#r data manipulation

extract year, month, day… from date

create vector of dates in standard format

my.dates <- as.Date(format(ISOdatetime(2000:2009,1:10,1:5,0,0,0),"%Y-%m-%d")) my.dates

[1] "2000-01-01" "2001-02-02" "2002-03-03" "2003-04-04" "2004-05-05" [6] "2005-06-01" "2006-07-02" "2007-08-03" "2008-09-04" "2009-10-05"

(make sure your object is of date fromat, check it with str(your.object)

extract years using format:

my.years <- format(my.dates,"%Y") # %y without century my.years

[1] "2000" "2001" "2002" "2003" "2004" "2005" "2006" "2007" "2008" "2009"

or months

my.months <- format(my.dates,"%m") my.months

[1] "01" "02" "03" "04" "05" "06" "07" "08" "09" "10"

or names of month (in current local)

my.months <- format(my.dates,"%b") # %B for long form my.months

[1] "Jan" "Feb" "Mär" "Apr" "Mai" "Jun" "Jul" "Aug" "Sep" "Okt"

or days (of month)

my.days <- format(my.dates,"%d") my.days

[1] "01" "02" "03" "04" "05" "01" "02" "03" "04" "05"

or week day (in current local)

my.days <- format(my.dates,"%a") # %A for long form, %w (0-6) of %u (1-7) for number my.days

[1] "Sa" "Fr" "So" "Fr" "Mi" "Mi" "So" "Fr" "Do" "Mo"

or day of year

my.days <- format(my.dates,"%j") my.days

[1] "001" "033" "062" "094" "126" "152" "183" "215" "248" "278"

or week of year (sunday as first day of the week)

my.weeks <- format(my.dates,"%U") # %W for monday = first day of the week my.weeks

[1] "00" "04" "09" "13" "18" "22" "27" "30" "35" "40"

local format of date

my.days <- format(my.dates,"%x") my.days

subscripting with positive and negative indices

character subscripting (vectors and lists)

logical subscripting

subsetting with logical indices

elements corresponding to a TRUE in the subscript vector are selected, those corresponding to a FALSE are excluded, NAs in the the logical subscript vector produce NAs

my.vector <- letters[1:10] # vector of a...j my.vector

[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"

my.index <- rep(c(TRUE,FALSE),5) # the subscript vector (of length 10) my.index

[1] TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE

my.vector[my.index]

[1] "a" "c" "e" "g" "i"

if the subscript vector consists of less elements than the subscripted vector, it is repeated as many times as necessary (warning if the the vector length is not a multiple of the length of the subscript vector)

short.index <- c(T,F) # T is the same as TRUE, F as FALSE short.index

[1] TRUE FALSE

my.vector[short.index] # has the same result as above

[1] "a" "c" "e" "g" "i"

if the subscript vector is longer than the vector for every TRUE or NA entry without target a NA is produced

short.vector <- letters[1:3] short.vector

[1] "a" "b" "c"

my.index <- rep(c(T,F,NA),2) my.index

[1] TRUE FALSE NA TRUE FALSE NA

short.vector[my.index]

subscripting with positive and negative indices

character subscripting (vectors and lists)

logical subscripting

subsetting with character indices - vectors and lists

extract elements of named vectors, lists

named.x <- 1:10 # produce a vector containing the numbers 1...10 named.x # has no names yet

[1] 1 2 3 4 5 6 7 8 9 10

if the object has no names, NAs are produced

named.x["one"]

[1] NA

name the vector:

names(named.x) <- c("one","two","three","four","five","six","seven","eight","nine","ten") named.x

one two three four five six seven eight nine ten 1 2 3 4 5 6 7 8 9 10

now it works

named.x["one"]

one 1

if names are duplicated only the first match is returned (the value with the lowest index)

names(named.x)<-rep(c("one","two","three","four","five"),rep(2,5)) named.x

one one two two three three four four five five 1 2 3 4 5 6 7 8 9 10

named.x[c("one","two","three")]

one two three 1 3 5

NA is returned if a name do not match any element

named.x["twenty"]

NA

NA is returned if a subscript is character NA

named.x[c("one",NA,"three")]

one three 1 NA 5

to find all occurences one can use %in% and then subset with the logical vector (you can do this in one step)

named.x[names(named.x) %in% c("one","two")]

subscripting with positive and negative indices

character subscripting (vectors and lists)

logical subscripting

subsetting

creating a list for examples

myl <- list(one=10,two=20,three=30) myi <- "one"

one

subsetting is carried out with three different operators:

dollar: $

square bracket: [

double square bracket: [[

these are generic functions

there are four basic types of subscript indices

positive integers

negative integers

logical vectors

character vectors

the first (left-most) index moves the fastest, the right-most index is the slowest (so a matrix is filled column by column)

the square bracket

the returned value is of the same type of the value it is applied to

myl[c(1,2)]

10 20

evaluate its second argument

myl[myi]

10

does not use partial matching when extracting named elements

myl["on"]

cannot be used for subscripting environments

the double square bracket

extract a single value

myl[[1]]

10

evaluate its second argument

myl[[myi]]

10

does not use partial matching when extracting named elements

myl[["on"]]

the dollar

extract a single value

myl$two

20

does not evaluate its second argument

myl$myi

uses partial matching extracting named elements

myl$on

10

cannot be used for subscripting atomic vectors

subsetting with positive indices

vector of positive integer values to indicate the set to be extracted

myv <- 1:10 myi <- c(2,4,7) myv[myi] myv[c(2,4,7)]

[1] 2 4 7 [1] 2 4 7

if the subscript is larger than the length of the vector, NAs are produce

myv[c(1,11,3)]

[1] 1 NA 3

NA as subscript produces also a NA

myv[c(1,11,NA)]

[1] 1 NA NA

zero as subscript will be ignored

myv[c(1,2,0,3)]

[1] 1 2 3

if the subscript vector has the length zero, the result is also of length zero

myi <- numeric() myv[myi]

integer(0)

example: select elements which have even numbered subscripts

v <- rnorm(20) myi <- seq(2,length(v),by=2) myi v[myi]

[1] 2 4 6 8 10 12 14 16 18 20 [1] 0.35380115 -0.02156840 -1.51804278 0.38278037 0.03867578 -1.25803279 [7] 0.62863255 0.07111270 -0.73416837 0.18966622

subsetting with negative indices

subscript vector consists of negative integer values indicating the indices not to be extracted

zero as subscript will be ignored

myv[-c(1,2,0,3)]

[1] 4 5 6 7 8 9 10

NAs are not allowed

myv[-c(1,NA,3)]

Error in myv[-c(1, NA, 3)] : nur Nullen dürfen mit negativen Indizes gemischt werden

zero length subscript produces zero length result

myv <- 1:10 myi <- numeric() myv[-myi]

integer(0)

positive and negatives subscripts cannot be mixed

recode() - from package lordif

library(lordif) # maybe you have first to install the package (install.packages("lordif"))

creating example data

z <- sample(0:1, 25, replace=T) original <- c(0,1) new <- c("boy", "girl") z original new

z: 1 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 0 original: 0 1 new: "boy" "girl"

so z is a vector consisting of 1 and 0

original contains the distinct values of z

and new contains the values which should replace the original values in the following way: the first item of original is replaced by the first value of new, the second by the second etc pp

z <- recode(z, original, new) z

[1] "girl" "girl" "girl" "boy" "boy" "boy" "girl" "girl" "girl" "girl" [11] "boy" "boy" "girl" "girl" "girl" "girl" "boy" "girl" "girl" "girl" [21] "girl" "girl" "girl" "girl" "boy"

First we create a data frame, on which we can work

x <- rnorm(100) y <- sample(1:3, 100, replace=T) z <- sample(1:2, 100, replace=T) df <- data.frame(group=y,sex=z,sds=x) head(df)

group sex sds 1 1 2 1.3412663 2 1 1 0.8000326 3 2 2 0.0371029 4 1 2 0.2064684 5 1 2 1.6429816 6 3 1 1.3271138

In the following I use ddply() from the plyr-package, so if you did not install and/or load it you have to do it now - the command library() without any argument gives you a list of the installed packages:

install.packages("plyr") # install the package library(plyr) # load it

Here is the function for doing the t.tests:

t.test.plyr <- function(x, var, mean=0 ){ y <- rep(NA,10) y[6] <- nrow(x)[1] # count observations if(nrow(x) < 2) return(y) # exits if too less observations res <- t.test(x[var], mu=mean) # doing the test y[1] <- res$statistic # extract values of interest y[2] <- res$p.value y[3] <- res$estimate y[4] <- res$conf.int[1] y[5] <- res$conf.int[2] y[7] <- res$parameter y[8] <- res$method y[9] <- res$alternative y[10] <- res$null.value names(y) <- c("statistic","p.value","estimate","conf.int1", "conf.int2", "nobs","dof","method","alternative","null.value") y }

where t.test.plyr() is a function of: - x - the data frame - var - the variable which should be testet - mean - the mean of the t-test now we can use ddply in the following way:

result <- ddply(df, .(group, sex), t.test.plyr, "sds") result

group sex statistic p.value estimate 1 1 1 0.774917157596946 0.445379429071851 0.132214420988956 2 1 2 -0.415594591752269 0.68359099812399 -0.0987341821354776 3 2 1 0.130257003222609 0.899578855830346 0.0602600950757679 4 2 2 -0.0579045286518569 0.954965406753824 -0.0150105343694587 5 3 1 1.18044920537142 0.256202807853187 0.406567809727671 6 3 2 1.49215606704641 0.15126489039304 0.294865619363907 conf.int1 conf.int2 nobs dof method alternative 1 -0.218494854049804 0.482923696027717 27 26 One Sample t-test two.sided 2 -0.605109702463049 0.407641338192093 16 15 One Sample t-test two.sided 3 -1.00655416438508 1.12707435453662 9 8 One Sample t-test two.sided 4 -0.592608791299465 0.562587722560548 11 10 One Sample t-test two.sided 5 -0.327541518602652 1.14067713805799 16 15 One Sample t-test two.sided 6 -0.117342538637965 0.70707377736578 21 20 One Sample t-test two.sided null.value 1 0 2 0 3 0 4 0 5 0 6 0

If you want to change the mu in you t.test, add the argument mean:

result <- ddply(df, .(group, sex), t.test.plyr, "sds", mean=1) result

If there is no need or you do not want a data frame as result you can just use tapply(). tapply results in a list with as many elements as subsets, each element contains the results of each t.test (as list).

res <- tapply(df$sds, list(df$sex,df$group), t.test) res

1 2 3 1 List,9 List,9 List,9 2 List,9 List,9 List,9

So the first element of res contains the t.test results for sex=1 and group=1 etc pp

res[[1]]

> format(ISOdatetime(2000,1:12,1,0,0,0),"%b")

gives:

[1] "Jan" "Feb" "Mrz" "Apr" "Mai" "Jun" "Jul" "Aug" "Sep" "Okt" "Nov" "Dez"

where the month names depend on local settings. If you use %B instead of %b you get the non abbreviated months. How ever, there is also an easier way if you want the months in English: the vector

> month.name

gives

[1] "January" "February" "March" "April" "May" "June" [7] "July" "August" "September" "October" "November" "December"

or

> month.abb

gives

ddply

dataframe in dataframe out

example

first we create a data frame with the following columns: id, date, value

to create the id column we use rep(), where the first argument 1:5 (the vector (1,2,3,4,5)) is repeated five times, so we have a vector with length 25

the vector dates is generated in the following way:

generate a vector of random numbers between 1 and 1500 and length 25 (sample.int())

use as.Date with the origin argument, this converts the integers into Date, where the the origin is day 0 (in the case 0 is convert to the origin i.e. 2000-01-01, 1 is mapped to 2000-01-02 etc pp, so we get a vector of dates between 2000 and beginning 2005

vals are generated by the random numbers function rnorm() (normal distributed with mean 5 and standard deviation 1

last but not least put the three together in one data frame (and let us show the first lines)

id <- rep(1:5,5) dates <- as.Date(sample.int(1500, 25, replace=T), origin="2000-01-01") vals <- rnorm(25, mean=5, sd=1) df <- data.frame(id=id, date=dates, val=vals) head(df)

id date val 1 1 2001-12-31 5.778680 2 2 2002-08-02 6.982799 3 3 2002-04-23 5.925903 4 4 2000-08-03 3.527375 5 5 2002-08-29 5.239211 6 1 2003-01-28 5.118337

id encodes a person, date contains the day of the measurement and val the values

now we want to add a column which contains the the first measurement in time

therefore we had to load plyr, then we use the function ddply() (the meaning of the first two letters is data frame in data frame out)

the first argument of the function is the data frame we pass to ddply

the second argument defines the groups (we want the min of each person so our grouping variable is id)

transform says we want to change existing data frame - like recode a variable or add a new (another choice would be summarise - if we want to aggregate the data) - so we add a variable named start and it should be the min() of our date per person

res <- ddply(df, .(id), transform, start=min(date)) res

iddatevalstart12001-12-315.778679592025192001-03-0712003-01-285.118336553288972001-03-0712003-11-143.750754634375052001-03-0712001-03-073.363050937734682001-03-0712001-03-076.61415837892332001-03-0722002-08-026.982798845791672000-05-2922003-12-305.44500191386462000-05-2922000-05-295.925679827166672000-05-2922000-11-264.601699565975442000-05-2922002-08-045.023958124582000-05-2932002-04-235.925903245936592000-02-0232003-04-286.592460569591672000-02-0232002-10-205.897804643006622000-02-0232000-02-024.05486188162692000-02-0232000-05-305.47226334023782000-02-0242000-08-033.527374580480372000-08-0342002-11-023.988538941622852000-08-0342002-11-134.543730885196552000-08-0342003-08-024.21442451843462000-08-0342003-12-194.987674782982792000-08-0352002-08-295.239211212098212000-04-2752000-04-274.204002414119612000-04-2752003-07-266.18219572726462000-04-2752003-05-184.535154046383622000-04-2752001-07-233.066952749405012000-04-27

if we change the function just a little we get the days elapsed from the first measurment:

res <- ddply(df, .(id), transform, dayselapsed=date-min(date)) res