Evaluation of text

Just to remember (once for all):
eval( parse( text="a <- c(1,2,3)" ) )

Mengen-Vergleiche

Interessanten Mengenvergleichs-Funktionen:

(A <- c(sort(sample(1:20, 9)),NA))

[1] 1 2 3 12 16 17 18 19 20 NA
(B <- c(sort(sample(3:23, 7)),NA))
[1] 8 10 15 16 18 21 22 NA


# alle Elemente aus A und B (KEINE Duplikate)
union(A, B)
[1] 1 2 3 12 16 17 18 19 20 NA 8 10 15 21 22

# alle Elemente die in A und in B vorkommen (Schnittmenge)
intersect(A, B)
[1] 16 18 NA

# Elemente in A, aber nicht in B vorkommen
setdiff(A, B)
[1] 1 2 3 12 17 19 20

# Elemente in B, aber nicht in A vorkommen
setdiff(B, A)
[1] 8 10 15 21 22


# Enthält A die gleichen Elemente wie B?
setequal(A, B)
[1] FALSE

Function call with a list of arguments

Use do.call to call a function with a previously built list of arguments:

# Define the arguments

args.legend <- list(
  x = "topleft"
  , legend = c(1,2)
  , fill = c("red","blue")
  , xjust = 1, yjust = 1)

plot(1:10, 1:10)
# call function with the arguments' list
do.call("legend", args.legend)

Get information about invisible built-in functions

How can we see the code of invisible functions, i.e. mosaicplot? 

> mosaicplot

yields only:

function (x, ...)
UseMethod("mosaicplot")


but the functions "methods" and "getAnywhere" do the trick:

> methods(mosaicplot)

[1] mosaicplot.default* mosaicplot.formula*

Non-visible functions are asterisked

> getAnywhere("mosaicplot.formula")
A single object matching ‘mosaicplot.formula’ was found
It was found in the following places
registered S3 method for mosaicplot from namespace graphics
namespace:graphics
with value

function (formula, data = NULL, ..., main = deparse(substitute(data)),
subset, na.action = stats::na.omit)
{
main
m <- match.call(expand.dots = FALSE)
.....

Layout-Muster

Split screens:

m <- matrix(0:3, 2, 2)
layout(m, c(1, 3), c(1, 3))
layout.show(3)

m <- matrix(1:4, 2, 2)
layout(m, widths=c(1, 3), heights=c(3, 1))
layout.show(4)

Date formats and functions

Date formats:

Remember the functions:

d <- as.Date(c("1937-01-10","1916-03-02","1913-09-19","1927-12-23","1947-07-28"))

quarters(d)
[1] "Q1" "Q1" "Q3" "Q4" "Q3"


# ... or as function
quarter <- function (x) {
  # Berechnet das Quartal eines Datums
  y <- as.numeric( format( x, "%Y") )
  paste(y, "Q", (as.POSIXlt(x)$mon)%/%3 + 1, sep = "")
}

# ... or alternatively with cut
cut(d, breaks="quarters")
[1] 1937-01-01 1916-01-01 1913-07-01 1927-10-01 1947-07-01
137 Levels: 1913-07-01 1913-10-01 1914-01-01 1914-04-01 ... 1947-07-01


months(d)
[1] "Januar" "März" "September" "Dezember" "Juli"

format(d,"%B") # months alternative
[1] "Januar" "März" "September" "Dezember" "Juli"

format(d,"%Y") # years
[1] "1937" "1916" "1913" "1927" "1947"

Legend variations

par(mar=c(5.1,4.1,4.1,11.1))
plot( x=1:5, y=1:5, type="n", xlab="x", ylab="y" )

legend( x=2, y=6, legend=c("A","B","C")
  , fill=c("red","blue","green")
  , density=30, bty="n", horiz=TRUE
  , xpd=TRUE )

legend( x=2, y=2, xjust=0.5, yjust=0
  , title=" My title:", title.col="grey40", title.adj=0
  , legend=c("A","B","C","D","E")
  , pch=c(22,22,22,45,45), pt.cex=c(1.2,1.2,1.2,2,2)
  , col=c(rep("black",3),"orange","red")
  , pt.bg=c("blue","green","yellow")
  , bg="grey95", cex=0.8
  , box.col="darkgrey", box.lwd=3, box.lty="dotted" )

legend("topright", inset=0.05, cex=0.8, bg="white"
  , legend=c("A-1","A-2","B-1", "B-2")
  , col=c("lightblue","blue","salmon","red"), pch=15, pt.cex=1.5
  , y.intersp=1.5, x.intersp=1.5 , ncol=2 )

windowsFonts("sans2"="Arial Black")
usr <- par(font=4, family="sans2" )
legend( x=5.5, y=3, legend=c("Label A","Label B","Label C")
  , fill=c("red","orange","yellow")
  , border="brown"
  , y.intersp=2, text.width=strwidth("Make larger")
  , text.col=c("red","orange","yellow")
  , xpd=TRUE )
par(usr)

legend( x="bottomleft", inset=0.02, legend=c("A","B","C","D")
  , lty=c("dashed","dotted",NA,"solid"), lwd=2, cex=0.8
  , pch=c(NA,NA,21,15)
  , col=c("red","blue","black","grey"), bg="white" )

merge

A simple merge example:

a <- data.frame(
  "id"=c(1,2,3,6,7,8),
  "name"=c("Anna","Berta","Claudia","Dora","Eliane","Frida"))
b <- data.frame(
  "id"=c(1,2,3,4,5,9),
  "ort"=c("Zürich","Davos","Zermatt","Chamonix","Verbier","Sedrun"))
c.inner <- merge(a,b)
c.left <- merge(a,b, all.x=TRUE)
c.right <- merge(a,b, all.y=TRUE)
c.full <-  merge(a,b, all.x=TRUE, all.y=TRUE)

Split - Apply - Combine

Another effort to make groupwise operations with a split - apply - combine solution (maybe we'll need it someday):

d.frm <- data.frame(
    name=c("Max","Max","Max","Max","Max","Moritz","Moritz","Moritz")
  , typ=c("rot","blau","grün","blau","grün","rot","rot","blau")
  , anz=c(5,4,5,8,3,2,9,1) )

# Split to list
groups <- split(d.frm, list(d.frm$name, d.frm$typ))

# Create result vector
results <- vector("list", length(groups))
# Apply
for(i in seq_along(groups)) {
  groups[[i]] <- transform(groups[[i]],
    rank = rank(-anz, ties.method = "first"))
  results[[i]] <- groups[[i]]
}

# Combine
result <- do.call("rbind", results)
result
     name  typ anz rank
2     Max blau   4    2
4     Max blau   8    1
8  Moritz blau   1    1
3     Max grün   5    1
5     Max grün   3    2
6     Max  rot   5    1
61 Moritz  rot   2    2
7  Moritz  rot   9    1

We might want to try here the elegant function ave as well:

d.frm$rank_g <- ave( -d.frm$anz, d.frm$name, d.frm$typ,
  FUN=function(x) rank(x, ties.method="first") )

... and not to forget the specially R-flavour:

split(x, g) <- lapply(split(x, g), FUN) 

Dotchart nach Cleveland

Alternativ zum Barplot, wenn Mittelwerte oder Mediane gruppenweise dargestellt werden sollen:

dotchart( tapply( ToothGrowth$len,  list(ToothGrowth$supp, ToothGrowth$dose), mean )
  , main="Guinea Pigs' Tooth Growth", cex=0.8
  , xlab="Vitamin C dose", ylab="tooth length")

Stripchart

Als Alternative zum Boxplot mit wenigen Datenpunkten eignet sich der Stripchart.

stripchart( len ~ dose, data=ToothGrowth, method="jitter", vertical=T
  , subset = supp == "VC", col="blue", main="Guinea Pigs' Tooth Growth"
  , xlab="Vitamin C dose", ylab="tooth length")

stripchart( len ~ dose, data=ToothGrowth, method="jitter", vertical=T
  , subset = supp == "OJ", col="red", add=T)

Grösse des Plot-Fensters bestimmen

Die Grösse des Plot-Fensters kann mit windows() festgelegt werden. Das Grafikfenster wird mit dev.off() wieder geschlossen.

windows(width=10, height=5)
plot( 1:5 )
dev.off()

Selektieren von Listenelementen

Listenelemente können systematisch mit lapply( list, "[", element ) extrahiert werden.

d.frm <- data.frame( id_name=c("1-Max","2-Maria","3-Steven","4-Jane"))

# split into list
lst <- strsplit( x=as.character(d.frm$id_name), split="-" )
# get first elements out of list
d.frm$id <- as.integer( unlist( lapply( lst, "[", 1) ))
d.frm$name <- unlist( lapply( lst, "[", 2) )

d.frm
   id_name   name id
1    1-Max    Max  1
2  2-Maria  Maria  2
3 3-Steven Steven  3
4   4-Jane   Jane  4

str(d.frm)
'data.frame':   4 obs. of  3 variables:
 $ id_name: Factor w/ 4 levels "1-Max","2-Maria",..: 1 2 3 4
 $ name   : chr  "Max" "Maria" "Steven" "Jane"
 $ id     : int  1 2 3 4

Reduce margin between plot region and axes with xaxs, yaxs

# get some data
x.i <- seq(0,1,length=5); y.i <- c( 0,0.1,0.2,0.8,1)
par(mfrow=c(1,2))

plot( y=y.i, x=x.i, type="s", panel.before=grid())
symbols( x=0, y=0, circles=0.12, inches=F, add=T, xpd=T, bg=rgb(0,0,1,0.2) )

plot( y=y.i, x=x.i, type="s", xaxs="i", yaxs="i")
grid(); box()
symbols( x=0, y=0, circles=0.12, inches=F, add=T, xpd=T, bg=rgb(0,0,1,0.2) )

Find most frequent elements

# the vector
x <- sample.int( n=10, size=20, replace=TRUE )
# the 3 most frequent elements
names( head( sort(-table(x)), 3 ) )
# the 3 most frequent elements with their frequencies
head( sort(-table(x)), 3 )

Groupwise boxplot

Groupwise boxplots can easily be created by means of the formula interface.

boxplot(len ~ supp*dose, data = ToothGrowth,
        main = "Guinea Pigs' Tooth Growth",
        xlab = "Vitamin C dose mg", ylab = "tooth length",
        col=c("yellow", "orange") 
        )

Why an outdated method is described in the boxplot help is however not directly clear. Maybe we are glad to know about the technique anyway someday...

boxplot(len ~ dose, data = ToothGrowth,
       boxwex = 0.25, at = 1:3 - 0.15,
       subset = supp == "VC", col = "yellow",
       main = "Guinea Pigs' Tooth Growth",
       xlab = "Vitamin C dose mg",
       ylab = "tooth length",
       xlim = c(0.5, 3.5), ylim = c(0, 35), yaxs = "i")

boxplot(len ~ dose, data = ToothGrowth, add = TRUE,
       boxwex = 0.25, at = 1:3 + 0.15,
       subset = supp == "OJ", col = "orange")

legend(2, 9, c("Ascorbic acid", "Orange juice"),
      fill = c("yellow", "orange"))

SQL-OLAP in R

How to generate SQL-OLAP functions in R:

d.frm <- data.frame( x=rep(1:4,3), g=gl(4,3,labels=letters[1:4]) )

# SQL-OLAP: sum() over (partition by g)
# (more than 1 grouping variables are enumerated like ave(..., g1,g2,g3, FUN=...)):
d.frm$sum_g <- ave( d.frm$x, d.frm$g, FUN=sum )


# same with rank (decreasing):
d.frm$rank_g <- ave( -d.frm$x, d.frm$g, FUN=rank )
d.frm


# get some more data
d.frm <- data.frame(
  id=c("p1","p1","p2","p2","p2","p3","p2","p3","p1","p1","p2"),
  A=c(0,1,1,1,0,0,0,0,0,0,0),
  B=c(1,0,0,0,0,0,0,0,0,0,0),
  C=c(0,0,0,0,1,1,1,0,1,1,1)
)

# get rownumber by group, based by original order
d.frm$rownr <- ave( 1:nrow(d.frm), d.frm$id, FUN=order )

# get some groupwise aggregation on more than one column 
d.frmby <- data.frame( lapply( d.frm[,-c(1,5)], tapply, d.frm$id, "max", na.rm=TRUE ))

# (see also 'Split - Apply - Combine' post)

Barplot mit Fehlerbalken

hh <- t(VADeaths)[, 5:1]
ci.l <- hh * 0.85
ci.u <- hh * 1.15


mb <- barplot(hh, beside = TRUE, ylim = c(0, 100)
  , col = c("lightblue", "mistyrose","lightcyan", "lavender")
  , main = "Death Rates in Virginia", font.main = 4
  , sub = "Faked 95 percent error bars", col.sub = "gray20"
  , cex.names = 1.5
  , legend.text = colnames(VADeaths), args.legend = list( bg="white" )
  , panel.before = {
      rect( xleft=par()$usr[1], ybottom=par()$usr[3], xright=par()$usr[2], ytop=par()$usr[4]
        , col="gray99" )
      grid( nx=NA, ny=NULL ) # horiz grid only
      box()
  }
  , xpd=F )


arrows( x0=mb, y0=ci.l, y1 = ci.u, angle=90, code=3, length=0.05 )


mtext( side = 1, at = colMeans(mb), line = 2,
text = paste("Mean", formatC(colMeans(hh))), col = "red" )

Simple textplot

Putting text on a plot is not that straight on, especially if there's more than one line of text.

data(iris)

# put the summary output into a variable
out <- capture.output( 
  summary(lm(Sepal.Length ~ Species + Petal.Width, iris)) )

cat( out, sep="\n" )

# create plot
plot.new()

# print text in mono font
text(labels=out, x=0
  , y=rev(1:length(out)) * strheight( "S", cex=0.8 ) * 1.3
  , adj=c(0,0), family="mono", cex=0.8 )



This is what I thought so far. Of course it is straight on to place bulk text on several lines, stupid...
Just collapse the text with newline as separator:

text(labels=paste(out, collapse="\n"), x=0, y=0
      , adj=c(0,0), family="mono", cex=0.8 )

Format

Tausender-Trennzeichen und Nachkommastellen
format( 1234, big.mark="'", nsmall=2 )
[1] "1'234.00"


leading zeros
sprintf("%04d", 15)
[1] "0015"

oder mit formatC:
formatC(1:5, width=2, flag="0")
[1] "01" "02" "03" "04" "05"


date
format( Sys.time(), "%Y-%m-%d %H:%M:%S" )
[1] "2010-05-19 10:31:57"

numers as hexnumbers
sprintf("%1$d %1$x %1$X", 0:255)

Codes unter R-Help: strptime 

Von R zu Word

Control MS Word from R, do some reporting. The package RDCOMClient is great for that.
The library DescTools (available on CRAN) uses this package and contains some wrapping functions to make things easier.

List in DataFrame

d.frm <- data.frame(name=c("Max","Max","Max","Max","Max","Moritz","Moritz","Moritz")
,typ=c("rot","blau","grün","blau","grün","rot","rot","blau")
,anz=c(5,4,5,8,3,2,9,1) )

d.frm <- d.frm[ order(d.frm$name, -d.frm$anz),]
d.frm


Zusammenfassung wie summaryBy:

> as.data.frame( do.call("rbind", tapply( d.frm$anz, d.frm$name, summary )))

       Min. 1st Qu. Median Mean 3rd Qu. Max.
Max       3     4.0      5    5    5.0     8
Moritz    1     1.5      2    4    5.5     9

Windows Schriftarten verwenden

Windows Schriftarten müssen definiert werden, bevor sie genutzt werden können:

# Type family examples - creating new mappings
plot(1:10,1:10,type="n")
windowsFonts(
A=windowsFont("Arial Black"),
B=windowsFont("Bookman Old Style"),
C=windowsFont("Comic Sans MS"),
D=windowsFont("Symbol")
)

text(3,3,"Hello World Default")
text(4,4,family="A","Hello World from Arial Black")
text(5,5,family="B","Hello World from Bookman Old Style")
text(6,6,family="C","Hello World from Comic Sans MS")
text(7,7,family="D", "Hello World from Symbol")

Add a time axis

Adding a time scaled x-axis to a plot can be tedious...

# add a time axis
axis(side = 1, at  =as.numeric(seq(as.Date("2000-01-01"), by = "year", length = 10))
  , labels = format( seq(as.Date("2000-01-01"), by = "year", length = 10) , "%Y")
  , cex.axis = 0.8, las = 1)

# plot the timegrid
# vertical gridlines according to chosen time-values
abline(v = seq(as.Date("2000-01-01"), by = "year", length = 10), col = "grey", lty = "dotted")
# where to plot the horizontal gridlines? get the coords by means of axTicks()
abline(h = axTicks(2), col = "grey", lty = "dotted")

Dummy-Kodierung

Gesucht ist die Dummy-Kodierung eines Faktors:

d.frm <- data.frame(
    name=c("Max","Max","Max","Max","Max","Moritz","Moritz","Moritz")
  , typ=c("rot","blau","grün","blau","grün","rot","rot","blau")
  , anz=c(5,4,5,8,3,2,9,1)
)

d.frm
    name  typ anz
1    Max  rot   5
2    Max blau   4
3    Max grün   5
4    Max blau   8
5    Max grün   3
6 Moritz  rot   2
7 Moritz  rot   9
8 Moritz blau   1

Gewünschtes liefert

model.matrix(~d.frm$typ)[,-1]
  d.frm$typgrün d.frm$typrot
1             0            1
2             0            0
3             1            0
4             0            0
5             1            0
6             0            1
7             0            1
8             0            0

oder alternativ die Funktion class.ind aus der library(nnet):

library(nnet)
class.ind( df$typ )

blau grün rot
[1,] 0 0 1
[2,] 1 0 0
[3,] 0 1 0
[4,] 1 0 0
[5,] 0 1 0
[6,] 0 0 1
[7,] 0 0 1
[8,] 1 0 0

Still another brilliant Ripley solution:

ff <- factor(sample(letters[1:5], 25, replace=TRUE))
diag(nlevels(ff))[ff,]

Read from/Write to clipboard

Daten lesen aus dem, resp. schreiben in das Clipboard ganz einfach:

d.x <- data.frame( x=c(1,2,3,3), y=c("A","B","A","A") )
write.table(d.x, file="clipboard", sep="\t", row.names=FALSE)
a <- read.table("clipboard", header=TRUE)

Modell-Formelnotation

---------------------------------------------------------------------------------------------
Operator Bedeutung
----------------------------------------------------------------------------------------------
+... Hinzunahme einer Variablen
-... Herausnahme einer Variablen (-1 für Achsenabschnitt)
:... Wechselwirkung/Interaktion von Variablen
*... Hinzunahme von Variablen und deren Wechselwirkungen
/... hierarchisch untergeordnet ("nested").
.... y/z bedeutet: z hat nur Wirkung innerhalb der Stufen von y, aber nicht global.
^... alle Interaktionen bis zum angegebenen Grad
.... alle Variablen aus dem Datensatz in das Modell aufnehmen
I(). innerhalb von I() behalten arithmetische Operatoren ihre ursprüngliche Bedeutung
.... ("Inhibit interpretation")
----------------------------------------------------------------------------------------------

Bsp:
y ~ x1 * x2 * x3 - x1:x2:x3 # ohne 3-fach Interaktion
y ~ (x1 + x2 + x3)^2 # alle Variablen inkl. Interaktion bis zum 2. Grad

Kombinationen tabellieren

Ausgangslage ist ein data.frame mit einer Zeile pro Element und n Indikatorvariablen. Es stellt sich die Frage nach den Häufigkeiten der Kombinationen der Indikatorvariablen.

d.frm <- data.frame(name=c("Franz", "Maria", "Claudine")
  , A=c(1,0,1), B=c(1,1,1), C=c(0,0,1), D=c(0,1,0))

Solution 1 (clumsy):

# Berechne die Summe der mit einer 2er-Potenz multiplizierten Indikatoren
d.frm$sum <- as.matrix(d.frm[,2:5]) %*% (2^(0:3))

# Erzeuge levels für eine Komb-Faktor
d.lvl <- expand.grid( c(0,"A"),c(0,"B"),c(0,"C"),c(0,"D") )

# Bilde eine neue Variable mit der Textbezeichnung der 2^n-Summe
d.frm$sum_x <- factor(d.frm$sum
  , levels=((d.lvl!=0)*1) %*% (2^(0:3))
  , labels=gsub( "0","", do.call( "paste", c( d.lvl, list(sep="") ))))

d.frm
......name A B C D sum sum_x
1 ...Franz 1 1 0 0 ..3 ...AB
2 ...Maria 0 1 0 1 .10 ...BD
3 Claudine 1 1 1 0 ..7 ..ABC

Solution 2 (smart):

d.frm$sum_xx <- apply( d.frm[,2:5], 1,
  function(x) paste(LETTERS[1:4][as.logical(x)], collapse="") )


Wenn einfach nur die Kombinationen gesucht sind, gibt's dafür die Bordmittel:

> combn( letters[1:4], 2 )
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "a" "a" "a" "b" "b" "c"
[2,] "b" "c" "d" "c" "d" "d"

Oder auch interessant mit den Funktionen outer und lower.tri (pairwise):

m <- outer(x, x, paste, sep="-" )
m[!lower.tri(m, diag=TRUE) ]

[1] "a-b" "a-c" "b-c" "a-d" "b-d" "c-d"

Get all binary combinations
The idea is to get a vector with n 0s and n 1s, chop it into n parts of c(0,1) (this is a list), and use expand.grid:

n <- 4="4">
expand.grid(split(rep(c(0,1), each=n), 1:n))

Gruppenweise Auswertung

Erstaunlich umständlich ist die gruppenweise Selektion von bestimmten Elementen eines dataframes. Zum Beispiel soll gruppenweise das erste Element nach einer vorgegebenen Sortierung zurückgegeben werden:

d.frm <- data.frame(
name=c("Max","Max","Max","Max","Max","Moritz","Moritz","Moritz")
,typ=c("rot","blau","grün","blau","grün","rot","rot","blau")
,anz=c(5,4,5,8,3,2,9,1) )

d.frm <- d.frm[ order(d.frm$name, -d.frm$anz),]
d.frm
....name .typ anz
4 ...Max blau ..8
1 ...Max .rot ..5
3 ...Max grün ..5
2 ...Max blau ..4
5 ...Max grün ..3
7 Moritz .rot ..9
6 Moritz .rot ..2
8 Moritz blau ..1


d.frm[ tapply( rownames(d.frm), d.frm$name, head, n=1), ]
. ..name. typ.anz
4....Max.blau...8
7 Moritz..rot...9

Noch einfacher ist allerdings:
d.frm <- d.frm[ order(d.frm$name, -d.frm$anz),]
d.frm[ !duplicated(d.frm$name),]

Wenn der zweite Wert gesucht wäre, ginge dies mit der Index-Funktion "[":
d.frm[ tapply( rownames(d.frm), d.frm$name, "[", 2),  ]

Umorganisation von Daten

Daten, die pivotisiert vorliegen, können mit reshape (siehe auch stack/unstack) wieder in die Faktor-Form gebracht werden.

# Ausgangslage: Breite Form
> d.wide <- data.frame( age=c(22,34,28,31), gender_c=c("m","w","w","m"), ZH=c(12,33,2,5), BE=c(5,6,3,1))

> d.wide

.age gender_c ZH BE
1 22 ....m 12 .5
2 34 ....w 33 .6
3 28 ....w .2 .3
4 31 ....m .5 .1

# ...und die lange Form:

> d.long <- reshape( d.wide, varying=3:4, times=names(d.tmp)[-c(1:2)], v.names="count", direction="long")

> d.long
age gender_c time count id
1.ZH 22 m ZH 12 1
2.ZH 34 w ZH 33 2
3.ZH 28 w ZH 2 3
4.ZH 31 m ZH 5 4
1.BE 22 m BE 5 1
2.BE 34 w BE 6 2
3.BE 28 w BE 3 3
4.BE 31 m BE 1 4

... und wieder zurück:
wide <- reshape( d.long, idvar=c("age","gender_c","id"), timevar="time", direction="wide" )
wide

Ränder-Problematik

Einstellungen für Ränder in Grafiken für: mar, mai, mgp, omi, oma.
mar=c(5,4,4,2)+0.1 gleich mar=c(u,li,o,re)+0.1,

mai=c(u,li,o,re) Angabe in inch,

mgp=c(3,1,0) gleich c(Titel, Label, Achse) in mex-Einheiten,

omi=c(0,0,0,0) gleich omi=c(u,li,o,re),

oma=c(0,0,0,0) gleich oma=c(u,li,o,re) in Textzeileneinheiten.

Quelle: http://www.geo.fu-berlin.de/geol/fachrichtungen/pal/mitarbeiter/plank/formeln_in_R.pdf

Gauss-Test

Beispiel
Das Gewicht von Briefumschlägen für Luftpost kann als normalverteilt angesehen werden, wobei das mittlere Gewicht weniger als 2 g betragen soll. Die Standardabweichung ist erfahrungsgemäss s=0.02 g. Ein Hersteller entnimmt der Produktion 100 Umschläge und stellt den Mittelwert mw_sp=1.98 g fest.

# Gausstest
n <- 100 # Anzahl Umschläge
MW <- 2 # Mittelwert der Grundgesamtheit
SD <- 0.02 # Standardabweichung der Grundgesamtheit
mw <- 1.98 # Mittelwert der Stichprobe
se <- SD/sqrt(n) # Standardfehler des Mittelwerts
alpha <- 0.05 # Signifikanzniveau

# Quantile der normalverteilung
qnorm(alpha/2, mean=MW, sd=se)
# p-Wert des zweiseitigen Gauss-Testes
2*min(pnorm(mw, mean=MW, sd=se), 1-pnorm(mw, mean=MW, sd=se))

Binomialtest

Beispiel
Jack Player wurde in einem Saloon in Texas erschossen, nachdem er beim Würfelspiel mit seinem Würfel in 60 Würfen 21mal die Sechs gewürfelt hatte und deshalb des Falschspiels bezichtigt wurde. War der Vorwurf seines (jähzornigen) Mitspielers berechtigt?
Man prüfe die Sachlage mit einem Binomialtest!

# Definition der Parameter
p <- 1/6
n <- 60
k <- 21
alpha <- 0.05

# Berechnung BinomialTest
binom.test(k, n, p, "greater")

dens <- dbinom(0:n, n, p)   # Berechnung der binomialen Dichte-Funktion
cols <- c(rep("lightgreen", green <- sum(cumsum(dens) < (1-alpha))), rep("red", n-green+1))

# Zeichnen des Säulen-Diagramms
plot(x=0:n, y=dens, col=cols, main="Tod eines Spielers",
     xlab="k", ylab="B(k,n,p)", las=1,
     cex.axis=0.8, type="h", lwd=3)

# ... und noch eine Textbemerkung
text(20, 0.08, "Binomialverteilung mit p=1/6, n=60", pos=4)

Kombiniertes Säulen-Linien-Diagramm

Ein etwas aufwendigeres Verfahren anhand eines Klimadiagramms:

# get some data
d.temp <- data.frame(
  month=c("Jan","Feb", "Mrz", "Apr", "Mai", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dez")
  ,nieder_96=c(9, 50, 41, 49, 141, 99, 161, 119, 52, 115, 123, 70)
  ,nieder_mittel=c(67, 65, 67, 85, 103, 135, 136, 130, 101, 81, 74, 76)
  ,temp_96=c(-1.9, -2.1, 3.8, 9.3, 11.8, 17.1, 17.3, 16.8, 10.2, 9.8, 5.4, 0.5)
  ,temp_mittel=c(-1, 0, 4.5, 7.3, 11.9, 15, 16.5, 15.5, 13.9, 8.1, 3.7, 0.2)
)

# define a few colors
hellblau <- rgb(red=204,green=255,blue=255, max=255)
dunkelblau <- rgb(red=51,green=204,blue=204, max=255)
dunkelgrau <- rgb(red=128,green=128,blue=128, max=255)
mittelgrau <- rgb(red=192,green=192,blue=192, max=255)
hellgrau <- rgb(red=227,green=227,blue=227, max=255)

# set the parameters
windows(width=7.2, height=5.5)
par(mar=c(5.1,4.1,7.1,16.1)) # set margins, default: c( 5.1, 4.1, 4.1, 2.1 )
par(bg=mittelgrau) # background color

# start plotting, we use barplot as basis
b <- barplot( t(d.temp[,c("nieder_mittel","nieder_96")])
  , col=c(dunkelgrau, hellblau)
  , beside=TRUE , xlab="Monate", cex.lab=0.8, mgp=c(2.2,0.7,0)
  , space=rep( c(0.3,-0.5), 12) # bars should overlap 50%
  , ylim=c(0,500), yaxt="n"
  , panel.first = {
      par(xpd=FALSE) # barplot paints over the whole figure region by default
      usr <- par("usr") # set background color lightgrey
      rect(xleft=usr[1], ybottom=usr[3], xright=usr[2], ytop=usr[4], col=hellgrau)
     grid(nx=NA, ny=10, col="white", lty="solid") # horiz grid only
     box()
     }
)

# find the centers of the bars and the gaps
barx <- apply(b, 2, FUN=mean)
run.mean <- filter( barx, filter=c(0.5,0.5))[-length(barx)]
gapx <- c(run.mean[1]-diff(barx)[1], run.mean, run.mean+diff(barx) )


# draw the vertikal gridlines
abline(v=gapx, col="white" )
box()

# design x-axis
axis(side=1, at=apply(b,2,FUN=mean), labels=d.temp$month, cex.axis=0.7
  , las=2, tck=-0.025  # no tickmarks for the x-axis
  , mgp=c(2.2,0.7,0) ) # decrease distance label to axis

# left y-axis
axis(side=2, at=seq(0,500,50), las=2, cex.axis=0.7)
rug( seq(0,500,10), side=2, ticksize=-0.01)
rug( seq(0,500,50), side=2, ticksize= 0.01)

# plot lines
par(new=TRUE)
matplot( x=barx, y=d.temp[,c("temp_96","temp_mittel")], col=c(dunkelblau,"grey60")
  , lwd=2, lty="solid", type="l", xaxt="n", yaxt="n", xlab="", ylab=""
  , xlim=par("usr")[1:2] # use the current xlim
  , ylim=c(-25, 25), xaxs="i", yaxs="i")

# design right axis
axis(side=4, labels=seq(-25,25,5), at=seq(-25,25,5), las=2, cex.axis=0.7)
rug( seq(-25,25,1), side=4, ticksize=-0.01)
rug( seq(-25,25,5), side=4, ticksize=0.01)

# write titles
mtext(text=c("Lufttemperatur [°C]","Niederschlag [mm]"), side=3, at=c(25,-3.2), adj=c(1,0)
  , las=1, line=1, cex=0.8 )
mtext(text="Klimadiagramm Zürich-SMA\n556 m. ü. M.", cex=1.2, font=2, side=3, line=3)

# plot legend
legend( x=30, y=27, xpd=TRUE
  , legend=c("Niederschlag 1996", "Niederschlag:\nlangjähriges Mittel", "Temperatur 1996", "Temperatur:\nlangjähriges Mittel" )
  , cex=0.7, bty="n", col=c(hellblau, dunkelgrau, dunkelblau, "black")
  , y.intersp=2.5, pt.cex=1.2, pch=c(15,15,45,45))

mtext("© Statistisches Amt des Kantons Zürich", side=1, line=3.5, at=-4, cex=0.7, las=1, adj=0)
mtext("Quelle: Schweizerische Meteorologische Anstalt (SMA)", side=1, line=3.5, at=41, cex=0.7, las=1, adj=1)

Plot-Zeichen

Was war das schon wieder für ein Zeichen?

parold <- par( mex=0.001, xaxt="n", yaxt="n", ann=F) # save graph sets
plot( x=1:25, y=rep(1,25), pch=1:25, cex=3, xlab="", ylab="", frame.plot=FALSE)
text( x=1:25, y=rep(0.8,25), labels=1:25 )
par(parold) # reset graph sets

Farben

Wie hiess doch schon wieder diese Farbe?

farben <- c( colors(), rep(NA,3) ) # 657 Farben um 3 NULL-Werte erweitern
zeilen <- 44; spalten <- 15 # 660 Farben
farben.zahlen <- matrix( 1:spalten ,nrow=zeilen, ncol=spalten, byrow=T ) # Matrix für Punkte

paralt <- par( mex=0.001, xaxt="n", yaxt="n", ann=F) # Grafikeinstellungen speichern
x_offset <- 0.5
x <- farben.zahlen[,1:spalten] # x-Werte (Zahlen)
y <- -rep(1:zeilen, spalten) # y-Werte (Zahlen)

plot( x, y
, pch=22 # Punkttyp Rechteck
, cex=3 # Vergrösserung Punkte
, bg=farben # Hintergrundfarben
, bty="n" # keine Box
, xlim=c(1,spalten+x_offset) # x-Wertebereich
)
text( x+0.1, y, farben, adj=0, cex=0.8 ) # Text Farben dazu

par(paralt) # Grafikeinstellungen zurücksetzen

Säulendiagramm mit gestapelten und überlappenden Säulen

# Prepare and organize some data
d.blue <- data.frame( grp1=c(5,4,3,5), grp2=c(3,3,2,1) )
rownames(d.blue) <- c( "t1","t2","t3","t4" )
set.seed(181)
d.red <- d.blue * runif(n=5)

# Barplot expects matrix instead of data.frame
m.blue <- t( as.matrix( d.blue ) )
m.red <- t( as.matrix( d.red ) )

# We need a function to prepare background of the barplot
mygrid <- function(){
  par(xpd=FALSE)     # barplot paints over the whole figure region by default  
  rect(xleft=-1, xright=199  # set background white, when bg has another col
  , ybottom=-99, ytop=99, col="white")
  grid(nx=NA, ny=NULL)       # horiz grid only
  par(xpd=TRUE)
  box()
}

# Set parameters
osp <- 0.5 # overlapping part in %
sp <- 1 # spacing between the bars
par( bg="grey90")            # set outer background

nbars <- dim(m.blue)[2]   # how many bars do we have?

# Create first barplot
b <- barplot( m.blue, col=c("lightblue","blue")
  , beside=FALSE, ylim=c(0,10)
  , axisnames=FALSE
  , main="My main title"
  , xlim=c(0, nbars*2-osp )      # enlarge x-Axis
  , space=c(0, rep(sp,nbars-1) ) # set spacing=1, starting with 0
  , panel.first=mygrid())

# Draw the red series
barplot( m.red, col=c("salmon","red"), beside=FALSE
  , space=c(1-osp, rep(1, nbars-1)) # shift to right by 1-osp
  , axisnames=FALSE, add=TRUE)

# Create axis separately, such that labels can be shifted to the left
axis(1, at=b+(1-osp)/2, labels=FALSE, tick=FALSE, las=1)

# Now draw the rotated textual axis labels
text( b+(1-osp)/2, par("usr")[3] - 0.2
  , labels = rownames(d.blue), srt = 45, pos = 1, xpd = TRUE)


# Add some other rotated text
text( x=b, y=1, labels="some text", srt=45 )

# Add legend
legend("topright", inset=0.05, bg="ivory1", cex=0.8
  , legend=c("A-grp 1","A-grp 2","B-gpr 1", "B-grp 2")
  , col=c("lightblue","blue","salmon","red"), pch=15, pt.cex=1.5
  , y.intersp=1.2, x.intersp=1 , ncol=2 )

Zeichenfunktionen

nchar
Count the number of characters, also auch Länge eines Strings:

> nchar( c("Lorem", "ipsum") )
[1] 5 5


tolower/toupper
Gross-/Kleinschreibung

> tolower( LETTERS[1:8] )
[1] "a" "b" "c" "d" "e" "f" "g" "h"
> toupper( letters[1:8] )
[1] "A" "B" "C" "D" "E" "F" "G" "H"


match
Positionen (Index) von Elementen in einer Liste finden sich mit match.

> match(c("d","p"),letters)
[1] 4 16


grep
suchen von Ausdrücken in einem Vektor, retourniert die Indizes der Elemente, in denen der Ausdruck vorkommt

> x <- c("This"," is a Andri 1,2"," and 3 * test sentence")
> grep(pattern="A",x=x)
[1] 2



regexpr
suchen von Ausdrücken in einem Vektor, retourniert pro Element die ERSTE Position des Vorkommens (-1 für nicht vorhanden) und die Länge der Übereinstimmung

> x <- c("This"," is a Andri 1,2"," and 3 * test sentence")
> regexpr(pattern="i",text=x)
[1] 3 2 -1
attr(,"match.length")
[1] 1 1 -1


gregexpr
suchen von Ausdrücken in einem Vektor,
retourniert pro Element ALLE Positionen und die Länge der Übereinstimmung

> x <- c("This"," is a Andri 1,2"," and 3 * test sentence")
> gregexpr(pattern="i",text=x)
[[1]]
[1] 3
attr(,"match.length")
[1] 1

[[2]]
[1] 2 11
attr(,"match.length")
[1] 1 1

[[3]]
[1] -1
attr(,"match.length")
[1] -1


sub
Ersetzt das ERSTE Vorkommen von pattern in einer Zeichenkette

> x <- c("This"," is a Andri 1,2"," and 3 * test sentence")
> sub(pattern="i", replacement="III", x=x)
[1] "ThIIIs" " IIIs a Andri 1,2" " and 3 * test sentence"


gsub
Ersetzt ALLE Vorkommen von pattern in einer Zeichenkette

> x <- c("This"," is a Andri 1,2"," and 3 * test sentence")
> gsub(pattern="i", replacement="III", x=x)
[1] "ThIIIs" " IIIs a AndrIII 1,2" " and 3 * test sentence"



substr
Teile eines Strings extrahieren (oder ersetzen, letzteres aber besser mit (g)sub):

> substr( x="Lorem ipsum dolor sit amet", start=7, stop=500 )
[1] "ipsum dolor sit amet"

substr( x<-"Lorem ipsum dolor sit amet", start=nchar(x)-3, stop=nchar(x) )


left, right und mid
Die guten alten BASIC-Zeichen-Funktionen mit regular expressions in Funktionen umgesetzt:

left <- function(x, n) {
# x...string, n...Anzahl Zeichen
sub( pattern=paste(".{",nchar(x)-n,"}$", sep=""), replacement="", x=x)
}
right <- function(x, n) {
sub( pattern=paste("^.{",nchar(x)-n,"}", sep=""), replacement="", x=x)
}
mid <- function(x, pos, n) {
x <- sub( pattern=paste("^.{",pos-1,"}", sep=""), replacement="", x=x)
x <- sub( pattern=paste(".{",nchar(x)-n,"}$", sep=""), replacement="", x=x)
return(x)
}


trim
Leerzeichen vor und hinter dem Text entfernen

sub(" *([^ ]+) *", "\\1", x) does not work..  where had I that from?

gsub( pattern="^ +| +$", replacement="", x=" trim links und rechts ")

oder wenn auch Tabulatoren gelöscht werden sollen:
gsub( pattern="^[ \t]+|[ \t]+$", replacement="", x=" trim links und rechts ")


Beispiel
Alle Vektorelemente, die eine Zahl enthalten:

> x[1] "This" " is a Andri 1,2" " and 3 * test sentence"
> x[ as.matrix( regexpr(pattern="[[:digit:]]",text=x) )>0 ]
[1] " is a Andri 1,2" " and 3 * test sentence"


paste
Spaltenweises Verbinden von Spalten eines data.frames:

do.call("paste", c(data.frame(letters, LETTERS), list(sep="/")))

[1] "a/A" "b/B" "c/C" "d/D" "e/E" "f/F" "g/G" "h/H" "i/I" "j/J" "k/K" "l/L" "m/M" "n/N" "o/O" "p/P" "q/Q" "r/R"[19] "s/S" "t/T" "u/U" "v/V" "w/W" "x/X" "y/Y" "z/Z"



und warum nicht direkt (?):  :-o
apply( data.frame(letters, LETTERS), 1, paste, collapse="/" )


und warum nicht nocht direkter (???):  :-o :-o :-o
data.frame(letters, LETTERS, paste(letters, LETTERS, sep="/"))

Als Faktor erkannte Datums-Variablen in Datum umwandeln

Datums-Angaben werden beim Laden oft als Faktor interpretiert.
Die Umwandlung in Datums-Format geht leichter, wenn die Feldnamen mit dem Suffix _d enden.

for( x in names(d.set)[ grep( "_d$", names(d.set) ) ] ) {
d.set[,x] <- as.Date( d.set[,x], "%d.%m.%Y" )
}

Plot-Basics

x.i <- 1:10; y.i <- 1/x.i

plot( y=y.i, x=x.i, cex.axis=0.8, yaxt="n", pch=15, col="red", ylab="", xlab="x-Werte", main="Titel" )

points( y=y.i/2, x=x.i, pch=1:10, col="blue" )

abline( h=seq(0,max(y.i),0.1), col="grey", lty="dotted")

segments( y0=y.i-0.1, y1=y.i+0.1, x0=x.i, x1=x.i )

axis( side=2, at=seq(0,1,0.1), labels=seq(0,1,0.1), cex.axis=0.8, las=2 )

legend( x="topright", inset=0.02, legend=c("y1-Werte", "y2-Werte"), col=c("red","blue"), pch=c(15,1), bg="white", cex=0.8 )

text( x=x.i+0.2, y=y.i, labels=paste("n=",round(y.i,1),sep=""), cex=0.8, adj=0 )

mtext( side=3, at=x.i, line=0.5, text=paste( "x=",x.i, sep=""), cex=0.8 )

savePlot( filename="d:/myplot.png", type="png" )

Faktoren kreuzen

Um ein data.frame mit dem kompletten Satz an Kombinationen von Ausprägungen mehrerer Faktoren zu erzeugen, kann expand.grid(...) verwendet werden.

sex_c <- c("m","w")
reg <- c("ZH","UR","VS","GR")
prod <- c("BASIS","SANA","HUV")

expand.grid(sex_c, reg, prod)
Var1 Var2 Var3
1 m ZH BASIS
2 w ZH BASIS
3 m UR BASIS
4 w UR BASIS
etc.

Um Kombinationen (ohne Beachtung der Reihenfolge) zu betrachten, gibt es die Funktion combn(...).

combn( letters[1:4], 2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "a" "a" "a" "b" "b" "c"
[2,] "b" "c" "d" "c" "d" "d"

Zweier Kombinationen können auch schnell gefunden werden mit:
m <- outer(letters[1:5], LETTERS[1:5], paste, sep="-")
m[lower.tri(m)]

Variationen eines Vektors können wie folgt bestimmt werden:

v <- 1:3
x <- expand.grid( v, v, v )
# entferne Zeilen aus x, in denen Elemente mehrfach vorkommen
x <- x[ apply( x, 1, FUN=function(x){ sum(duplicated(x)) } )==0, ]
# .. und sortiere noch sinnvoll
x <- x[ order(x[,1],x[,2],x[,3]), ]

x
Var1 Var2 Var3
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1