# All my life

Here is a little script to show you your life. In weeks. Each point is a week. Each black point is a week that you have already spent. The number of weeks corresponds to 90 years, which is higher than the current life expectancy anywhere in the world.

Have fun.

birthdate <- "1973-05-25"
seq1 <- 1:(as.numeric((Sys.Date() - as.Date(birthdate)))/7) - 1
seq2 <- 1:(90*52) - 1
plot(NULL, xlim=c(0,53), ylim=c(91, 0), bty="n", xlab="Week of the year", ylab="Age")
points(seq2 %% 52, floor(seq2 / 52), pch=15, col="grey")
points(seq1 %% 52, floor(seq1 / 52), pch=15)


# Copy text from images

All the elements were there. Algorithms freely available; implementations ready to download, for free. It just took one clever person to FINALLY make it: a seamless way to copy text from images. At least in Chrome, for now, but since it is Open Source, I guess it is just a matter of time and we will see it as a build-in feature in Firefox and other browsers. For now, use the Project Naphta Chrome browser extension. The main project page is also an interesting read.

Words on the web exist in two forms: there’s the text of articles, emails, tweets, chats and blogs— which can be copied, searched, translated, edited and selected— and then there’s the text which is shackled to images, found in comics, document scans, photographs, posters, charts, diagrams, screenshots and memes. Interaction with this second type of text has always been a second class experience, the only way to search or copy a sentence from an image would be to do as the ancient monks did, manually transcribing regions of interest.

(from the Naphta web site)

It works really nice, assuming that you select the “English tesseract” option from the “Languages” sub-menu — the off-line javascript implementation of the OCRad is not very effective, in contrast to the cloud-based tesseract service.

# Rotating movies with RGL

Learned another thing today: it is very simple to create animated GIFs in the rgl package with the build-in functions spin3d, play3d and movie3d.

library( pca3d )
data(metabo)
pca <- prcomp( metabo[,-1], scale.= TRUE )
pca3d( pca, group= metabo[,1] )
rot <- spin3d( axis= c( 0, 1, 0 ) )
movie3d( rot, duration= 12 )


Here is the result:

# Check credit card numbers using the Luhn algorithm

You can find a nice infographic explaining how to understand the credit card number here.

Credit card numbers include, as the last digit, a simple checksum calculated using the Luhn algorithm. In essence, we add up the digits of the card number in a certain way, and if the resulting sum divided by 10 is not an integer (i.e., the reminder of dividing by 10 is zero), then the number is invalid (the reverse is not true; a number divisible by 10 can still be invalid).

This can be easily computed in R, although I suspect that there might be an easier way:

checkLuhn <- function( ccnum ) {

# helper function
sumdigits <- function( x )
sapply( x, function( xx )
sum( as.numeric( unlist( strsplit( as.character( xx ), "" ))))
)
# split the digits
ccnum <- as.character( ccnum )
v <- as.numeric( unlist( strsplit( ccnum, "" ) ) )
v <- rev( v )

# indices of every second digit
i2 <- seq( 2, length( v ), by= 2 )
v[i2] <- v[i2] * 2
v[ v > 9 ] <- sumdigits( v[ v > 9 ] )

if( sum( v ) %% 10 == 0 ) return( TRUE )
else return( FALSE )

}



# Sample size / power calculations for Kaplan-Meier survival curves

The problem is simple: we have two groups of animals, treated and controls. Around 20% of the untreated animals will die during the course of the experiment, and we would like to be able to detect effect such that instead of 20%, 80% of animals will die in the treated group, with power 0.8 and alpha=0.05. Group sizes are equal and no other parameters are given.

What is the necessary group size?

I used the ssizeCT.default function from the powerSurvEpi R package. Based on the explanation in the package manual, this calculates (in my simple case) the required sample size in a group as follows:

$n = \frac{m}{p_E + p_C}$

where $p_E$ and $p_C$ are, respectively, probabilities of failure in the E(xpermiental) and C(ontrol) groups. I assume that in my case I should use 0.8 and 0.2, respectively, so $n=m$. The formulas here are simplified in comparison with the manual page of ssizeCT.default, simply because the group sizes are identical.

$m$ is calculated as

$m=\big(\frac{RR+1}{RR-1}\big)^2(z_{1-\alpha/2}+z_{1-\beta})^2$

RR is the minimal effect size that we would like to be able to observe with power 0.8 and at alpha 0.05. That means, if the real effect size is RR or greater, we have 80% chance of getting a p-value smaller than 0.05 if the group sizes are equal to $m$. To calculate RR, I first calculate $\theta$, the hazard ratio, and for this I use the same approximate, expected mortality rates (20% and 80%):

$\theta = \log(\frac{\log(0.8)}{\log(0.2)}) = -1.98$

Since $RR=exp(\theta)=0.139$; thus $m=18.3$. This seems reasonable (based on previous experience).

# Copy-paste and R interactive sessions

I wanted to have a way of easily copying R commands from my interactive sessions. I often work interactively, while creating a pipeline or lab-book like readme in a separate file.

Of course, you can just copy anything that is on the screen and paste it. However, I want to have my readme file to be directly executable, so generally I want only to copy the commands. The solution was to write a small function, called xclip, which copies an arbitrary number of lines from history into the X window system primary clipboard. Naturally, it is not compatible with Windows or MacOS. However, I can copy now a fragment of recent history with one command and one mouse click (since the primary X window system clipboard runs without shortcuts).

The function actually is only a wrapper around the little xclip command utility, available for Linux.

xclip <- function (n = 5)
{
t <- tempfile()
savehistory(file = t)
system(paste0("tail -n ", n, " ", t, "|xclip"))
}


# riverplot

Prompted by this cross-validated discussion, I have created the riverplot package. Here is a minimal gallery of the graphics produced by the package:

Here is an example which recreates the famous Minard plot:

So, how to do these figures:

First, you need to create a specific riverplot object that can be directly plotted. (Use riverplot.example to generate an example object). Here, I show how to recreate the Minard plot using the provided data. makeRiver, the function that will create the object necessary for plotting, will use data frames to input information about nodes and edges, but we must use specific naming of the columns:

library( riverplot )
data( minard )
nodes <- minard$nodes edges <- minard$edges
colnames( nodes ) <- c( "ID", "x", "y" )
colnames( edges ) <- c( "N1", "N2", "Value", "direction" )


Now we can add some style information to the “edge” columns, to mimick the orignal Minard plot:

# color the edges by troop movement direction
edges$col <- c( "#e5cbaa", "black" )[ factor( edges$direction ) ]
# color edges by their color rather than by gradient between the nodes
edges$edgecol <- "col" # generate the riverplot object river <- makeRiver( nodes, edges )  The makeRiver function reads any columns that match the style information (like colors of the nodes) and uses it to create the river object. The river object is just a simple list, you can easily view and manipulate it — or create it with your own functions. The point about makeRiver is to make sure that the data is consistent. Once you have created a riverplot object with one of the above methods (or manually), you can plot it either with plot(x) or riverplot(x). I enforce the use of lines, and I also tell the plotting function to use a particular style. The default edges look curvy. style <- list( edgestyle= "straight", nodestyle= "invisible" ) # plot the generated object plot( river, lty= 1, default.style= style ) # Add cities with( minard$cities, points( Longitude, Latitude, pch= 19 ) )
with( minard\$cities, text( Longitude, Latitude, Name, adj= c( 0, 0 ) ) )