Introduction to Using Regular Expressions in R

R and Regular Expressions

Some people, when confronted with a problem, think “I know, I’ll use regular expressions.” Now they have two problems. – Jamie Zawinski, courtesy of Jeffrey Friedl’s blog

This is a blog post on using regular expressions in R. I am sure there are plenty of others out there with the same information. However, this is also an exercise for me to see how hard it is to change my knitr .Rnw files into markdown and then into HTML. It turns out that most of the work can be done by running pandoc over the LaTeX file I get from knitting my .Rnw file. The rest I did manually.

What are regular expressions?

Regular expressions provide a powerful and sophisticated way of matching patterns in text. For example with regular expressions I can:

  • Find a word or a series of letters

  • Do wild-card searches

  • Match patterns at the start or the end of a line

  • Make replacements in text based on the match

A simple example

Very often when people read about regular expressions they do not grasp the power, thinking “I could do that without using regular expressions at all.”

Here is something I did the other day. I had a file whose lines consisted of the names of files containing R code. That is, I had twenty lines of text with the end of each line ending in .R. I needed to

  1. insert export( at the start of each line, and

  2. replace the .R at end of each line with )

You can probably think of a way of doing this with a mixture of find and replace, and manual insertion. That is all well and good if you only have 20 lines. What if you had 20,000 lines?

I did this in Notepad++ by matching the regular expression ^(.*)\.[rR]$ and replacing it with export(\1)

Before After
#onewayPlot.R# \export(onewayPlot)
autocor.plot.R \export(autocor.plot)
boxqq.r \export(boxqq)
boxqq.r~ \export(boxqq)
ciReg.R \export(ciReg)
cooks20x.R \export(cooks20x)
crossFactors.R \export(crossFactors)
crossFactors.R~ \export(crossFactors)
crosstabs.R \export(crosstabs)
eovcheck.R \export(eovcheck)
estimateContrasts.R \export(estimateContrasts)
estimateContrasts1.R \export(estimateContrasts1)
estimateContrasts2.R \export(estimateContrasts2)
freq1way.r \export(freq1way)
freq1way.r~ \export(freq1way)
getVersion.R \export(getVersion)
interactionPlots.R \export(interactionPlots)
layout20x.R \export(layout20x)
levene.test.R \export(levene.test)
levene.test.R~ \export(levene.test)
multipleComp.R \export(multipleComp)
normcheck.R \export(normcheck)
onewayPlot.R \export(onewayPlot)
onewayPlot.R~ \export(onewayPlot)
pairs20x.R \export(pairs20x)
pairs20x.R~ \export(pairs20x)
predict20x.R \export(predict20x)
predict20x.R~ \export(predict20x) \export(
residPlot.R \export(residPlot)
rowdistr.r \export(rowdistr)
  • Regular expressions are powerful way of describing patterns that we might search for or replace in a text document

  • In some sense they are an extension of the wildcard search and replace operations you might carry out in Microsoft word or a text editor.

  • To the untrained eye they look like gobbledygook!

  • Most programming languages have some form of regular expression library

  • Some text editors, such as Emacs, Notepad++, RStudio, also have regular expressions

  • This is very useful when you don't need to write a programme

  • The file utility grep uses regular expressions to find occurrences of a pattern in files

  • Mastering regular expressions could take a lifetime, however you can achieve a lot with a good introduction

  • A couple of very good references are:

    • Jeffery Freidl's Mastering Regular Expressions, (2006), 3rd Edition, O'Reilly Media, Inc.
    • Paul Murrell's Introduction to Data Technologies, (2009), Chapman & Hall/CRC Computer Science & Data Analysis.
    • This entire book is on the web:, but if you use it a lot you really should support Paul and buy it
    • Chapter 11 is the most relevant for this post

Tools for regular expressions in R

We need a set of tools to use regular expressions in something we understand — i.e. R. The functions I will make the most use of are

  • grepl and grep

  • gsub

  • gregexpr

Functions for matching

  • grep and grepl are the two simplest functions for pattern matching in R

  • By pattern matching I mean being able to either

    (i) Return the elements of, or indices of, a vector that match a set of characters (or a pattern)

    (ii) Return TRUE or FALSE for each element of a vector on the basis of whether it matches a set of characters (or a pattern)

grep does (i) and grepl does (ii).

Very simple regular expressions

At their simplest, a regular expression can just be a string you want to find. For example the commands below look for the string James in the vector of names names. This may seem like a silly example, but it demonstrates a very simple regular expression called a string literal, informally meaning match this string — literally!

names = c('James Curran', 'Robert Smith', 
          'James Last')
grep('James', names)
## [1] 1 3

Wild cards and other metacharacters

  • At the next level up, we might like to search for a string with a single wild card character

  • For example, my surname is sometimes (mis)spelled with an e or an i

  • The regular expression symbol/character for any character is the the full stop or period

  • So my regular expression would be Curr.n, e.g.

surnames = c('Curran', 'Curren', 'Currin',
grepl('Curr.n', surnames)
  • The character . is the simplest example of a regular expression metacharacter

  • The other metacharacters are [ ], [^ ], \, ?, *. +,{,}, ^, $, \<, \>, | and ()

  • If a character is a regular expression metacharacter then it has a special meaning to the regular expression interpreter

  • There will be times, however, when you want to search for a full stop (or any of the other metacharacters). To do this you can escape the metacharacter by preceding it with a double backslash \\.

  • Note that we only use two backslashes in R – nearly every other language uses a single backslash

  • Note that \\ followed by a digit from 0 to 9 has special meaning too, e.g. \\1


  • Whilst this example obviously works, there is a more sensible way to do this and that is to use the alternation or or operator |.

  • E.g.

grepl('Curr(a|e|i)n', c('Curran', 'Curren', 
                        'Currin', 'Curin'))
  • This regular expression contains two metacharacters ( and |

  • The round bracket ( has another meaning later on, but here it delimits the alternation.

  • We read (a|e|i) as a or e or i.

A bigger example – counting words

In this example we will use the text of Moby Dick. The R script below does the following things

  1. Opens a file connection to the text

  2. Reads all the lines into a vector of lines

  3. Counts the number of lines where the word whale is used

  4. Counts the number of lines where the word Ahab is used

## open a read connection to the Moby Dick 
## text from Project Gutenberg
mobyURL = ''
f1 = url(mobyURL, 'r')

## read the text into memory and close
## the connection
Lines = readLines(f1)

Note: The code above is what I would normally do, but if you do it too often you get this nice message from Project Gutenberg

Don't use automated software to download lots of books. We have a limit on how fast you can go while using this site. If you surpass this limit you get blocked for 24h.

which is fair enough, so I am actually using a version I stored locally.

## Throw out all the lines before 
## 'Call me Ishmael'
i = grep('Call me Ishmael', Lines)
Lines = Lines[-(1:(i - 1))]

numWhale = sum(grepl('(W|w)hale', Lines))
numAhab = sum(grepl('(A|a)hab', Lines))

cat(paste('Whale:', numWhale, ' Ahab:', 
           numAhab, '\n'))
## Whale: 1487  Ahab: 491

Note:I am being explicit here about the capitals. In fact, as my friend Mikkel Meyer Andersen points out, I do not have to be. grep, grepl, regexpr and gregexpr all have an argument which can be set to TRUE. However I did want to highlight that generally regular expressions are case sensitive.

This programme will count the number of lines containing the words whale or Ahab but not the number of occurrences. To count the number of occurrences, we need to work slightly harder. The gregexpr function globally matches a regular expression. If there is no match, then gregpexr returns -1. However, if there is one or match, then gregexpr returns the position of the match, and the length of the match for each matching instance. For example, we will look for the word the in the following three sentences:

s1 = 'James is a silly boy'
s2 = 'The cat is hungry'
s3 = 'I do not know about the cat but the dog is stupid'

## Set the regular expression
## Note: This would match 'There' and 'They' as
## well as others for example 
pattern = '[Tt]he'
## We will store the matches so that we 
## can examine them in turn
m1 = gregexpr(pattern, s1)
m2 = gregexpr(pattern, s2)
m3 = gregexpr(pattern, s3)

There are no matches in the first sentence so we expect gregexpr to return -1

## [[1]]
## [1] -1
## attr(,"match.length")
## [1] -1
## attr(,"useBytes")
## [1] TRUE

which it does.

In the second sentence, there is a single match at the start of the sentence

## [[1]]
## [1] 1
## attr(,"match.length")
## [1] 3
## attr(,"useBytes")
## [1] TRUE

This result tells us that there is a single match at character position 1, and that the match is 3 characters long.

In the third example there are two matches, at positions 21 and 33, and they are both 3 characters long

## [[1]]
## [1] 21 33
## attr(,"match.length")
## [1] 3 3
## attr(,"useBytes")
## [1] TRUE

So in order to count the number of occurences of a word we need to use gregexpr and keep all of those instances where the result is not -1 and then count, using the length function the number of matches, e.g.

## count the number of occurences of whale
pattern = '(W|w)hale'
matches = gregexpr(pattern, Lines)

## if gregexpr returns -1, then the number of 
## matches is 0 if not, then the number of 
## matches is given by length
counts = sapply(matches, 
                  if(x[1] == -1) 

cat(paste('Whale:', sum(counts), '\n'))
## Whale: 1564

Character classes/sets or the [ ] operator

  • Regular expression character sets provide a simple mechanism for matching any one of a set of characters

  • For example [Tt]he will match The and the

  • The real strength of character sets is in its special range and set operators

  • For example the regular expressions:

    • [0-9] will match any digit from 0 to 9
    • [a-z] will match any lower case letter from a to z
    • [A-Z0-9] will match any upper case letter from A to Z or any digit from 0 to 9 and so on
  • You may initially think that character sets are like alternation, but they are not. Character sets treat their sets as an unordered list of characters

    • So (se|ma)t will (fully) match set
    • But mat, [sema]t will not
    • Alternatively [sema]t will match st,at, mt and at
  • The POSIX system defines a set of special character classes which are supported in R and can be very useful. These are

[:alpha:] Alphabetic (only letters)
[:lower:] Lowercase letters
[:upper:] Uppercase letters
[:digit:] Digits
[:alnum:] Alphanumeric (letters and digits)
[:space:] White space
[:punct:] Punctuation
  • The regular expression [[:lower:]] will help you capture accented lower case letters like &agrave, &eacute, or &ntilde whereas [a-z] would miss all of them

  • You may think this is uncommon, but optical character recognition (OCR) text often has characters like this present

Negated character sets – [^...]

  • Negated character sets provide a way for you to match anything but the characters in this set

  • A very common example of this is when you want to match everything between a set of (round) brackets

  • E.g. The regular expression ([^)]) would match any single character between a pair of round brackets

Matching more than one character — quantifiers

Another common thing we might do is match zero, one, or more occurrences of a pattern. We have four ways to do this

  1. ? means match zero or one occurrences of the previous pattern
  2. * means match zero or more occurrences of the previous pattern

  3. + means match one or more occurrences of the previous pattern

  4. {a,b} means match from a to b occurrences of the previous pattern*

  5. b may be omitted so that {a,} means match a or more occurrences of the previous pattern

  6. b and the comma may be omitted so that {a} means match exactly a occurences of the previous pattern

Continuing with our misspelling example, I would like a way of picking up all of the possibilities of misspelling my surname. Variants I've seen are Curren, Currin, Curin, Curan, Curen, Curn and even Karen!

If I wanted to construct a regular expression to match all of these possibilities I need to match (in this order):

  1. a C or a K

  2. a u or an a

  3. one or two occurence of r

  4. zero or more occurrences of e or i

  5. and finally an n

This is how I do this with regular expressions

pattern = '[CK](u|a)r{1,2}(e|i)*n';
badNames = c('Curren', 'Currin', 'Curin', 
             'Curan', 'Curen', 'Curn', 'Karen')
grepl(pattern, badNames)

Notice how the regular expression didn't match Curan. To fix the code so that it does match we need to change the set of possible letters before the n from (e|i) to allow a as a possibility, i.e. (a|e|i) or alternatively [aei]

pattern1 = '[CK](u|a)r{1,2}(a|e|i)*n'
pattern2 = '[CK](u|a)r{1,2}[aei]*n'
badNames = c('Curren', 'Currin', 'Curin', 'Curan', 
             'Curen', 'Curn', 'Karen')
grepl(pattern1, badNames)
grepl(pattern2, badNames)

Anchors — matching a position

  • The metacharacters ^, $, \< and \> match positions

  • ^ and $ match the start and the end of a line respectively

  • \< and \> match the start and end of a word respectively

  • I find I use ^ and $ more often

  • For example

    • ^James will match "James hates the cat" but not "The cat does not like James"
    • cat$ will match "James hates the cat" but not "The cat does not like James"

Summary — R functions for matching regular expressions

The functions grep and regexpr are the most useful. Loosely,

  • grep tells you which elements of your vector match the regular expression

  • whereas regexpr tells you which elements match, where they match, and how long each match is

  • gregexpr matches every occurrence of the pattern, whereas regexpr only matches the first occurrence

The example below shows the difference

poss = c('Curren', 'Currin', 'Curin', 'Curan', 
         'Curen', 'Curn', 'Karen')
pattern = '[CK](u|a)r{1,2}(i|e)*n'
grep(pattern, poss)
## [1] 1 2 3 5 6 7
gregexpr(pattern, poss)
## [[1]]
## [1] 1
## attr(,"match.length")
## [1] 6
## attr(,"useBytes")
## [1] TRUE
## [[2]]
## [1] 1
## attr(,"match.length")
## [1] 6
## attr(,"useBytes")
## [1] TRUE
## [[3]]
## [1] 1
## attr(,"match.length")
## [1] 5
## attr(,"useBytes")
## [1] TRUE
## [[4]]
## [1] -1
## attr(,"match.length")
## [1] -1
## attr(,"useBytes")
## [1] TRUE
## [[5]]
## [1] 1
## attr(,"match.length")
## [1] 5
## attr(,"useBytes")
## [1] TRUE
## [[6]]
## [1] 1
## attr(,"match.length")
## [1] 4
## attr(,"useBytes")
## [1] TRUE
## [[7]]
## [1] 1
## attr(,"match.length")
## [1] 5
## attr(,"useBytes")
## [1] TRUE

String substitution

  • Finding or matching is often only one half of the equation

  • Quite often we want to find and replace

  • This process is called string substitution

  • R has two functions sub and gsub

  • The difference between them is that sub only replaces the first occurrence of the pattern whereas gsub replaces every occurrence

  • Normal usage is quite straight forward. E.g.

poss = 'Curren, Currin, Curin, Curan, Curen, Curn and Karen'
pattern = '[CK](u|a)r{1,2}(i|e)*n'

sub(pattern, 'Curran', poss)
## [1] "Curran, Currin, Curin, Curan, Curen, Curn and Karen"
gsub(pattern, 'Curran', poss)
## [1] "Curran, Curran, Curran, Curan, Curran, Curran and Curran"

Back substitution

  • Abnormal usage is probably quite unlike anything you have seen before

  • One of the most powerful features of regular expressions is the ability to re-use something that you matched in a regular expression

  • This idea is called back substitution

  • Imagine that I have a text document with numbered items in it. E.g.

    1. James
    2. David
    3. Kai
    4. Corinne
    5. Vinny
    6. Sonika
  • How would I go about constructing a regular expression that would take each of the lines in my document and turn them into a nice LaTeX itemized list where the numbers are the list item markers?

  • The trick is to capture the numbers at the start of each line and use them in the substitution

  • To do this we use the round brackets to capture the match of interest

  • And we use the \\1 and \\2 backreference operators to retrieve the information we matched. E.g.

Lines = c('1. James', '2. David', '3. Kai', 
          '4. Corinne', '5. Vinny', 
          '6. Sonika')
pattern = '(^[0-9]\\.)[[:space:]]+([[:upper:]][[:lower:]]+$)'
gsub(pattern, '\\\\item[\\1]{\\2}', Lines)
## [1] "\\item[1.]{James}"   "\\item[2.]{David}"  "\\item[3.]{Kai}"    
## [4] "\\item[4.]{Corinne}" "\\item[5.]{Vinny}"   "\\item[6.]{Sonika}"

Note the double backslash will become a single backslash when written to file.

I actually used a regular expression with back substitution to format output for LaTeX in the file name example at the start of this post. My regular expression was the following:


and this was my back substitution expression

 \\verb!\1!  &  \\verb!\\export(\2)! \\\\

There is only a single \ in the back references because I just did this in the RStudio editor, not in R. Note how there are two back references, corresponding to two capture groups, one of which is nested inside the other. In nesting situtations like this, the capture groups are labelled in order from the outermost inwards.

String manipulation

We need two more things to finish this section

  • The ability to extract smaller strings from larger strings

  • The ability to construct strings from smaller strings

  • The first is called extracting substrings

  • The second is called string concatenation

  • We use the functions substr and paste for these tasks respectively


  • substr is very simple

  • Its arguments are

    • the string, x,
    • a starting position, start,
    • and an stopping position, stop.
  • It extracts all the characters in x from start to stop

  • If the alias substring is used then stop is optional

  • If stop is not provided then substring extracts all the characters from start to the end of x


substr('abcdef', 2, 4)
## [1] "bcd"
substring('abcdef', 2)
## [1] "bcdef"


paste is a little more complicated:

  • paste takes 1 or more arguments, and two optional arguments sep and collapse

  • sep is short for separator

  • For the following discussion I am going to assume that I call paste with two arguments x and y

  • If x and y are both scalars thenpaste(x,y) will join them together in a single string separated bya space, e.g.

paste(1, 2)
## [1] "1 2"
paste('a', 'b')
## [1] "a b"
paste('a =', 1)
## [1] "a = 1"
  • If x and y are both scalars and you define sep to be "," say then paste(x, y, sep = ",") will join them together in a single string separated by a comma,
paste(1, 3, sep = ',')
## [1] "1,3"
paste(TRUE, 3, sep = ',')
## [1] "TRUE,3"
paste('a', 3, sep = ',')
## [1] "a,3"
  • If If x and y are both vectors and you define sep to be "," say then paste(x, y , sep = ",") will join each element of x with each element of y into a set of strings where the elements are separated by a comma, e.g.
x = 1:3
y = LETTERS[1:3]
paste(x, y, sep = ',')
## [1] "1,A" "2,B" "3,C"
  • If collapse is not NULL, e.g “-” then each of the strings will be pasted together into a single string.
x = 1:3
y = LETTERS[1:3]
paste(x, y, sep = ',', collapse = '-')
## [1] "1,A-2,B-3,C"
  • And of course you can take it as far as you like 🙂
x = 1:3
y = LETTERS[1:3]
paste('(', paste(x, y, sep = ','), ')', 
       sep = '', collapse = '')
## [1] "(1,A)-(2,B)-(3,C)"


Mikkel also points out that paste0 is a shortcut to avoid specifying sep = "" everytime

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If you are not a statistician, this cartoon is mocking statistical significance used in most frequentist hypothesis tests. If we use a threshold of \(\)P<0.05[/latex] then we accept that on average we will falsely declare there to be a significant departure from the null approximately one time in twenty ([latex]1/20 = 0.05[/latex]). In lay terms we will decide there is a difference when there really is not one time in twenty. Thanks Randall.

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New PS/4 game console disappointments

Last year around August, during all the hype about the new Xbox One and Sony’s PlayStation 4, I decided to pre-order both consoles. However, as the release date approached I realized that the number of titles available for both consoles was extremely limited with many studios aiming for a 2014 release date. Based on this I decided that I did not need approximately $NZD1,300 of electronics that were glorified Blue ray players. I should inform readers (he says optimistically) who are not based in New Zealand or Australia that we do not have access to Netflix in New Zealand, and nor do we have cable TV that integrates with either box. Voice activation and control of the Xbox One is also currently disabled in Australia and New Zealand. Microsoft won’t say why but one strongly suspects that the software simply cannot deal with the Australasian accent. The voice features of Android used to be hopeless for anyone who did not have an American accent. Google has spent considerable time working to remedy this, and now it mostly works.

So I held off until May of this year, and finally purchased the PS/4. I had originally thought I would buy the Xbox first because I have been a much heaver Xbox 360 gamer, but the only two new games that interested me were Watchdogs, available on all platforms, and Sucker Punch’s inFamous: Second Son, which is a PS/4 exclusive. The latter swung the decision as I have enjoyed playing both inFamous and inFamous 2, and, to be fair, with the exception of the idiotic spray-painting side “quests”, I enjoyed this iteration as well.

Watch Dogs

Watch Dogs, from Ubisoft Montreal, was majorly hyped at 2013’s E3 event. Billed as open world action-adventure title with driving, shooting, running and climbing (parkour), and hacking all being part of the game. In reality, they should have just stuck with driving. This game reminds me so much of Grand Theft Auto it is not funny. Many of the integral parts of the story line require the player to drive at high-speed, either in pursuit of someone, or to get to a specific location within a time limit. These are the exact things I loathe about GTA, but can forgive in GTA, because hey that game is called Grand Theft Auto. It is about cars, and you expect cars to be an integral part of the game. Personally, I like in this sort of game to jump, climb and precision shoot. Watch Dog’s lead protagonist, Aiden Pearce, can run, and climb in a very limited fashion, but not jump — he can fall down to a lower level, or vault a fence which is really just an extension of the running, but he cannot jump from say a boat to a pier.

The hacking part of the game is sort of interesting in that it can provide you a way of carrying out a mission without entering a building, but mostly it doesn’t. It lets you steal money from unsuspecting citizens, which Eurogamer points out you can’t even give to the numerous beggars around the city. It does not really help you when you are driving, except in the odd case where you can cause bollards to rise out of the road to stop a car, or raise a bridge. Eurogamer states, again correctly, that it is a game about hacking where hacking quickly becomes incidental.

A number of side missions require you to interfere with, or stop, criminal activity. Fine, I have no problem with that, but then why do most of these missions require me to tackle someone who is shooting at me with an automatic weapon? Overall, I am extremely disappointed with this game and will probably not bother to complete it.

No DLNA support, no third party media

Both the Xbox 360 and Sony PlayStation 3 could see and play all the media content hosted on my Windows box in the basement. This was a nice feature because it meant I could play media (music, video, browse photos) without buggering around with discs, portable hard drives, or flash drives. So you might say, okay minor inconvenience, stick it on a flash drive and away you go. No dice. Sony, in its infinite wisdom, has decided that “Unfortunately as noted pre-launch of the PS4, you cannot play movies on the PS4 off of an external hard drive or memory stick. However this is something that is looking to be added in the future with an update.” One question – “Why?” It is hardly a revolutionary feature — my TV does it. The inability to insert an audio disc seems like a kick in the face to the consumer, and overall Sony, you are just pissing off a lot of people who give your company considerable revenue.

Final thoughts

Its possible that Sony may fix some of these issues, but I am not holding my breath. I am hoping that Far Cry 4, releasing in November, is as good as the previous titles in the series.

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Python and statistics – is there any point?

This semester I gave my graduate student class a project. The brief was relatively simple: implement the iteratively reweighted least squares (IRLS) algorithm to perform a simple (single covariate) logistic regression in Python. Their programmes were supposed to be able to read data in from a text file, perform the simple matrix algebra and math needed to carry out the IRLS computation and return some formatted output – similar to that you would get from R’s summary.glm function. Of course, you do not need matrix algebra to do this, but the idea was for the students to learn a bit of mathematical statistics that they had not seen before.  On the IRLS front, they were allowed to use a simple least squares routine like numpy’s linalg.lstsq and some of numpy’s simple matrix operators, but expressly forbidden from simply loading pandas or statsmodels and using the generalized linear models functions contained therein.

I thought this sounded like a straightforward enough task. The students divided themselves into pairs to work on it, and they had 13 weeks to complete the task.

The kicker was that I did not provide any instruction, either in Python or in the IRLS algorithm. An aim of the project was to simulate the situation where someone asks you to solve a problem, and you have to go and do some research to do it. Their first task was to complete 100 exercises on as a reasonable introduction to a language none of them had seen before.

Problems – versions

There are two major versions of Python in the wild, 2.7 and 3.4. Codeacademy teaches using version 2.7. One fundamental difference between 2.7 and 3.4 is the syntax of the print function. All of my students are users of R, to varying levels of skill. When they go to install R at home, they know to go to the CRAN website, or a mirror, and download the current, stable release of R. If they followed this policy, as I did myself, then they would have installed Python 3.4 and found that the way they were taught to use print by Codeacademy does not work, without any sort of helpful “That syntax has been depricated. Python 3 onwards uses the syntax…” This is not the only issue, with the way Python 3.4 handles execution of loops over numbered ranges being another example of a fundamental difference.

Problems – platform issues

Most students at my institution use Windows, especially at home. There is some Mac penetration, and Linux is virtually non-existent (these are statistics students, not computer science remember). The official Python installers work perfectly well under Windows in my experience. However, then we come to the issue of installing numpy. The official advice from the numpy website seems to be “download a third party version of Python which already has it.” For students who come from a world where a package can be installed by going to a menu, this is less than useful. The common advice from the web is that “there is no official release of numpy 1.8.1 for Python 2.7 or higher for Windows” but that you can download it and install it from a the builds very thoughtfully provided by Christoph Gohlke at UC Irvine here. Christoph’s builds work fine, but again, for something that seems, at least from the outside, very mainstream in the Python community should the user have to go to this level of effort?

Problems – local installations

Like any instructor, I face the issue that a number of my students have no option but to use the computer laboratories provided for them by the university. This means that we encounter the issue of local installation of libraries for users. Most, if not all, R packages from CRAN can be installed in a local library. As far as I can tell, this is not true for a Windows installation of Python. I am happy to be corrected on this point. The aforementioned Python binaries come with proper Windows installers, which want to install into the Python root directory, something students do not have permission to do. If I had realized this problem in December of last year, I could have asked the admins to pre-install it for all users, however, given I only formulated the problem in February, it was just a tad too late.

Would I do it again?

I might, but there would have to be serious efforts to resolve the problems listed above on my part. It also would not solve problems of students trying to set up Python at home, and I do not feel like hand-holding people through an installation process. My initial plan had been to try Javascript. I may return to this idea.

I would be the first to admit that I am not a Python user, but I am an experienced programmer with over thirty years of experience in at least a dozen different languages, and on multiple platforms. I know many people find Python a very useful language for their scientific computing, and I am not attempting to bad mouth the language – it seems a decent enough language with the constructs and functionality that I would expect to find in any modern language – but I do not think there is much incentive for a statistician to move away from R, or an R/C++ combination when raw compute power is required.

I am glad that my students experienced programming in a non-vectorized language. R does give a distorted perspective on programming with regards to its handling of vectors, and I think it is beneficial for students to learn about flow structures for element-wise computation.


Nat Dudley has made the suggestion I used on online IDE like

Second updates

Despite the difficulties, nearly all of my students have managed to complete the task, and some have done an exceptionally good job, even adding in the ability to parse R-like formulae.

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Bayesian modelling of left-censored data using JAGS

Bayesian modelling of left-censored data using JAGS

Martyn Plummer's JAGS very helpfully provides us with a way to model censored data through the use of the dinterval distribution. However, almost all of the examples that one finds on the web are for right censored data. The changes to model left censored data are not major, but I do think they warrant a) a post/page of their own and b) hopefully an easy-to-understand example.

Left-censored data arises very commonly when dealing with detection limits from instrumentation. In my own work, I often end up involved in the modelling of data derived from electropherograms.

I will start by generating some left censored data. For simplicity I am going to assume that my data is exponentially distributed, with a true rate of 1.05 (\(\lambda = 1.05\)), and a detection/censoring threshold at log(29). This means that approximately 97.1% of my data (on average) will not exceed my detection threshold. This may seem extreme, but it is the kind of setup that is common in my own work.

x = rexp(10000, rate = 1.05)

## set all the censored values to NA's
x[x < log(29)] = NA

This gives us a data set of size 10,000 with 9,691 values that fall below the detection threshold.

It can be a useful check, if feasible, to see what the maximum likelihood estimate is. We can do this in R using the optim function

## define the log-likelihood
logLik = function(lambda){
  isCensored =
  nCensored = sum(isCensored)
  LOD = log(29)

  ll = sum(dexp(x[!isCensored], rate = lambda, 
                log = TRUE)) + 
       nCensored  * pexp(LOD, rate = lambda, 
                         log = TRUE)

  ## return the -ve value because optim 
  ## is a minimizer

fit = optim(0.5, logLik, method = "Brent", 
            lower = 0, upper = 10, 
            hessian = TRUE)
fisherInfo = 1/fit$hessian
sigma = sqrt(fisherInfo)
upper = fit$par + 1.96 * sigma
lower = fit$par - 1.96 * sigma
interval = c(lower, fit$par, upper)
names(interval) = c("lower", "MLE", "upper")
## lower   MLE upper 
## 1.001 1.032 1.064

So the 95% confidence interval contains our true value which is always a good start!

The trick, if there is any, to dealing with left-censored data in JAGS is to make sure that your indicator variable tells JAGS which variables are above the detection threshold.

So in the next step I will set up the list that contains my data.

bugsData = list(N = length(x),
                isAboveLOD = ifelse(!, 
                                    1, 0),
                x = x,
                LOD = rep(log(29), length(x)))

There are two points to make about the preceding code. Firstly the variable isAboveLOD uses the NA status of the data. If you have not recoded your censored values to NA then obviously this will not work. Secondly, there is a limit of detection vector LODregardless of whether the observation is below the limit of detection or not.

Next we need to set up some intial values. The key here is setting not only an initial value for \(\lambda\), but initial values for the observations that have been censored.

bugsInits = list(list(lambda = 0.5, 
                      x = rep(NA, length(x))))

## set the missing values to 
## random variates from
## U(0, log(29))
nMissing = sum(!bugsData$isAboveLOD)
bugsInits[[1]]$x[!bugsData$isAboveLOD] = 
           runif(nMissing, 0, log(29))

I have chosen uniform random variates between zero and the limit of detection (log(29)) as initial values for my censored data.

Now we need to set up our JAGS model. I will store this in a string, and then use writeLines to write this to disk.

modelString = "
    for(i in 1:N){
      isAboveLOD[i] ~ dinterval(x[i], LOD[i])
      x[i] ~ dexp(lambda)

    lambda ~ dgamma(0.001, 0.001)

writeLines(modelString, con = "model.bugs.R")

Note that I have used a vague gamma prior for \(\lambda\). This is not especially noteworthy, except from the point of view about being explicit about what I have done. dinterval returns 0 if x[i] < LOD[i] and a 1 otherwise. Many people who try to use dinterval often get this cryptic error Observed node inconsistent with unobserved parents at initialization. This can happen for two reasons. Firstly the indicator variable can be incorrectly set, that is the observations above the limit of detection have been coded with a 0 instead of a 1 and vice versa for the censored observations. Secondly, the error can occur because the initial values for the censored observations are outside of the censored interval.

We now have the three components necessary to fit our model: a list containing the data, a list of initial values, and a BUGS model. Firstly we initialize the model

## Loading required package: coda
## Loading required package: lattice
## Linked to JAGS 3.3.0
## Loaded modules: basemod,bugs
jagsModel = jags.model(file = "model.bugs.R", data = bugsData, inits = bugsInits)
## Compiling model graph
##    Resolving undeclared variables
##    Allocating nodes
##    Graph Size: 30003
## Initializing model

Next, we let the model burn-in for an arbitrary period. I will use a burn-in period of 1,000 iterations.

update(jagsModel, n.iter = 1000)

And finally, we take a sample (hopefully) from the posterior distribution of the parameter of interest, \(\lambda\)

parameters = c("lambda")
simSamples = coda.samples(jagsModel, 
                          variable.names = parameters, 
                          n.iter = 10000)
stats = summary(simSamples)

We can obtain a 95% confidence interval by using the posterior mean and standard error, i.e.

mx = stats$statistics[1]
sx = stats$statistics[2]
ci = mx + c(-1, 1) * 1.96 * sx
## [1] 1.002 1.065

or we can get a 95% credible interval by using the posterior quantiles, i.e.

ci = stats$quantiles
##  2.5% 97.5% 
## 1.002 1.065

Both intervals contain the true value of 1.05 and have a fairly reasonable estimate of it as well. They also are very close to the ML interval.

I hope this clears things up for someone.

Credit where credit is due

I could not have written this without following the right censored example provided by John Kruschke here, and from reading Martyn Plummer's presentation here.

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