# Modeling Pandemics (3)

In Statistical Inference in a Stochastic Epidemic SEIR Model with Control Intervention, a more complex model than the one we’ve seen yesterday was considered (and is called the SEIR model). Consider a population of size $N$, and assume that $S$ is the number of susceptible, $E$ the number of exposed, $I$ the number of infectious, and $R$ for the number recovered (or immune) individuals, \displaystyle{\begin{aligned}{\frac {dS}{dt}}&=-\beta {\frac {I}{N}}S\\[8pt]{\frac {dE}{dt}}&=\beta {\frac {I}{N}}S-aE\\[8pt]{\frac {dI}{dt}}&=aE-b I\\[8pt]{\frac {dR}{dt}}&=b I\end{aligned}}Between $S$ and $I$, the transition rate is $\beta I$, where $\beta$ is the average number of contacts per person per time, multiplied by the probability of disease transmission in a contact between a susceptible and an infectious subject. Between $I$ and $R$, the transition rate is $b$ (simply the rate of recovered or dead, that is, number of recovered or dead during a period of time divided by the total number of infected on that same period of time). And finally, the incubation period is a random variable with exponential distribution with parameter $a$, so that the average incubation period is $a^{-1}$.

Probably more interesting, Understanding the dynamics of ebola epidemics suggested a more complex model, with susceptible people $S$, exposed $E$, Infectious, but either in community $I$, or in hospitals $H$, some people who died $F$ and finally those who either recover or are buried and therefore are no longer susceptible $R$.

Thus, the following dynamic model is considered\displaystyle{\begin{aligned}{\frac {dS}{dt}}&=-(\beta_II+\beta_HH+\beta_FF)\frac{S}{N}\\[8pt]\frac {dE}{dt}&=(\beta_II+\beta_HH+\beta_FF)\frac{S}{N}-\alpha E\\[8pt]\frac {dI}{dt}&=\alpha E+\theta\gamma_H I-(1-\theta)(1-\delta)\gamma_RI-(1-\theta)\delta\gamma_FI\\[8pt]\frac {dH}{dt}&=\theta\gamma_HI-\delta\lambda_FH-(1-\delta)\lambda_RH\\[8pt]\frac {dF}{dt}&=(1-\theta)(1-\delta)\gamma_RI+\delta\lambda_FH-\nu F\\[8pt]\frac {dR}{dt}&=(1-\theta)(1-\delta)\gamma_RI+(1-\delta)\lambda_FH+\nu F\end{aligned}}In that model, parameters are $\alpha^{-1}$ is the (average) incubation period (7 days), $\gamma_H^{-1}$ the onset to hospitalization (5 days), $\gamma_F^{-1}$ the onset to death (9 days), $\gamma_R^{-1}$ the onset to “recovery” (10 days), $\lambda_F^{-1}$ the hospitalisation to death (4 days) while $\lambda_R^{-1}$ is the hospitalisation to recovery (5 days), $\eta^{-1}$ is the death to burial (2 days). Here, numbers are from Understanding the dynamics of ebola epidemics (in the context of ebola). The other parameters are $\beta_I$ the transmission rate in community (0.588), $\beta_H$ the transmission rate in hospital (0.794) and $\beta_F$ the transmission rate at funeral (7.653). Thus

epsilon = 0.001 Z = c(S = 1-epsilon, E = epsilon, I=0,H=0,F=0,R=0) p=c(alpha=1/7*7, theta=0.81, delta=0.81, betai=0.588, betah=0.794, blambdaf=7.653,N=1, gammah=1/5*7, gammaf=1/9.6*7, gammar=1/10*7, lambdaf=1/4.6*7, lambdar=1/5*7, nu=1/2*7)

If $\boldsymbol{Z}=(S,E,I,H,F,R)$, if we write $$\frac{\partial \boldsymbol{Z}}{\partial t} = SEIHFR(\boldsymbol{Z})$$where $SEIHFR$ is

SEIHFR = function(t,Z,p){ S=Z[1]; E=Z[2]; I=Z[3]; H=Z[4]; F=Z[5]; R=Z[6] alpha=p["alpha"]; theta=p["theta"]; delta=p["delta"] betai=p["betai"]; betah=p["betah"]; gammah=p["gammah"] gammaf=p["gammaf"]; gammar=p["gammar"]; lambdaf=p["lambdaf"] lambdar=p["lambdar"]; nu=p["nu"]; blambdaf=p["blambdaf"] N=S+E+I+H+F+R dS=-(betai*I+betah*H+blambdaf*F)*S/N dE=(betai*I+betah*H+blambdaf*F)*S/N-alpha*E dI=alpha*E-theta*gammah*I-(1-theta)*(1-delta)*gammar*I-(1-theta)*delta*gammaf*I dH=theta*gammah*I-delta*lambdaf*H-(1-delta)*lambdaf*H dF=(1-theta)*(1-delta)*gammar*I+delta*lambdaf*H-nu*F dR=(1-theta)*(1-delta)*gammar*I+(1-delta)*lambdar*H+nu*F dZ=c(dS,dE,dI,dH,dF,dR) list(dZ)}

We can solve it, or at least study the dynamics from some starting values

library(deSolve) times = seq(0, 50, by = .1) resol = ode(y=Z, times=times, func=SEIHFR, parms=p)

For instance, the proportion of people infected is the following

plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="",col="red") lines(resol[,"time"],resol[,"H"],col="blue")

# Modeling pandemics (2)

When introducing the SIR model, in our initial post, we got an ordinary differential equation, but we did not really discuss stability, and periodicity. It has to do with the Jacobian matrix of the system. But first of all, we had three equations for three function, but actually$$\displaystyle{{\frac{dS}{dt}}+{\frac {dI}{dt}}+{\frac {dR}{dt}}=0}$$so it means that our problem is here simply in dimension 2. Hence\displaystyle {\begin{aligned}&X={\frac {dS}{dt}}=\mu(N-S)-{\frac {\beta IS}{N}},\\[6pt]&Y={\frac {dI}{dt}}={\frac {\beta IS}{N}}-(\mu+\gamma)I\end{aligned}}and therefore, the Jacobian of the system is$$\begin{pmatrix}\displaystyle{\frac{\partial X}{\partial S}}&\displaystyle{\frac{\partial X}{\partial I}}\\[9pt]\displaystyle{\frac{\partial Y}{\partial S}}&\displaystyle{\frac{\partial Y}{\partial I}}\end{pmatrix}=\begin{pmatrix}\displaystyle{-\mu-\beta\frac{I}{N}}&\displaystyle{-\beta\frac{S}{N}}\\[9pt]\displaystyle{\beta\frac{I}{N}}&\displaystyle{\beta\frac{S}{N}-(\mu+\gamma)}\end{pmatrix}$$We should evaluate the Jacobian at the equilibrium, i.e. $$S^\star=\frac{\gamma+\mu}{\beta}=\frac{1}{R_0}$$and$$I^\star=\frac{\mu(R_0-1)}{\beta}$$We should then look at eigenvalues of the matrix.

Our very last example was

times = seq(0, 100, by=.1) p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="")

We can compute values at the equilibrium

mu=p["mu"]; beta=p["beta"]; gamma=p["gamma"] N=1 S = (gamma + mu)/beta I = mu * (beta/(gamma + mu) - 1)/beta

and the Jacobian matrix

J=matrix(c(-(mu + beta * I/N),-(beta * S/N), beta * I/N,beta * S/N - (mu + gamma)),2,2,byrow = TRUE)

Now, if we look at the eigenvalues,

eigen(J)$values [1] -0.024975+0.6318831i -0.024975-0.6318831i or more precisely $2\pi/b$ where $a\pm ib$ are the conjuguate eigenvalues 2 * pi/(Im(eigen(J)$values[1])) [1] 9.943588

we have a damping period of 10 time lengths (10 days, or 10 weeks), which is more or less what we’ve seen above,

The graph above was obtained using

p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[1:1e5,"time"],resol[1:1e5,"I"],type="l",xlab="time",ylab="",lwd=3,col="red") yi=resol[,"I"] dyi=diff(yi) i=which((dyi[2:length(dyi)]*dyi[1:(length(dyi)-1)])&lt;0) t=resol[i,"time"] arrows(t[2],.008,t[4],.008,length=.1,code=3)

If we look carefully. at the begining, the duration is (much) longer than 10 (about 13)… but it does converge towards 9.94

plot(diff(t[seq(2,40,by=2)]),type="b") abline(h=2 * pi/(Im(eigen(J)values[1])) So here, theoretically, every 10 weeks (assuming that our time length is a week), we should observe an outbreak, smaller than the previous one. In practice, initially it is every 13 or 12 weeks, but the time to wait between outbreaks decreases (until it reaches 10 weeks). # Modeling pandemics (1) The most popular model to model epidemics is the so-called SIR model – or Kermack-McKendrick. Consider a population of size $N$, and assume that $S$ is the number of susceptible, $I$ the number of infectious, and $R$ for the number recovered (or immune) individuals, \displaystyle {\begin{aligned}&{\frac {dS}{dt}}=-{\frac {\beta IS}{N}},\\[6pt]&{\frac {dI}{dt}}={\frac {\beta IS}{N}}-\gamma I,\\[6pt]&{\frac {dR}{dt}}=\gamma I,\end{aligned}}so that $$\displaystyle{{\frac{dS}{dt}}+{\frac {dI}{dt}}+{\frac {dR}{dt}}=0}$$which implies that $S+I+R=N$. In order to be more realistic, consider some (constant) birth rate $\mu$, so that the model becomes\displaystyle {\begin{aligned}&{\frac {dS}{dt}}=\mu(N-S)-{\frac {\beta IS}{N}},\\[6pt]&{\frac {dI}{dt}}={\frac {\beta IS}{N}}-(\gamma+\mu) I,\\[6pt]&{\frac {dR}{dt}}=\gamma I-\mu R,\end{aligned}}Note, in this model, that people get sick (infected) but they do not die, they recover. So here, we can model chickenpox, for instance, not SARS. The dynamics of the infectious class depends on the following ratio:$$\displaystyle{R_{0}={\frac {\beta }{\gamma +\mu}}}$$ which is the so-called basic reproduction number (or reproductive ratio). The effective reproductive ratio is $R_0S/N$, and the turnover of the epidemic happens exactly when $R_0S/N=1$, or when the fraction of remaining susceptibles is $R_0^{-1}$. As shown in Directly transmitted infectious diseases:Control by vaccination, if $S/N the disease (the number of people infected) will start to decrease. Want to see it ? Start with mu = 0 beta = 2 gamma = 1/2 for the parameters. Here, $R_0=4$. We also need starting values epsilon = .001 N = 1 S = 1-epsilon I = epsilon R = 0 Then use the ordinary differential equation solver, in R. The idea is to say that $\boldsymbol{Z}=(S,I,R)$ and we have the gradient $$\frac{\partial \boldsymbol{Z}}{\partial t} = SIR(\boldsymbol{Z})$$where $SIR$ is function of the various parameters. Hence, set p = c(mu = 0, N = 1, beta = 2, gamma = 1/2) start_SIR = c(S = 1-epsilon, I = epsilon, R = 0) The we must define the time, and the function that returns the gradient, times = seq(0, 10, by = .1) SIR = function(t,Z,p){ S=Z[1]; I=Z[2]; R=Z[3]; N=S+I+R mu=p["mu"]; beta=p["beta"]; gamma=p["gamma"] dS=mu*(N-S)-beta*S*I/N dI=beta*S*I/N-(mu+gamma)*I dR=gamma*I-mu*R dZ=c(dS,dI,dR) return(list(dZ))} To solve this problem use library(deSolve) resol = ode(y=start_SIR, times=times, func=SIR, parms=p) We can visualize the dynamics below par(mfrow=c(1,2)) t=resol[,"time"] plot(t,resol[,"S"],type="l",xlab="time",ylab="") lines(t,resol[,"I"],col="red") lines(t,resol[,"R"],col="blue") plot(t,t*0+1,type="l",xlab="time",ylab="",ylim=0:1) polygon(c(t,rev(t)),c(resol[,"R"],rep(0,nrow(resol))),col="blue") polygon(c(t,rev(t)),c(resol[,"R"]+resol[,"I"],rev(resol[,"R"])),col="red") We can actually also visualize the effective reproductive number is $R_0S/N$, where R0=p["beta"]/(p["gamma"]+p["mu"]) The effective reproductive number is on the left, and as we mentioned above, when we reach 1, we actually reach the maximum of the infected, plot(t,resol[,"S"]*R0,type="l",xlab="time",ylab="") abline(h=1,lty=2,col="red") abline(v=max(t[resol[,"S"]*R0&gt;=1]),col="darkgreen") points(max(t[resol[,"S"]*R0&gt;=1]),1,pch=19) plot(t,resol[,"S"],type="l",xlab="time",ylab="",col="grey") lines(t,resol[,"I"],col="red",lwd=3) lines(t,resol[,"R"],col="light blue") abline(v=max(t[resol[,"S"]*R0&gt;=1]),col="darkgreen") points(max(t[resol[,"S"]*R0&gt;=1]),max(resol[,"I"]),pch=19) And when adding a $\mu$ parameter, we can obtain some interesting dynamics on the number of infected, times = seq(0, 100, by=.1) p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="") # Richesse et espérance de vie Ce matin, je découvrais un graphique de l’INSEE qui présentait les taux de mortalité par sexe, âge et niveau de vie, avec entre autres le graphique suivant Comme souvent avec l’INSEE, on a accès aux données… pas celles au niveau individuel (malheureusement) mais au moins on peut retravailler la visualisation. En fait, les données sont même encore plus fines, puisque les niveaux de richesses sont définis avec des tranches de 5%, et en plus, on a le détail entre hommes et femmes b = read.csv2("MORT-RICHESSE.csv") plot(b[,1],b[,2]/1000,col="red",type="l",ylab="% de survivants",xlab="Age") lines(b[,1],b[,3]/1000,col="red",type="l",lty=2) lines(b[,1],b[,4]/1000,col="blue",type="l",lty=1) lines(b[,1],b[,5]/1000,col="blue",type="l",lty=2) legend("bottomleft",c("Femmes 95-100%","Hommes 95-100%","Femmes 0-5%","Hommes, 0-5%"), bty="n",col=c("red","blue","red","blue"),lty=c(2,2,1,1)) Je me demandais si on ne pouvait pas tenter une lecture inverse de ce graphique : sur ce graphique, assez naturellement, on regarde à pourcentage donné l’écart entre la courbe en trait plein (les pauvres) et celle en trait pointillé (les riches). Si c’est cette information qu’on veut avoir, on peut alors tenter de la visualiser. Pour ça, il faut inverser notre fonction de survie (je l’ai fait rapidement, avec une simple interpolation linéaire… je pense qu’on peut faire mieux) inversef = function(p,k=2){ y=1-b[,k]/100000 idx=sum(y&lt;=p) y1=y[idx-1] y2=y[idx] w1=(y1-p)/(y1-y2) w2=(p-y2)/(y1-y2) w2*b[idx-1,1]+w1*b[idx,1] } Ensuite, on peut construire les inverses, et mieux, les différences entre les courbes des riches et des pauvres diffF = function(p) inversef(p,3)-inversef(p,2) diffH = function(p) inversef(p,5)-inversef(p,4) u = seq(.01,.99,by=.01) vF = Vectorize(diffF)(u) vH = Vectorize(diffH)(u) plot(u*100,vF,col="red",type="l",xlab="Probabilité (%)",ylab="Nombre d'années",ylim=c(0,max(vF,vH))) lines(u*100,vH,col="blue",type="l",lty=1) legend("topright",c("Femmes","Hommes"), bty="n",col=c("red","blue"),lty=c(1,1)) Je ne suis pas très à l’aise avec le graphique. Tout d’abord parce que je ne sais pas ce que la richesse indique (un pauvre de 20 ans peut devenir un riche de 50 ans, non ?), la richesse était souvent liée à l’âge. Après l’axe des abscisses me semble aussi avoir une interprétation compliquée : quand on regarde à 10%, on regarde les pauvres et les riches qui sont morts relativement jeunes (relativement car je regarde le quantile de la fonction de survie à tranche de richesse donnée). Autrement dit, si je me place à 10%, je compare les jeunes riches morts très jeunes (10% des riches seulement sont morts plus jeunes – c’est l’interprétation d’un quantile) et les jeunes pauvres (mort à un âge que seulement 90% des pauvres ont dépassé), on observe une différence de 20 ans environ. J’ai aussi l’impression qu’on pourrait dire que pour la majorité des hommes, les riches vivent 12 ans de plus que les pauvres, soit le double des femmes (de l’ordre de 6 ans). On pourrait bien sûr se contenter de calculer les différences entre les aires, ce qui donne une différence entre les espérances de vie à la naissance des pauvres et des riches (comme le fait l’INSEE) et qu’on peut visualiser sur le graphique suivant plot(b[,1],b[,2]/1000,col="white",type="l",ylab="% de survivants",xlab="Age") polygon(c(b[,1],rev(b[,1])),c(b[,3]/1000,rev(b[,2]/1000)),col="red",border=NA) et un calcul donne une différence de l’ordre de 8 ans sum(b[,3]-b[,2])/100000 [1] 8.239346 mais la visualisation raconte bien plus qu’un simple calcul d’aire. Par exemple, le graphique ci-dessous donne exactement le même écart entre les espérances de vie des pauvres et des riches diff = sum(b[,3]-b[,2])/1000 y1 = b[,2]/1000 for(i in 1:100){ y1[i] = b[i,2]/1000+min(100-b[i,2]/1000,diff) diff = diff-(y1[i]-b[i,2]/1000) } plot(b[,1],b[,2]/1000,col="white",type="l",ylab="% de survivants",xlab="Age") polygon(c(b[,1],rev(b[,1])),c(y1,rev(b[,2]/1000)),col="red",border=NA) sum(b[,3]-b[,2])/100000 lines(b[,1],b[,2]/1000,col="red") Ici, on dit que la moitié des femmes riches meurent vers 81 ans, et les autres meurent au même âge qu’une femmes pauvre (mais une femme pauvre qui vivrait longtemps. Les distributions sont vraiment différentes, et c’est ça que je cherche à visualiser… Parce que la densité de l’âge au décès ne me semble pas forcément très simple à analyser… Comme toujours, les commentaires sont ouverts si certains ont des idées quant à la visualisation de ces données… # Proportion of people alive in 1945 that are still alive In demography, we like to use life tables to estimate the probability that someone born in 1945 (say) is still alive nowadays. But another interesting quantity might be the probability that someone alive in 1945 is still alive nowadays. The main difference is that we do not know when that person, alive in 1945, was born. Someone who was old in 1945 is very unlikely still alive in 2017. To compute those probabilities, we can use datasets from http://www.mortality.org/hmd/. More precisely, we need both death and birth data. I assume that datasets (text files) were downloaded (it is necessary to register – for free – to get the data). D=read.table("FRDeaths_1x1.txt",skip=1,header=TRUE) B=read.table("FRBirths.txt",skip=1,header=TRUE) In the death dataset, there is a “110+” for people older than 110 years. For convenience, let us cap our observations at 110 years old, DAge=as.numeric(as.character(D$Age)) D$Age[is.na(D$Age)]=110 Consider now a first function that will return, for people born in 1930 (say) two informations • the number of people (here, let us consider women only) born in 1930 (from the birth database) • the number of death of people of age 0 in 1930, people of age 1 in 1931, people of age 2 in 1932, etc… The code is simple nb=function(y=1930){ debut=1816 MatDFemale=matrix(D$Female,nrow=111) colnames(MatDFemale)=debut+0:198 cly=y-debut+1:111 deces=diag(MatDFemale[,cly[cly%in%1:199]]) return(c(B$Female[B$Year==y],deces))}

We have a single number for the number of births, and then a vector for the number of deaths. Consider now another function. Consider the people born in 1930. We want to get two numbers : the number of people still alive in 1945 (say), and the number of people still alive nowadays. The ratio will be the proportion of people born in 1930 that were alive in 1945, that are still alive in 2015.

pop=function(ne=1930,an=1945){ comptage=nb(ne) s=0 if(an>ne) s=sum(comptage[seq(2,1+an-ne)]) p1=max(comptage[1]-s,0) p2=max(p1-sum(comptage[seq(2+an-ne,length(comptage))]),0) c(p1,p2) }

Then, for a given year (say 1945), to get the proportion of people alive in 1945 that are still alive today, we need to count how many people born in 1944 were still alive in 1945, and in 2015, but also born in 1943, 1942, etc, And we simply consider the ratio of the total number of people alive in 2015 over the total number of people alive in 1945

ptn=function(y=1945){ V=Vectorize(function(x) pop(x,y))(1816:y) sum(V[2,!is.na(V[2,])])/sum(V[1,!is.na(V[1,])]) }

Hence, 22% of those alive in 1945 are still alive in 2015,

> ptn(1945) [1] 0.2209435

Actually, instead of looking only at 1945, it is possible to get a plot

P=Vectorize(ptn)(1900:2010) plot(1900:2010,P,type="l",ylim=0:1)

For instance,

> ptn(1975) [1] 0.6377413

i.e. 63.7% of those alive in 1975 are stil alive 40 years after. That is a rather interesting function, I was surprised that I couldn’t find it is standard demographical R package…

# Dynamique de la Pyramide des Ages

Très joli billet sur blog.revolutionanalytics.com avec un code de @kyle_e_walker permettant, très simplement (moyennant une inscription pour avoir une clé permettant d’utiliser l’API du census) de construire une pyramide des âges dynamiques.

> devtools::install_github('walkerke/idbr')
> library(idbr)
> library(ggplot2)
> library(animation)
> library(dplyr)
> library(ggthemes)
> idb_api_key("mykey1239F2f324zf9GGZgege32R2ii4")

On importe alors les données pour les hommes et les femmes,

> male <- idb1('FR', 2010:2050, sex = 'male') %>%
+   mutate(POP = POP * -1,
+   SEX = 'Male')
> female <- idb1('FR', 2010:2050, sex = 'female') %>% mutate(SEX = 'Female')

et on stocke le tout

> france <- rbind(male, female) %>%
+   mutate(abs_pop = abs(POP))

Ensuite, on crée l’animation,

> saveGIF({
+
+   for (i in 2010:2050) {
+
+     title <- as.character(i)
+
+     year_data <- filter(france, time == i)
+
+     g1 <- ggplot(year_data, aes(x = AGE, y = POP, fill = SEX, width = 1)) +
+       coord_fixed() +
+       coord_flip() +
+       annotate('text', x = 98, y = -800000,
+       label = 'Data: US Census Bureau IDB; idbr R package', size = 3) +
+       geom_bar(data = subset(year_data, SEX == "Female"), stat = "identity") +
+       geom_bar(data = subset(year_data, SEX == "Male"), stat = "identity") +
+       scale_y_continuous(breaks = seq(-1000000, 1000000, 500000),
+       labels = paste0(as.character(c(seq(1, 0, -0.5), c(0.5, 1))), "m"),
+       limits = c(min(france$POP), max(france$POP))) +
+       theme_economist(base_size = 14) +
+       scale_fill_manual(values = c('#ff9896', '#d62728')) +
+       ggtitle(paste0('Population structure of France, ', title)) +
+       ylab('Population') +
+       xlab('Age') +
+       theme(legend.position = "bottom", legend.title = element_blank()) +
+       guides(fill = guide_legend(reverse = TRUE))
+     print(g1)
+   }
+ }, movie.name = 'france_pyramid.gif', interval = 0.1, ani.width = 700, ani.height = 600)

Et le résultat est vraiment joli, non ?

# Mortality by Weekday and Age

A few days ago, I did mention on Twitter a nice graph, with

My colleague Jean-Philippe was extremely sceptical, so I tried to reproduce that graph. The good thing is that we have the Social Security Death Master File, for data in the US. To be more specific, I have three big files on my hard drive, and in order to reproduce that graph, we’ll load the data by chunks. But before, because we have the day of birth, and the day of death, I need a function to compute the age. So here it is

> age_years <- function(earlier, later)
+ {
+   lt <- data.frame(earlier, later)
+   age <- as.numeric(format(lt[,2],format="%Y")) - as.numeric(format(lt[,1],format="%Y"))
+   dayOnLaterYear <- ifelse(format(lt[,1],format="%m-%d")!="02-29",
+                            as.Date(paste(format(lt[,2],format="%Y"),"-",format(lt[,1],format="%m-%d"),sep="")),
+                            ifelse(as.numeric(format(later,format="%Y")) %% 400 == 0 | as.numeric(format(later,format="%Y")) %% 100 != 0 & as.numeric(format(later,format="%Y")) %% 4 == 0,
+                                   as.Date(paste(format(lt[,2],format="%Y"),"-",format(lt[,1],format="%m-%d"),sep="")),
+                                   as.Date(paste(format(lt[,2],format="%Y"),"-","02-28",sep=""))))
+   age[which(dayOnLaterYear > lt$later)] <- age[which(dayOnLaterYear > lt$later)] - 1
+   age
+ }

from github.com/nzcoops. Now, it is possible to create a similar table, based on that huge file (we have almost 50 million observations)

> cols <- c(1,9,20,4,15,15,1,2,2,4,2,2,4,2,5,5,7)
> noms_col <- c ("code","ssn","last_name","name_suffix","first_name","middle_name","VorPCode","date_death_m","date_death_d","date_death_y","date_birth_m","date_birth_d","date_birth_y","state","zip_resid","zip_payment","blanks")
> library(LaF)

> TABLE_AGE_DAY=function(temp = "ssdm3"){
+ ssn <- laf_open_fwf( temp,column_widths = cols,column_types=rep("character",length(cols) ),column_names = noms_col,trim = TRUE)
+ object.size(ssn)
+ go_through <- seq(1,nrow(ssn),by = 1e05 )
+ if(go_through[ length(go_through)] != nrow( ssn)) go_through <- c(go_through,nrow( ssn))
+ go_through <- cbind(go_through[-length(go_through)],c(go_through[-c(1,length(go_through)) ]-1,go_through [ length(go_through)]))
+ go_through
+
+ pb <- txtProgressBar(min = 0, max = nrow( go_through), style = 3)
+ count_birthday <- function(s){
+   #print(s)
+   setTxtProgressBar(pb, s)
+   data <- ssn[ go_through[s,1]:go_through[s,2],c("date_death_y","date_death_m","date_death_d",
+                                                  "date_birth_y","date_birth_m","date_birth_d")]
+   date1=as.Date(paste(data$date_birth_y,"-",data$date_birth_m,"-",data$date_birth_d,sep=""),"%Y-%m-%d") + date2=as.Date(paste(data$date_death_y,"-",data$date_death_m,"-",data$date_death_d,sep=""),"%Y-%m-%d")
+   idx=which(!(is.na(date1)|is.na(date2)))
+   date1=date1[idx]
+   date2=date2[idx]
+   itg=try(age<-age_years(date1,date2),silent=TRUE)
+   if(inherits(itg, "try-error")) age=trunc((date2-date1)/365.25)
+   w=weekdays(date2)
+   T=table(age,w)
+   Tab=matrix(0,106,7)
+   for(i in 1:nrow(T)) if(as.numeric(rownames(T)[i])<106) Tab[as.numeric(rownames(T)[i]),]=T[i,]
+   return(Tab)
+ }
+ D <- lapply( seq_len(nrow( go_through)),count_birthday)
+ T=D[[1]]
+ for(s in 2:length(D)) T=T+D[[s]]
+ return(T)
+ }

If we run that function on the three files

> D1=TABLE_AGE_DAY("ssdm1")
|========================================| 100%
> D2=TABLE_AGE_DAY("ssdm2")
|========================================| 100%
> D3=TABLE_AGE_DAY("ssdm3")
|========================================| 100%

we can visualize not percentages, as on the figure above, but counts

> D=D1+D2+D3
> colnames(D)=
c("Sun","Thu","Mon","Tue","Wed","Sat","Fri")
> D=D1[,
c("Sun","Mon","Tue","Wed","Thu","Fri","Sat")]

and we have here (I remove the Saturday to get a better output)

> D[,1:6]
Sun    Mon    Tue    Wed    Thu    Fri
[1,]   2843   2888   2943   3020   2979   3038
[2,]   2007   1866   1918   1974   1990   2137
[3,]   1613   1507   1532   1530   1515   1613
[4,]   1322   1256   1263   1259   1207   1330
[5,]   1155   1061   1092   1128   1112   1171
[6,]   1067    985    950   1082   1009   1055
[7,]   1129    901    915    954    941   1044
[8,]   1026    927    944    935    911   1005
[9,]   1029   1012    871    908    939    998
[10,]   1093   1011    974    958    928   1018
[11,]   1106   1031   1019   1036   1087   1122
[12,]   1289   1219   1176   1215   1141   1292
[13,]   1618   1455   1487   1484   1466   1633
[14,]   2121   2000   1900   1941   1845   2138
[15,]   2949   2647   2519   2499   2524   2748
[16,]   4488   3885   3798   3828   3747   4267
[17,]   5709   4612   4520   4422   4443   5005
[18,]   7280   5618   5400   5271   5344   5986
[19,]   8086   6172   5833   5820   6004   6628
[20,]   8389   6507   6166   6055   6430   6955
[21,]   8794   7038   6794   6628   6841   7572
[22,]   8578   6528   6512   6472   6757   7342
[23,]   8345   6750   6483   6469   6714   7338
[24,]   8361   6859   6589   6623   6854   7369
[25,]   8398   6974   6892   6766   6964   7613
[26,]   8432   7210   7012   7175   7343   7801
[27,]   8757   7641   7526   7352   7674   7950
[28,]   9190   8041   7843   7851   7940   8268
[29,]   9495   8409   8555   8400   8469   8934
[30,]   9876   9041   9015   9166   9106   9641
[31,]  10567   9952   9506   9634   9770  10212
[32,]  11417  10428  10402  10275  10455  11169
[33,]  11992  11306  11124  11095  11243  11749
[34,]  12665  12327  11760  12025  12137  12443
[35,]  13629  13135  13179  13037  12968  13724
[36,]  14560  14009  13927  13822  14105  14436
[37,]  15660  14990  15013  15009  15101  15700
[38,]  16749  16504  16148  16091  15912  16863
[39,]  17815  17760  17519  17144  17553  17943
[40,]  19366  19057  18918  18517  18760  19604
[41,]  20770  20458  20154  20339  20349  21238
[42,]  21962  22194  22020  21499  21690  22347
[43,]  23803  23922  23701  23681  23437  24227
[44,]  25685  26133  25559  25209  25287  26115
[45,]  27506  28110  27363  27042  27272  28228
[46,]  29366  29744  29555  29245  29678  30444
[47,]  31444  32193  31817  31504  31753  32302
[48,]  33452  34719  33529  33954  33441  34618
[49,]  36186  37150  36005  36064  36226  37138
[50,]  38401  39244  38813  38465  38506  39884
[51,]  40331  41830  41168  41110  40937  42014
[52,]  43181  44351  43975  43949  43579  44734
[53,]  45307  47134  46522  46149  46089  47286
[54,]  47996  49441  49139  48678  48629  49903
[55,]  50635  52424  51757  51433  51477  52550
[56,]  53509  55337  54556  54482  54406  55906
[57,]  55703  58482  58016  57400  57097  58758
[58,]  59016  61453  60652  61024  60557  62473
[59,]  62475  65651  64169  63824  63829  65592
[60,]  66621  69185  68885  68217  68752  69963
[61,]  69759  73144  72421  71784  71745  73414
[62,]  80346  84253  83044  83177  82416  83833
[63,]  86851  90059  89002  88985  89245  90334
[64,]  91839  95465  94602  93985  94154  96195
[65,]  98461 102846 101348 101328 101306 103170
[66,] 104569 108722 107768 107711 107729 109350
[67,] 111230 115477 114418 114743 113935 116356
[68,] 116999 122053 120727 120342 119782 122926
[69,] 123695 128339 127184 126822 126639 129037
[70,] 129956 136123 134555 135120 133842 137390
[71,] 137984 142964 141316 142855 141419 143620
[72,] 145132 150708 148407 149345 149448 151910
[73,] 152877 157993 155861 156349 155924 158725
[74,] 159109 164652 162722 163499 163157 165744
[75,] 165848 172121 170730 170482 170585 173431
[76,] 172457 179036 177185 177328 177392 180215
[77,] 179936 185015 183223 183932 183237 186663
[78,] 185900 191053 189986 189730 189639 193038
[79,] 191498 196694 194246 194810 195246 197812
[80,] 195505 201289 199684 199561 198968 203226
[81,] 199031 204927 202204 202622 202951 205792
[82,] 201589 207928 204929 204001 204396 208224
[83,] 201665 206743 205194 204676 205256 207980
[84,] 200965 205653 203422 202393 203422 206012
[85,] 197445 202692 199498 199730 200075 201728
[86,] 192324 195961 193589 194754 193800 196102
[87,] 183732 188063 185153 186104 186021 188176
[88,] 174258 177474 175822 176078 176761 177449
[89,] 163180 166706 162810 164367 164281 166436
[90,] 149169 151738 150148 150212 150535 152435
[91,] 134218 136866 134959 134922 135027 136381
[92,] 118936 121106 119591 119509 119793 120998
[93,] 102734 104955 102944 102865 103345 104776
[94,]  87418  88885  88023  86963  87546  87872
[95,]  72023  72698  72151  71579  71530  72287
[96,]  56985  58238  57478  57319  57163  57615
[97,]  44447  45058  44607  44469  43888  44868
[98,]  33457  34132  33022  33409  33454  33642
[99,]  24070  24317  24305  24089  24020  24383
[100,]  17165  17295  16755  17115  16957  17207
[101,]  11799  12125  11709  11816  11824  11719
[102,]   7714   7741   7959   7691   7648   7633
[103,]   5024   5012   4822   4792   4882   4916
[104,]   2987   3101   2978   3049   3093   2906
[105,]   1781   1894   1811   1756   1734   1834

So clearly, for young people, the number of deaths is rather small…

And to visualize it, as above, we can use

> P=D/apply(D,1,sum)*100
> range(P)
[1] 12.34857 17.59386
> dP=trunc((P-min(P))/(max(P)+.01-min(P))*11)
> library(RColorBrewer)
> CLR=rev(brewer.pal(name="RdYlBu", 11))

> plot(0:1,0:1,ylim=c(55,110),xlim=c(-1,7))
> for(i in 1:106){
+   for(j in 1:7){
+  rect(j-1,108-i,j,107-i,col=CLR[dP[i,j]])
+   }}
> text(rep(-.5,106),107.5-1:106,0:105,cex=.4)

As above, we observe a strong difference among weekdays for the date of death for young people (below 30) which disappear after (even if there is still a sunday effect)

# Démographie québécoise

Cet automne, j’étais allé voir Mummy, de Xavier Dolan. Et j’avoue avoir adoré le film ! J’ai tout aimé ! La musique, le cadrage, le rythme, les acteurs (même si j’avais du mal à croire Anne Dorval qui restera toujours pour moi la maman des Parent, surtout que j’ai passé mon temps à Montréal à croiser Daniel Brière – l’acteur, pas le joueur des Avalanches – qui habitait à côté de la maison, et qui faisait souvent le marché en même temps que moi… pour moi les deux étaient attachés à tout jamais…. et la mère de la série et celle du film de Xavier Dolan sont assez différentes).

Le seul reproche que j’aurais pu faire, ce sont les sous-titres (qui nous étaient imposés, en France). Au Québec, on avait pris l’habitude de vivre sans sous-titres, et on a appris une langue en la parlant, au quotidien. Par exemple, j’ai appris ce que voulait dire câlisser sans avoir à ouvrir un dictionnaire, ou sans demander de traduction. De même que j’ai découvert qu’il existait des variantes, comme décâlisser (et j’ai même fini par comprendre le sens). Mais je n’avais jamais eu besoin de visualiser la traduction du mot. Voir des traductions de mots que j’avais fini par découvrir et comprendre m’a déstabilisé. Je reverrais d’autant mieux le film en DVD, sans les sous-titres qui m’étaient imposés au cinéma.

C’est amusant car pendant les vacances, j’en ai profité pour finir Magasin Général de Régis Loisel et Jean-Louis Tripp. Et j’ai eu plaisir à retrouver des expressions québécoises tout au long de la lecture. Même si l’histoire se déroule à Notre Dame du Lac, situé sur le bas Saint Laurent (même si le nom ne semble plus exister depuis quelques années), on est loin des dialogues que l’on peut entendre en région, et on a une version agréable de ce qu’on pourrait entendre à Montréal…

En fait, comme cela est indiqué (trop) discrètement, c’est Jimmy Beaulieu (dont j’avais déjà souligné le travail admirable dans un précédant billet) qui a fait la “traduction” des dialogues, pour avoir (comme le disent les premières pages du livres) des dialogues en québécois “qui soient compréhensibles des deux côtés de l’Atlantique”. Ce qui montre bien qu’un français de France peut comprendre le québécois (moyennant un peu de bonne volonté). Je crois que j’aurais bien aimé avoir des sous-titres pour Mummy dans la même langue que celle parlée dans Magasin Général.

En lisant en particulier les deux derniers tomes de Magasin Général (oui, j’avais un peu de retard), et les histoires de natalité, je me suis souvenu du travail que l’on avait fait il y a un peu plus d’un an avec Julie, qui était venu faire un stage à l’UQAM, et avec qui on avait découvert la démographie du Lac Saint Jean (certes, par rapport à Magasin Général, on est de l’autre côté du Saint Laurent).

les slides (de la soutenance de stage) sont en ligne,

# Reinterpreting Lee-Carter Mortality Model

Last week, while I was giving my crash course on R for insurance, we’ve been discussing possible extensions of Lee & Carter (1992) model. If we look at the seminal paper, the model is defined as follows

# Mais que s’est-il passé pendant la Première Guerre Mondiale?

La réponse courte est que des gens sont morts. Beaucoup. Cela étant dit, on ne dit pas grand chose. On peut comparer les pyramides des âges pour mieux comprendre ce qui a pu se passer. Juste avant la guerre (en 1913), la pyramide des âges ressemblait à ça (en utilisant les données de mortality.org)

> EXPO  <- read.table(
> EM=EXPO$Male > EF=EXPO$Female
> Y= EXPO$Year > A= EXPO$Age
> I=which(A=="110+")
> base=data.frame(Female=EF,Male=EM,Y=Y,Ages=A)
> base=base[-I,]
> France1913=base[base$Y==1913,] > France1919=base[base$Y==1919,]
> France1913$Ages=as.numeric( + as.character(France1913$Ages))
> France1919$Ages=as.numeric( + as.character(France1919$Ages))
> France1913=France1913[,c("Male","Female",
+ "Ages")]
> library(pyramid)
> plot(c(0,100), c(0,100), type="n",
+ frame=FALSE, axes=FALSE, xlab="", ylab="",
+ main="Pyramide des Ages, France, 1913")
> pyramidf(France1913, frame=c(10, 75, 0, 90),
+ Clab="", Lcol="skyblue", Rcol="pink",
+ Cstep=10, Laxis=0:4*60000, AxisFM="d")

En revanche, juste après la guerre (en 1919), la pyramide des âges des âges ressemblait à celle là Continue reading Mais que s’est-il passé pendant la Première Guerre Mondiale?

# Men set to live as long as women by 2030?

A few months ago, in Men set to live as long as women, figures show, it was mentioned that (in the U.K.)

the gap between male and female life expectancy is closing and men could catch up by 2030, according to an adviser for the Office for National Statistics.

(the slides are available online http://cass.city.ac.uk/…).

This week, I discovered a picture on http://waitbutwhy.com/, which represent a (so-called) typical human life, in weeks,

I found that interesting. But the first problem is that I don’t understand the limit, below: 90 years, that’s not the average life length. That’s not what you should expect to live when you get born. The second problem is that it cannot be as static as it might seem, when you look at the picture. I mean, life expectancy at age 0 is not the same as life expectancy at age 30, or 50. So I did try to make an animated graph, using prospective life tables. Here a code to generate life tables, at different period, for a French population (I distinguish, here male and female)

library(demography)
france.fcast <- forecast(france.LC1,h=100)
L2 <- lifetable(france.fcast)
ex2=L2$ex L1=lifetable(fr.mort,series="female") ex1=L1$ex
exF=cbind(ex1,ex2)
france.fcast <- forecast(france.LC1,h=100)
L2 <- lifetable(france.fcast)
ex2=L2$ex L1=lifetable(fr.mort,series="male") ex1=L1$ex
exM=cbind(ex1,ex2)
Y=colnames(exF)

Based on those lifetables, we can extract remaining life expectancy, at various ages (say, for instance 50, 51, 52, etc), for someone born on some given year (say 1950). Based on those expected remaining lifetimes, we can plot

picture=function(yearborn=1950,age=50){
k=which(Y==yearborn)
M=diag(exM[,k+0:100])
F=diag(exF[,k+0:100])
par(mfrow=c(1,2))
va=0:(52*100-1)
plot(va%%52,va%/%52,cex=.6,pch=15,col=c("light yellow","light blue","white")[1+
(va>=age*52)*1+(va>(age+M[age+1])*52)*1],ylim=c(100,0),axes=FALSE,xlab="Week",
ylab="Age",main=paste("Man, born on ",yearborn,
", age ",age,sep=""))
axis(1)
axis(2)
plot(va%%52,va%/%52,cex=.6,pch=15,col=c("light yellow","pink","white")[1+
(va>=age*52)*1+(va>(age+F[age+1])*52)*1],ylim=c(100,0),axes=FALSE,xlab="Week",
ylab="Age",main=paste("Woman, born on ",yearborn,
", age ",age,sep=""))
axis(1)
axis(2)}

For instance, if we want the graph above, for someone age 30, born in 1980, we use

picture(1980,30)

Now, if we run a code to get an animated gif, we can get, for someone born in 1950,

and for someone born in 2000

Now, if I could get historical datasets, with the average time spent in schools, ages of retirement, etc, I guess I could add it on the graph. But that’s another story…

# Smoothing mortality rates

This morning, I was working with Julie, a student of mine, coming from Rennes, on mortality tables. Actually, we work on genealogical datasets from a small region in Québec, and we can observe a lot of volatiliy. If I borrow one of her graph, we get something like

Since we have some missing data, we wanted to use some Generalized Nonlinear Models. So let us see how to get a smooth estimator of the mortality surface.  We will write some code that we can use on our data later on (the dataset we have has been obtained after signing a lot of official documents, and I guess I cannot upload it here, even partially).

DEATH <- read.table(
"http://freakonometrics.free.fr/Deces-France.txt",
"http://freakonometrics.free.fr/Exposures-France.txt",
library(gnm)
D=DEATH$Male E=EXPO$Male
A=as.numeric(as.character(DEATH$Age)) Y=DEATH$Year
I=(A<100)
base=data.frame(D=D,E=E,Y=Y,A=A)
subbase=base[I,]
subbase=subbase[!is.na(subbase$A),] The first idea can be to use a Poisson model, where the mortality rate is a smooth function of the age and the year, something like $D_{x,t}\sim\mathcal{P}(E_{x,t}\cdot \exp[{\color{blue}s(x,t)}])$that can be estimated using library(mgcv) regbsp=gam(D~s(A,Y,bs="cr")+offset(log(E)),data=subbase,family=quasipoisson) predmodel=function(a,y) predict(regbsp,newdata=data.frame(A=a,Y=y,E=1)) vX=trunc(seq(0,99,length=41)) vY=trunc(seq(1900,2005,length=41)) vZ=outer(vX,vY,predmodel) persp(vZ,theta=-30,col="green",shade=TRUE,xlab="Ages (0-100)", ylab="Years (1900-2005)",zlab="Mortality rate (log)") The mortality surface is here It is also possible to extract the average value of the years, which is the interpretation of the $a_x$ coefficient in the Lee-Carter model, predAx=function(a) mean(predict(regbsp,newdata=data.frame(A=a, Y=seq(min(subbase$Y),max(subbase$Y)),E=1))) plot(seq(0,99),Vectorize(predAx)(seq(0,99)),col="red",lwd=3,type="l") We have the following smoothed mortality rate Recall that the Lee-Carter model is $D_{x,t}\sim\mathcal{P}(E_{x,t}\cdot \exp[{\color{blue}a_x+b_x\cdot k_t}])$ where parameter estimates can be obtained using regnp=gnm(D~factor(A)+Mult(factor(A),factor(Y))+offset(log(E)), data=subbase,family=quasipoisson) predmodel=function(a,y) predict(regnp,newdata=data.frame(A=a,Y=y,E=1)) vZ=outer(vX,vY,predmodel) persp(vZ,theta=-30,col="green",shade=TRUE,xlab="Ages (0-100)", ylab="Years (1900-2005)",zlab="Mortality rate (log)") The (crude) mortality surface is with the following $a_x$ coefficients. plot(seq(1,99),coefficients(regnp)[2:100],col="red",lwd=3,type="l") Here we have a lot of coefficients, and unfortunately, on a smaller dataset, we have much more variability. Can we smooth our Lee-Carter model ? To get something which looks like $D_{x,t}\sim\mathcal{P}(E_{x,t}\cdot \exp[{\color{blue}s_a(x)+s_b(x)\cdot s_k(t)}])$ Actually, we can, and the code is rather simple library(splines) knotsA=c(20,40,60,80) knotsY=c(1920,1945,1980,2000) regsp=gnm(D~bs(subbase$A,knots=knotsA,Boundary.knots=range(subbase$A),degre=3)+ Mult(bs(subbase$A,knots=knotsA,Boundary.knots=range(subbase$A),degre=3), bs(subbase$Y,knots=knotsY,Boundary.knots=range(subbase$Y),degre=3))+ offset(log(E)),data=subbase, family=quasipoisson) BpA=bs(seq(0,99),knots=knotsA,Boundary.knots=range(subbase$A),degre=3)
BpY=bs(seq(min(subbase$Y),max(subbase$Y)),knots=knotsY,Boundary.knots= range(subbase$Y),degre=3) predmodel=function(a,y) predict(regsp,newdata=data.frame(A=a,Y=y,E=1)) v Z=outer(vX,vY,predmodel) persp(vZ,theta=-30,col="green",shade=TRUE,xlab="Ages (0-100)", ylab="Years (1900-2005)",zlab="Mortality rate (log)") The mortality surface is now and again, it is possible to extract the average mortality rate, as a function of the age, over the years, BpA=bs(seq(0,99),knots=knotsA,Boundary.knots=range(subbase$A),degre=3)
Ax=BpA%*%coefficients(regsp)[2:8]
plot(seq(0,99),Ax,col="red",lwd=3,type="l")

We can then play with the smoothing parameters of the spline functions, and see the impact on the mortality surface

knotsA=seq(5,95,by=5)
knotsY=seq(1910,2000,by=10)
regsp=gnm(D~bs(A,knots=knotsA,Boundary.knots=range(subbase$A),degre=3)+ Mult(bs(A,knots=knotsA,Boundary.knots=range(subbase$A),degre=3),
bs(Y,knots=knotsY,Boundary.knots=range(subbase\$Y),degre=3))
+offset(log(E)),data=subbase,family=quasipoisson)
predmodel=function(a,y) predict(regsp,newdata=data.frame(A=a,Y=y,E=1))
vZ=outer(vX,vY,predmodel)
ylab="Years (1900-2005)",zlab="Mortality rate (log)")

We now have to use those functions our our small data sample ! That should be fun….

# Job for life ? Bishop of Rome ?

The job of Bishop of Rome – i.e. the Pope – is considered to be a life-long commitment. I mean, it usually was. There have been 266 popes since 32 A.D. (according to http://oce.catholic.com/…): almost all popes have served until their death. But that does not mean that they were in the job for long… One can easily extract the data from the website,

> L2=scan("http://oce.catholic.com/index.php?title=List_of_Popes",what="character")
> index=which(L2=="</td><td>Reigned")
> X=L2[index+1]
> Y=strsplit(X,split="-")

But one should work a little bit because sometimes, there are inconsistencies, e.g. 911-913 and then 913-14, so we need some more lines. Further, we can extract from this file the years popes started to reign, the year it ended, and the length, using those functions

> diffyears=function(x){
+ s=NA
+ if(sum(substr(x,1,1)=="c")>0){x[substr(x,1,1)=="c"]=substr(x[substr(x,1,1)=="c"],3,nchar(x[substr(x,1,1)=="c"]))}
+ if(length(x)==1){s=1}
+ if(length(x)==2){s=diff(as.numeric(x))}
+ return(s)}
> whichyearsbeg=function(x){
+ s=NA
+ if(sum(substr(x,1,1)=="c")>0){x[substr(x,1,1)=="c"]=substr(x[substr(x,1,1)=="c"],3,nchar(x[substr(x,1,1)=="c"]))}
+ if(length(x)==1){s=as.numeric(x)}
+ if(length(x)==2){s=as.numeric(x)[1]}
+ return(s)}
> whichyearsend=function(x){
+ s=NA
+ if(sum(substr(x,1,1)=="c")>0){x[substr(x,1,1)=="c"]=substr(x[substr(x,1,1)=="c"],3,nchar(x[substr(x,1,1)=="c"]))}
+ if(length(x)==1){s=as.numeric(x)}
+ if(length(x)==2){s=as.numeric(x)[2]}
+ return(s)}

On our file, we have

> Years=unlist(lapply(Y,whichyearsbeg))
> YearsB=c(Years[1:91],752,Years[92:length(Years)])
> YearsB[187]=1276
> Years=unlist(lapply(Y,whichyearsend))
> YearsE=c(Years[1:91],752,Years[92:length(Years)])
> YearsE[187]=1276
> YearsE[266]=2013
> YearsE[122]=914
> W=unlist(lapply(Y,diffyears))
> W=c(W[1:91],1,W[92:length(W)])
> W[W==-899]=1
> which(is.na(W))
[1] 187 266
> W[187]=1
> W[266]=2013-2005

If we plot it, we have the following graph,

> plot(YearsB,W,type="h")

and if we look at the average length, we have the following graph,

> n=200
> YEARS = seq(0,2000,length=n)
> Z=rep(NA,n)
> for(i in 2:(n-1)){
+ index=which((YearsB>YEARS[i]-50)&(YearsE<YEARS[i]+50))
+ Z[i] = mean(W[index])}
> plot(YEARS,Z,type="l",ylim=c(0,30))
> n=50
> YEARS = seq(0,2000,length=n)
> Z=rep(NA,n)
> for(i in 2:(n-1)){
+ index=which((YearsB>YEARS[i]-50)&(YearsE<YEARS[i]+50))
+ Z[i] = mean(W[index])}
> lines(YEARS,Z,type="l",col="grey")

which does not reflect mortality improvements that have been observed over two millenniums. It might related to the fact that the average age at time of election has  increased over time (for instance, Benedict XVI was elected at 78 – one of the oldest to be elected). Actually, serving a bit more than 7 years is almost the median,

> mean(W>=7.5)
[1] 0.424812

(42% of the Popes did stay at least 7 years in charge) or we can look at the histogram,

> hist(W,breaks=0:35)

Unfortunately, I could not find more detailed database (including the years of birth for instance) to start a life-table of Popes.

# Les députés sont-ils à l’image de la population

Beaucoup de choses ont été écrites sur le fait que les députés ne sont pas vraiment le reflet de la population, que ce soit en terme de profession, de sexe, d’origine, d’age, etc. La liste pourrait être longue. Il y a plusieurs mois, j’avais commencé à regarder le profil des députés, par age. En effet, le site http://assemblee-nationale.fr/ permet d’accéder à des données sur tous les députés, depuis la Révolution. Y compris leur date de naissance. En croisant ces données avec des données de population, par exemple via http://www.mortality.org/, on peut comparer la répartition des ages des députés, avec la répartition des ages de la population.

Pour les amateurs, le code pour récupérer les données (ou au moins les dates de naissance des députés) ressemble à

N=2002
URL=paste("http://www.assemblee-nationale.fr/
regle_nom=est&Nom=&departement=&
choixdate=intervalle&D%C3%A9butMin=01%2F01%2F",
N,"&FinMin=31%2F12%2F",N,"&Dateau=&legislature=",
s,"&choixordre=chrono&Rechercher=
Lancer+la+recherche",sep="")
HTML=scan(URL,what="character")

k=which(HTML=="class=\"titre\">Né")
vHTML=HTML[k:length(HTML)]
vk=which(substr(vHTML,1,7)==">&nbsp;")
liste=vHTML[vk]
naissance=liste[seq(1,length(liste),by=2)]
NAISSANCE=as.Date(substr(naissance,8,17),
"%d/%m/%Y")

Maintenant, pour être tout à fait honnête, je ne suis pas certain de ce qui est vraiment renvoyé, et j’ai des doutes que cela correspondent réellement à la requête faite. En effet, même si je demande à avoir la liste des députés après l’élection, j’ai trop de monde… mais peut-être est-ce du aux décès éventuels, et il est possible que l’ensemble des députés qui ont siégé pendant la mandature apparaissent dans le résultat de la requête.

Sur la figure suivante on voit, sur plusieurs élections depuis plus de 100 ans, comment les deux distributions se déforment, avec en rouge la distribution de l’age des députés, et en bleu, la distribution de la population française, dans son ensemble (population de plus de 18 ans)

Si on veut tout suivre sur un graphique, au lieu de se regarder une animation, on peut représenter les différents quantiles (10%, 25%, 75% et 90%, retenus sur la population de plus de 18 ans, et l’age médian, au centre), avec la population française l’année de l’élection,

et l’ensemble des élus au parlement,

Si on veut faciliter la comparaison, on peut se contenter de visualiser l’évolution des ages moyens,

ou encore, du ratio (en % de différence) entre l’age moyen des députés, et celui de l’ensemble de la population.

Sur ce graphique, on voit que depuis 30 ans, l’age moyen des députés croit plus vite que celui de la population: la population français vieilli, mais moins que ses députés… La gérontocratie perdure donc en France. En espérant que cela ne débouche pas sur le clash générationnel que l’on semble observer ces temps-ci au Québec…