# R0 and the exponential growth of a pandemic, an update

A few days ago, I wrote a blog post – R0 and the exponential growth of a pandemic – where I was trying to generate some visualization of some exponential growth, in the context of a pandemic. After giving some thoughts, the previous graph might not be the best one to see an exponential based contagion.

Having graphs evolving, from the left to the right, gives us the (false) idea of some temporal evolution. Which is no necessarily correct. It simply means that contaminated people will contaminate other people, and we look at the number of iterations here. So maybe some concentric dots would look better.

And from a technical perspective, what I did was fun, but probably too complicated. In my previous post, I wanted to pack optimally $k$ identical disks intro a unit circle. On http://hydra.nat.uni-magdeburg.de/packing, it was possible to get the “best known packings of equal circles in a circle”, with the coordinates. But as we will see, we can use something much more simple here.

My idea is now to create some picture like one below, with concentric colored dot. In the center, we have the first people that were contaminated, and then, we can see the transmission, somehow

From a technical perspective, here, I use a different strategy. I decided to draw random points, uniformly. The problem with randomness is the natural high discrepancy, with monte carlo methods: it is very likely that some disks will overlap. It is not a major issue, but it might distort the message. So I decided to use some low-discrepancy sequences, such as Halton‘s sequence.

library(randtoolbox) S = halton(n=5000, dim = 2)*2-1

Here, I have disk coordinates in $[-1,+1]^2$. Then, to get disks in a circle, I simply compute the distance to the origin $(0,0)$,

D0 = S[,1]^2+S[,2]^2

and take the ranks. If I want to visualize $k=200$ people, I consider the 200 smaller ranks. To get concentric circles, each part having $k_i$ individuals, I use as thresholds $R_0^{\bar k_{i-1}},R_0^{\bar k_{i}},R_0^{\bar k_{i+1}}$, etc, where $\bar k_i=\bar k_{i-1}+k_i$,

R0 = rank(D0,ties.method = "random") C0 = as.numeric(cut(R0,c(0,cumsum(k)+.5)),100000)

where

R0=1.8 k=round(R0^(seq(1,9,by=2)))

Then we can plot the dots, with appropriate colors,

points(S,pch=19,col=colrpal[C0],cex=.75)

And of course, we can try that with different values, for $R_0$

R0=2.2 k=round(R0^(seq(1,9,by=2))) kmax=max(k) S = halton(n=5000, dim = 2)*2-1 plot(S,col="light yellow",axes=FALSE,xlab="",ylab="",xlim=c(-1.3,1),ylim=c(-1,1),cex=.75,pch=19) D0 = S[,1]^2+S[,2]^2 R0 = rank(D0,ties.method = "random") C0 = as.numeric(cut(R0,c(0,cumsum(k)+.5)),100000) points(S,pch=19,col=colrpal[C0],cex=.75)

# Inter-relationships in a matrix

Last week, I wanted to displaying inter-relationships between data in a matrix. My friend Fleur, from AXA, mentioned an interesting possible application, in car accidents. In car against car accidents, it might be interesting to see which parts of the cars were involved. On https://www.data.gouv.fr/fr/, we can find such a dataset, with a lot of information of car accident involving bodily injuries (in France, a police report is necessary, and all of them are reported in a big dataset… actually several dataset, with information of people involved, cars, locations, etc). For 2014 claims, the dataset is

> base = read.csv("https://www.data.gouv.fr/s/resources/base-de-donnees-accidents-corporels-de-la-circulation-sur-6-annees/20150806-153355/vehicules_2014.csv")


Let us keep only claims involving two vehicules,

> T=table(base$Num_Acc) > idx=names(T)[which(T==2)] For 2014, we have 32,222 claims. > length(idx) [1] 32222 In this dataset, we have information about where cars were hit, plus ‘9’ for multiple hot (in rollover accidents) and ‘0’ should be missing information. > nom=c("NA","Front","Front R",'Front L',"Back","Back R","Back L","Side R","Side L","Multiple")  Now, we simply have to go through our dataset, and get the matrix. My first idea was to get a symmetric one, > B=base[base$Num_Acc %in% idx,]
> B=B[order(B$Num_Acc),] > M=matrix(0,10,10) > for(i in seq(1,nrow(B),by=2)){ + a=B$choc[i]+1
+   b=B$choc[i+1]+1 + M[a,b]=M[a,b]+1 + M[b,a]=M[b,a]+1 + } > rownames(M)=nom > colnames(M)=nom The problem, when we ask for a symmetric chord diagram, is that we cannot have Front – Front claims (since values on the diagonal are removed) > library(circlize) > chordDiagramFromMatrix(M,symmetric=TRUE) So let’s pretend that there could be some possible distinction in the dataset, between the first and the second row. Like the first one is the ‘responsible’ driver. Or like, for insurer, the first one is your insured. Just to avoid this symmetry problem > M=matrix(0,10,10) > for(i in seq(1,nrow(B),by=2)){ + a=B$choc[i]+1
+   b=B\$choc[i+1]+1
+ M[a,b]=M[a,b]+1
+ }
> rownames(M)=paste("A",nom,sep=" ")
> colnames(M)=paste("B",nom,sep=" ")

If we visualize the chord diagram, this time it is more complex to analyze,

> chordDiagram(M)

Below we have the first row (say our driver, letter A) and on top, the second row (say the other driver, letter B),

In bodily injury claims, we observe a large proportion of Front – Front claims, as well as Front – Back. And as expected Back-Back are not that common….