OBJEK DI R

Vector

a1<- c(2,4,7,3)
assign("a2",c("2","4","7","3"))
a1
## [1] 2 4 7 3

Baris Bilangan

a3<-seq(1,10, by=0.5)
a3
##  [1]  1.0  1.5  2.0  2.5  3.0  3.5  4.0  4.5  5.0  5.5  6.0  6.5  7.0  7.5  8.0
## [16]  8.5  9.0  9.5 10.0
a4<-seq(1,10,length.out=12)
a4
##  [1]  1.000000  1.818182  2.636364  3.454545  4.272727  5.090909  5.909091
##  [8]  6.727273  7.545455  8.363636  9.181818 10.000000

Bilangan Berulang

a5<- rep(1,3)
a5
## [1] 1 1 1
a6<-rep(1:3,3)
a6
## [1] 1 2 3 1 2 3 1 2 3
a7<-rep(1:3,1:3)
a7
## [1] 1 2 2 3 3 3
a8<-rep(1:3,rep(2,3))
a8
## [1] 1 1 2 2 3 3
a9 <-rep(1:3,each=2)
a9
## [1] 1 1 2 2 3 3

Karakter Berpola

a10<-paste("A",1:10,sep="")
a10
##  [1] "A1"  "A2"  "A3"  "A4"  "A5"  "A6"  "A7"  "A8"  "A9"  "A10"
a11 <-paste0("A",1:10)
a11
##  [1] "A1"  "A2"  "A3"  "A4"  "A5"  "A6"  "A7"  "A8"  "A9"  "A10"
a12<-paste("A",1:10,sep="-")
a12
##  [1] "A-1"  "A-2"  "A-3"  "A-4"  "A-5"  "A-6"  "A-7"  "A-8"  "A-9"  "A-10"
a13<-paste0("A",a8)
a13
## [1] "A1" "A1" "A2" "A2" "A3" "A3"

Akses Vector

a2[3]
## [1] "7"
a3[10:15]
## [1] 5.5 6.0 6.5 7.0 7.5 8.0
a10[c(4,7,9)]
## [1] "A4" "A7" "A9"
a13[-c(1:2)]
## [1] "A2" "A2" "A3" "A3"
length(a4)
## [1] 12

Latihan 1

#Tentukan output syntax berikut:
c("la","ye")[rep(c(1,2,2,1),times=4)]
##  [1] "la" "ye" "ye" "la" "la" "ye" "ye" "la" "la" "ye" "ye" "la" "la" "ye" "ye"
## [16] "la"
c("la","ye")[rep(rep(1:2,each=3),2)]             
##  [1] "la" "la" "la" "ye" "ye" "ye" "la" "la" "la" "ye" "ye" "ye"

Latihan 2

#Buatlah syntax agar dihasilkan output berikut:
#X1 Y2 X3 Y4 X5 Y6 X7 Y8 X9 Y10
#1 4 7 10 13 16 19 22 25 28

latihan2a<-(paste0(rep(c("X","Y"),5),1:10))
latihan2a
##  [1] "X1"  "Y2"  "X3"  "Y4"  "X5"  "Y6"  "X7"  "Y8"  "X9"  "Y10"
latihan2b<-seq(1,28,by=3)
latihan2b
##  [1]  1  4  7 10 13 16 19 22 25 28

Matriks

Matriks 1

a14 <-1:12
a14
##  [1]  1  2  3  4  5  6  7  8  9 10 11 12
b1<-matrix(a14,3,4)
b1
##      [,1] [,2] [,3] [,4]
## [1,]    1    4    7   10
## [2,]    2    5    8   11
## [3,]    3    6    9   12
b2<-matrix(a14,3,4,byrow=TRUE)
b2
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
b3<-matrix(1:16,4,4)
b3
##      [,1] [,2] [,3] [,4]
## [1,]    1    5    9   13
## [2,]    2    6   10   14
## [3,]    3    7   11   15
## [4,]    4    8   12   16
b4<-matrix(1:20,4,5)
b4
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    5    9   13   17
## [2,]    2    6   10   14   18
## [3,]    3    7   11   15   19
## [4,]    4    8   12   16   20
b5 <-a14
dim(b5)<-c(6,2)
b5
##      [,1] [,2]
## [1,]    1    7
## [2,]    2    8
## [3,]    3    9
## [4,]    4   10
## [5,]    5   11
## [6,]    6   12

Matriks 2

b6 <-matrix(1:4,2)
b6
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
b7 <-matrix(6:9,2)
b7
##      [,1] [,2]
## [1,]    6    8
## [2,]    7    9
b8 <-rbind(b6,b7)
b8
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## [3,]    6    8
## [4,]    7    9
b9 <-cbind(b7,b6)
b9
##      [,1] [,2] [,3] [,4]
## [1,]    6    8    1    3
## [2,]    7    9    2    4
dim(b8)
## [1] 4 2
dim(b9)
## [1] 2 4
dim(a14)
## NULL
length(b3)
## [1] 16

Matriks 3

b2
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
b2[2,3]
## [1] 7
b2[2,2:4]
## [1] 6 7 8
b2[1:2,]
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
b2[c(1,3),-2]
##      [,1] [,2] [,3]
## [1,]    1    3    4
## [2,]    9   11   12
b2[10]
## [1] 4

Array

c1<-array(a14,dim=c(2,2,3))
c1
## , , 1
## 
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## 
## , , 2
## 
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8
## 
## , , 3
## 
##      [,1] [,2]
## [1,]    9   11
## [2,]   10   12
c2<-array(a14,dim=c(2,1,2,3))
c2
## , , 1, 1
## 
##      [,1]
## [1,]    1
## [2,]    2
## 
## , , 2, 1
## 
##      [,1]
## [1,]    3
## [2,]    4
## 
## , , 1, 2
## 
##      [,1]
## [1,]    5
## [2,]    6
## 
## , , 2, 2
## 
##      [,1]
## [1,]    7
## [2,]    8
## 
## , , 1, 3
## 
##      [,1]
## [1,]    9
## [2,]   10
## 
## , , 2, 3
## 
##      [,1]
## [1,]   11
## [2,]   12
c3 <-array(a14,dim=c(1,2,4,2))
c3
## , , 1, 1
## 
##      [,1] [,2]
## [1,]    1    2
## 
## , , 2, 1
## 
##      [,1] [,2]
## [1,]    3    4
## 
## , , 3, 1
## 
##      [,1] [,2]
## [1,]    5    6
## 
## , , 4, 1
## 
##      [,1] [,2]
## [1,]    7    8
## 
## , , 1, 2
## 
##      [,1] [,2]
## [1,]    9   10
## 
## , , 2, 2
## 
##      [,1] [,2]
## [1,]   11   12
## 
## , , 3, 2
## 
##      [,1] [,2]
## [1,]    1    2
## 
## , , 4, 2
## 
##      [,1] [,2]
## [1,]    3    4
c4 <-array(a14,dim=c(3,4))
c4
##      [,1] [,2] [,3] [,4]
## [1,]    1    4    7   10
## [2,]    2    5    8   11
## [3,]    3    6    9   12
c2[,,1,]
##      [,1] [,2] [,3]
## [1,]    1    5    9
## [2,]    2    6   10
c2[,,,2]
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8
c2[,,1,3] 
## [1]  9 10

Factor

a15 <-c("A","B","AB","O")
a15
## [1] "A"  "B"  "AB" "O"
d1 <-factor(a15)
d1
## [1] A  B  AB O 
## Levels: A AB B O
d2 <-factor(a15,levels=c("O","A","B","AB"))
levels(d2)
## [1] "O"  "A"  "B"  "AB"
d2
## [1] A  B  AB O 
## Levels: O A B AB
a16 <-c("SD","SMP","SMA")
a16
## [1] "SD"  "SMP" "SMA"
d3 <-ordered(a16) 
d3
## [1] SD  SMP SMA
## Levels: SD < SMA < SMP
d4 <-ordered(a16, levels=a16)
d4
## [1] SD  SMP SMA
## Levels: SD < SMP < SMA
d5 <-factor(a16, levels=a16, ordered=TRUE)
d5
## [1] SD  SMP SMA
## Levels: SD < SMP < SMA
levels(d4)
## [1] "SD"  "SMP" "SMA"
d1[2]
## [1] B
## Levels: A AB B O
d4[2:3]
## [1] SMP SMA
## Levels: SD < SMP < SMA

List

a1; b2; c1; d2
## [1] 2 4 7 3
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
## , , 1
## 
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## 
## , , 2
## 
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8
## 
## , , 3
## 
##      [,1] [,2]
## [1,]    9   11
## [2,]   10   12
## [1] A  B  AB O 
## Levels: O A B AB
e1 <-list(a1,b2,c1,d2)
e1
## [[1]]
## [1] 2 4 7 3
## 
## [[2]]
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
## 
## [[3]]
## , , 1
## 
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## 
## , , 2
## 
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8
## 
## , , 3
## 
##      [,1] [,2]
## [1,]    9   11
## [2,]   10   12
## 
## 
## [[4]]
## [1] A  B  AB O 
## Levels: O A B AB
e2 <-list(vect=a1,mat=b2,array=c1,fac=d2)
e2
## $vect
## [1] 2 4 7 3
## 
## $mat
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
## 
## $array
## , , 1
## 
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## 
## , , 2
## 
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8
## 
## , , 3
## 
##      [,1] [,2]
## [1,]    9   11
## [2,]   10   12
## 
## 
## $fac
## [1] A  B  AB O 
## Levels: O A B AB
e1[[1]]
## [1] 2 4 7 3
e2$fac
## [1] A  B  AB O 
## Levels: O A B AB
e2[2]
## $mat
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
e1[c(2,4)]
## [[1]]
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    5    6    7    8
## [3,]    9   10   11   12
## 
## [[2]]
## [1] A  B  AB O 
## Levels: O A B AB
dim(e2)
## NULL
length(e2)
## [1] 4
names(e2)
## [1] "vect"  "mat"   "array" "fac"

Data Frame

a17 <-11:15
a17
## [1] 11 12 13 14 15
d5 <-factor(LETTERS[6:10])
d5
## [1] F G H I J
## Levels: F G H I J
f1 <-data.frame(d5,a17)
f1
##   d5 a17
## 1  F  11
## 2  G  12
## 3  H  13
## 4  I  14
## 5  J  15

Akses Data Frame

f1[1,2]
## [1] 11
f1[3,]
##   d5 a17
## 3  H  13
f1$d5
## [1] F G H I J
## Levels: F G H I J
f1[,"a17"]
## [1] 11 12 13 14 15
colnames(f1)
## [1] "d5"  "a17"
str(f1)
## 'data.frame':    5 obs. of  2 variables:
##  $ d5 : Factor w/ 5 levels "F","G","H","I",..: 1 2 3 4 5
##  $ a17: int  11 12 13 14 15
summary(f1)
##  d5         a17    
##  F:1   Min.   :11  
##  G:1   1st Qu.:12  
##  H:1   Median :13  
##  I:1   Mean   :13  
##  J:1   3rd Qu.:14  
##        Max.   :15

Latihan 3

Seorang peneliti merancang sebuah perancangan percobaan RAKL dengan 4 perlakuan dan 3 kelompok (anggaplah respon percobaan berupa baris bilangan). Bantulah peneliti tersebut untuk membuat raw data seperti output sebagai berikut!

Perl <- paste("P",rep(1:4,each=3), sep="")
Kel <- factor(rep(1:3,4))
Resp <- seq(1,23,by=2)
data1 <- data.frame(Perl,Kel,Resp)
data1
##    Perl Kel Resp
## 1    P1   1    1
## 2    P1   2    3
## 3    P1   3    5
## 4    P2   1    7
## 5    P2   2    9
## 6    P2   3   11
## 7    P3   1   13
## 8    P3   2   15
## 9    P3   3   17
## 10   P4   1   19
## 11   P4   2   21
## 12   P4   3   23

PENGOLAHAN OBJEK DAN STRUKTUr KENDALI

Pengolahan Objek

x1<- c(2,6,9,5)
x2<- 1:4
x3<- x1 + 1:2
x4<- x1 + 1:3
## Warning in x1 + 1:3: longer object length is not a multiple of shorter object
## length
x5<- x1*x2
x6<- x1 %*% x1
x7<- x1 %o% x1

Operasi Dasar Vektor Karakter

y1 <- c("Institut Pertanian Bogor")
y1
## [1] "Institut Pertanian Bogor"
n1 <- nchar(y1)
n1
## [1] 24
y2 <- c("Adam","Pramesti ","Fathi","Ririn")
y2
## [1] "Adam"      "Pramesti " "Fathi"     "Ririn"
n2 <- nchar(y2)
n2
## [1] 4 9 5 5
y3 <- substr(y1,15,18) #”nian”
y3
## [1] "nian"
y4 <- substring(y1,15) #”nian Bogor”
y4
## [1] "nian Bogor"
y5 <- substring(y1,4,8) #”titut”
y5
## [1] "titut"

Operasi Dasar Matriks

z1<- matrix(1:6,2,3)
z1
##      [,1] [,2] [,3]
## [1,]    1    3    5
## [2,]    2    4    6
z2<- matrix(1:6,3,2,byrow=T)
z2
##      [,1] [,2]
## [1,]    1    2
## [2,]    3    4
## [3,]    5    6
z3<- matrix(6:9,2,2)
z3
##      [,1] [,2]
## [1,]    6    8
## [2,]    7    9
z4<- z1 %*% z2
z4
##      [,1] [,2]
## [1,]   35   44
## [2,]   44   56
z5<- z3*z4
z5
##      [,1] [,2]
## [1,]  210  352
## [2,]  308  504
invz <- solve(z4) 
invz %*% z4
##      [,1]         [,2]
## [1,]    1 2.842171e-14
## [2,]    0 1.000000e+00
invz
##           [,1]      [,2]
## [1,]  2.333333 -1.833333
## [2,] -1.833333  1.458333
h <- c(5,11)
h
## [1]  5 11
p <- solve(z4,h) 
p
## [1] -8.500  6.875
e <- eigen(z4) 
e
## eigen() decomposition
## $values
## [1] 90.7354949  0.2645051
## 
## $vectors
##           [,1]       [,2]
## [1,] 0.6196295 -0.7848945
## [2,] 0.7848945  0.6196295
e$values
## [1] 90.7354949  0.2645051
e[[2]]
##           [,1]       [,2]
## [1,] 0.6196295 -0.7848945
## [2,] 0.7848945  0.6196295
e
## eigen() decomposition
## $values
## [1] 90.7354949  0.2645051
## 
## $vectors
##           [,1]       [,2]
## [1,] 0.6196295 -0.7848945
## [2,] 0.7848945  0.6196295

Struktur Kendali

•Eksekusi bersyarat • if (kondisi) ekspresi else ekspresi • ifelse(kondisi,ekspresi benar,ekspresi salah) • switch(“kondisi”=ekspresi,…) •Pengulangan (loops) • for (objek in sekuens) ekspresi • while (kondisi) ekspresi • repeat ekspresi (untuk menghentikan gunakan perintah break) •Tanpa pengulangan • apply(array, margin, function, function args)

for (i in 1:5) print(i^2)
## [1] 1
## [1] 4
## [1] 9
## [1] 16
## [1] 25
i<-1
while (i<=5) {
print(i^2)
i=i+1
}
## [1] 1
## [1] 4
## [1] 9
## [1] 16
## [1] 25
y=runif(20)
for (i in y) {
if(i<0.5){
print(100*i)
} else print(i/100)
}
## [1] 1.020809
## [1] 7.197112
## [1] 27.1151
## [1] 0.005583264
## [1] 0.008218678
## [1] 0.00889217
## [1] 38.25674
## [1] 0.009073083
## [1] 21.49073
## [1] 0.006291529
## [1] 9.388005
## [1] 0.008924986
## [1] 10.42855
## [1] 0.006946286
## [1] 0.005411837
## [1] 28.17348
## [1] 39.66309
## [1] 9.979162
## [1] 12.43514
## [1] 40.46978
z=0
while(z<=10) {
y=runif(20)
z=sum(y)
print(z)
}
## [1] 12.00503
acak <-sample(1:5,1)
switch(EXPR=acak, "1" = "a", "2" = "z",
"3" = "m", "4" = "h", "5" = "t")
## [1] "t"
Z6 <-matrix(1:25,5,5)
apply(Z6,1,sum)
## [1] 55 60 65 70 75
apply(Z6,2,sd)
## [1] 1.581139 1.581139 1.581139 1.581139 1.581139

Latihan

a<-0
for(i in 1:5) { b<-a+i
print(b)
a<-b
}
## [1] 1
## [1] 3
## [1] 6
## [1] 10
## [1] 15
i<-1
z<-1
while(z<15)
{y<-z+i
z<-y
print(z)
i<-i+1
}
## [1] 2
## [1] 4
## [1] 7
## [1] 11
## [1] 16
i<-1
m<-2
repeat
{m<-m+i
print(m)
i<-i+1
if(m>15)
break
}
## [1] 3
## [1] 5
## [1] 8
## [1] 12
## [1] 17

MUNGING/WRANGLING DATA FRAME

Membuat peubah baru dalam Data Frame

Perl <- paste("P",rep(1:4,each=3), sep="")
Kel <- factor(rep(1:3,4))
Resp <- seq(1,23,by=2)
data1 <- data.frame(Perl,Kel,Resp)
print(data1)
##    Perl Kel Resp
## 1    P1   1    1
## 2    P1   2    3
## 3    P1   3    5
## 4    P2   1    7
## 5    P2   2    9
## 6    P2   3   11
## 7    P3   1   13
## 8    P3   2   15
## 9    P3   3   17
## 10   P4   1   19
## 11   P4   2   21
## 12   P4   3   23
Latihan 1

Pada data1, buatlah peubah baru1 yang berisi nilai dari 12 sampai 1 secara berurutan.

data1$baru1<-12:1
data1
##    Perl Kel Resp baru1
## 1    P1   1    1    12
## 2    P1   2    3    11
## 3    P1   3    5    10
## 4    P2   1    7     9
## 5    P2   2    9     8
## 6    P2   3   11     7
## 7    P3   1   13     6
## 8    P3   2   15     5
## 9    P3   3   17     4
## 10   P4   1   19     3
## 11   P4   2   21     2
## 12   P4   3   23     1

Subsetting Data

• Dilakukan untuk akses sebagian data

• Membuat ide logic untuk diterapkan dalam vektor logic yang diinginkan

• Fungsi yang digunakan :==, !=, >, >=, <, <=, %in%, duplicated(…), which(…), is.na(…), is.null (…), is.numeric(…), dll. ##### Latihan 2 Dari data1 tersebut ambillah yang termasuk kelompok 1

indeks1 <-data1$Kel == 1
data2 <-data1[indeks1,]
data2
##    Perl Kel Resp baru1
## 1    P1   1    1    12
## 4    P2   1    7     9
## 7    P3   1   13     6
## 10   P4   1   19     3
Latihan 3

Dari data1 tersebut ambillah yang termasuk kelompok 1 atau perlakuan P2

indeks2 <-data1$Kel == 1 | data1$Perl == "P2"
data3 <-data1[indeks2,]
data3
##    Perl Kel Resp baru1
## 1    P1   1    1    12
## 4    P2   1    7     9
## 5    P2   2    9     8
## 6    P2   3   11     7
## 7    P3   1   13     6
## 10   P4   1   19     3
Latihan 4

Dari data1 tersebut ambillah amatan yang responnya berupa bilangan prima

indeks3 <-data1$Resp %in% c(2,3,5,7,11,13,17,19,23)
data4 <-data1[indeks3,]
data4
##    Perl Kel Resp baru1
## 2    P1   2    3    11
## 3    P1   3    5    10
## 4    P2   1    7     9
## 6    P2   3   11     7
## 7    P3   1   13     6
## 9    P3   3   17     4
## 10   P4   1   19     3
## 12   P4   3   23     1

Sorting Data

•Dilakukan untuk mengurutkan data berdasarkan beberapa peubahtertentu

•Dilakukan dengan membuat vektor logika untuk melakukan pengurutan data

•Fungsi yang sering digunakan order(…), sort(…), rev(…), unique(…) ##### Latihan 5 Urutkan data1 tersebut berdasarkan kelompok secara ascending

indeks4 <-order(data1$Kel)
data5 <-data1[indeks4,]
data5
##    Perl Kel Resp baru1
## 1    P1   1    1    12
## 4    P2   1    7     9
## 7    P3   1   13     6
## 10   P4   1   19     3
## 2    P1   2    3    11
## 5    P2   2    9     8
## 8    P3   2   15     5
## 11   P4   2   21     2
## 3    P1   3    5    10
## 6    P2   3   11     7
## 9    P3   3   17     4
## 12   P4   3   23     1
Latihan 6

Urutkan data1 tersebut berdasarkan kelompok dan respon secara descending

indeks5 <-order(data1$Kel, data1$Resp, decreasing=TRUE)
data6 <-data1[indeks5,]
data6
##    Perl Kel Resp baru1
## 12   P4   3   23     1
## 9    P3   3   17     4
## 6    P2   3   11     7
## 3    P1   3    5    10
## 11   P4   2   21     2
## 8    P3   2   15     5
## 5    P2   2    9     8
## 2    P1   2    3    11
## 10   P4   1   19     3
## 7    P3   1   13     6
## 4    P2   1    7     9
## 1    P1   1    1    12
Latihan 7

Urutkan data1 tersebut berdasarkan kelompok secara ascending dan respon secara descending

indeks6 <-order(data1$Resp, decreasing=TRUE)
data7 <-data1[indeks6,]
indeks7 <-order(data7$Kel)
data8 <-data7[indeks7,]
data8
##    Perl Kel Resp baru1
## 10   P4   1   19     3
## 7    P3   1   13     6
## 4    P2   1    7     9
## 1    P1   1    1    12
## 11   P4   2   21     2
## 8    P3   2   15     5
## 5    P2   2    9     8
## 2    P1   2    3    11
## 12   P4   3   23     1
## 9    P3   3   17     4
## 6    P2   3   11     7
## 3    P1   3    5    10
data8$Resp
##  [1] 19 13  7  1 21 15  9  3 23 17 11  5
sort(data8$Resp)
##  [1]  1  3  5  7  9 11 13 15 17 19 21 23
rev(data8$Resp)
##  [1]  5 11 17 23  3  9 15 21  1  7 13 19
order(data8$Resp)
##  [1]  4  8 12  3  7 11  2  6 10  1  5  9
rank(data8$Resp)
##  [1] 10  7  4  1 11  8  5  2 12  9  6  3
data8$Resp>10
##  [1]  TRUE  TRUE FALSE FALSE  TRUE  TRUE FALSE FALSE  TRUE  TRUE  TRUE FALSE
which(data8$Resp>10)
## [1]  1  2  5  6  9 10 11
data8$Resp[data8$Resp>10]
## [1] 19 13 21 15 23 17 11
data8$Resp[which(data8$Resp>10)]
## [1] 19 13 21 15 23 17 11

Recording Data

•Digunakan untuk membuat nilai baru dari nilai peubahyang sudah ada

•Dapatdilakukansecara logical, fungsi ifelse(…), dan fungsi recode(…) ##### Latihan 8 Lakukanlah recoding pada data8 untuk variabel respon dengan kondisi jika respon < 15 maka Code = 1, selainnya Code = 0

#dengan logical
data8$Code1<-0*(data8$Resp>=15) + 1*(data8$Resp<15)
data8$Code1
##  [1] 0 1 1 1 0 0 1 1 0 0 1 1
#denganfungsiifelse
data8$Code2 <-ifelse(data8$Resp<15,1,0)
data8$Code2
##  [1] 0 1 1 1 0 0 1 1 0 0 1 1
#dengan fungsi recode
library(car)
## Loading required package: carData
data8$Code3 <-recode(data8$Resp,'1:14=1; else=0')
data8$Code3
##  [1] 0 1 1 1 0 0 1 1 0 0 1 1

Merging Data

•Bisa dilakukan dengan rbind(…)atau cbind(…)

•Lebih mudah dengan fungsi merge(…) ##### Latihan 9 Gabungkanlah data1 dengan tabel1 berdasarkan peubah pertamanya

tabel1<-data.frame( Tr=c("P4","P2","P5"), k1=c(50,100,200))
tabel1
##   Tr  k1
## 1 P4  50
## 2 P2 100
## 3 P5 200
data9<-merge(data1, tabel1, by.x=1, by.y=1, all=FALSE)
data9
##   Perl Kel Resp baru1  k1
## 1   P2   3   11     7 100
## 2   P2   1    7     9 100
## 3   P2   2    9     8 100
## 4   P4   1   19     3  50
## 5   P4   2   21     2  50
## 6   P4   3   23     1  50
data10<-merge(data1, tabel1, by.x="Perl",by.y="Tr",all=TRUE)
data10
##    Perl  Kel Resp baru1  k1
## 1    P1    1    1    12  NA
## 2    P1    2    3    11  NA
## 3    P1    3    5    10  NA
## 4    P2    3   11     7 100
## 5    P2    1    7     9 100
## 6    P2    2    9     8 100
## 7    P3    2   15     5  NA
## 8    P3    3   17     4  NA
## 9    P3    1   13     6  NA
## 10   P4    1   19     3  50
## 11   P4    2   21     2  50
## 12   P4    3   23     1  50
## 13   P5 <NA>   NA    NA 200

Reshaping Data

•Membentuk data baru dengan cara :

•Long towideformat

•Widetolongformat

•Menggunakan fungsi reshape(…) ##### Latihan 10 Ubahlah data1 menjadi data dengan setiap barisnya merupakan masing-masing perlakuan

data1
##    Perl Kel Resp baru1
## 1    P1   1    1    12
## 2    P1   2    3    11
## 3    P1   3    5    10
## 4    P2   1    7     9
## 5    P2   2    9     8
## 6    P2   3   11     7
## 7    P3   1   13     6
## 8    P3   2   15     5
## 9    P3   3   17     4
## 10   P4   1   19     3
## 11   P4   2   21     2
## 12   P4   3   23     1
#long to wide
data11 <-reshape(data1[,-4], idvar="Perl", timevar="Kel", direction="wide")
data11
##    Perl Resp.1 Resp.2 Resp.3
## 1    P1      1      3      5
## 4    P2      7      9     11
## 7    P3     13     15     17
## 10   P4     19     21     23
#wide to long
data12 <-reshape(data11, idvar="Perl", timevar="Kel", direction="long")
data12
##      Perl Kel Resp.1
## P1.1   P1   1      1
## P2.1   P2   1      7
## P3.1   P3   1     13
## P4.1   P4   1     19
## P1.2   P1   2      3
## P2.2   P2   2      9
## P3.2   P3   2     15
## P4.2   P4   2     21
## P1.3   P1   3      5
## P2.3   P2   3     11
## P3.3   P3   3     17
## P4.3   P4   3     23