribution With sufficient data, a cross-validation analysis may indicate strong support for layers that each contribute very little to overall covariance. In such a case, the input from cross-validation and layer contributions are at odds, with the former arguing for the inclusion of more layers, and the latter arguing against. What to do with that situation? Well, the specifics will vary from dataset to dataset, but we encourage users to distinguish between statistical and biological significance, and not get too caught up in the first at the expense of the second. ## Advanced options Below, we include information on advanced topics that will not be of use for the average user. ### Specifying data partitions In many cases, there will be no genome assembly available for the focal species in a conStruct analysis, and the genotyped loci will have no known genomic location. When genomic positions are known, advanced users may wish to specify their own data partitions to maximize the efficacy of the cross-validation procedure. This is because the cross-validation results are most trustworthy when the testing data partition is independent from but still representative of the training data. Because coalescent histories tend to be shared by adjacent loci on the genome, if neighboring loci are split (one in the training dataset, the other in the testing dataset), the training/testing partitions might not be truly independent. In this case, the model parameterized by the training dataset will be fitting coalescent “noise” that’s also present in the testing dataset, the most likely result of which is overfitting. Another concern is that different regions of the genome have different properties (e.g., centromeres vs. non-centromeric DNA), so to keep the training and testing partitions representative of each other, it’s best to try to match by genomic properties. Our recommendation: If genomic position and LD information is available, we recommend divvying the genome up into blocks of length equal to twice the scale of LD, then randomly assigning 90% of those blocks to a training partition, and the remaining 10% to the testing partition for each replicate. To facilitate this type of custom data partitioning, users can specify their own data partitions for a x.validation analysis using the data.partitions argument. There is no function in the package for generating custom a custom data partitions object, as the details of the data format and specifics of the desired partitioning scheme will vary from user to user and genome to genome. Instead, we describe the structure of the data.partitions object in detail below so that users can create it for themselves. The data.partitions object must be a list of length n.reps as specified in the x.validation function (one partitioning scheme per cross-validation replicate). Each of the n.reps elements of the list must contain two elements, one named training and one named testing, which contain the training and testing data partitions, respectively. Each training and testing element of the list must contain three named elements: data, n.loci, and varMeanFreqs. The data element contains the allelic covariance matrix for that partition of the data; n.loci gives the number of loci in that partition; and varMeanFreqs gives the variance in mean allele frequencies across loci (averaged over choice of counted allele). Peeking under the hood of how conStruct creates this data.partitions object when none is specified, the relevant functions are: • conStruct:::make.data.partitions • conStruct:::xval.process.data • conStruct:::calc.covariance • conStruct:::get.var.mean.freqs Users attempting to specify their own data.partitions object are encouraged to use these functions as guides for what operations are being carried out to generate the data partitions list for a cross-validation analysis. The structure of an example data.partitions object with 3 partitioning schemes (for 3 cross-validation replicates) is shown below: # In this dataset, there are 36 samples and 1e4 loci total, # and the data partitions are generated # with a 90% training 10% testing split str(data.partitions,max.level=3,give.attr=FALSE,vec.len=3) #> List of 3 #>$ rep1:List of 2
#>   ..$training:List of 3 #> .. ..$ data        : num [1:16, 1:16] 0.25 0.224 0.215 0.217 ...
#>   .. ..$n.loci : int 7803 #> .. ..$ varMeanFreqs: num 0.13
#>   ..$testing :List of 3 #> .. ..$ data        : num [1:16, 1:16] 0.25 0.226 0.217 0.218 ...
#>   .. ..$n.loci : int 867 #> .. ..$ varMeanFreqs: num 0.134
#>  $rep2:List of 2 #> ..$ training:List of 3
#>   .. ..$data : num [1:16, 1:16] 0.25 0.224 0.215 0.217 ... #> .. ..$ n.loci      : int 7803
#>   .. ..$varMeanFreqs: num 0.131 #> ..$ testing :List of 3
#>   .. ..$data : num [1:16, 1:16] 0.25 0.223 0.216 0.219 ... #> .. ..$ n.loci      : int 867
#>   .. ..$varMeanFreqs: num 0.126 #>$ rep3:List of 2
#>   ..$training:List of 3 #> .. ..$ data        : num [1:16, 1:16] 0.25 0.224 0.215 0.218 ...
#>   .. ..$n.loci : int 7803 #> .. ..$ varMeanFreqs: num 0.13
#>   ..$testing :List of 3 #> .. ..$ data        : num [1:16, 1:16] 0.25 0.222 0.214 0.214 ...
#>   .. ..$n.loci : int 867 #> .. ..$ varMeanFreqs: num 0.133