a1<- c(2,4,7,3)
assign("a2",c("2","4","7","3"))
a1
## [1] 2 4 7 3
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
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
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"
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
#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"
#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
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
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
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
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
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
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]
## $mat
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 5 6 7 8
## [3,] 9 10 11 12
e1[c(2,4)]
## [[1]]
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 5 6 7 8
## [3,] 9 10 11 12
##
## [[2]]
## [1] A B AB O
## Levels: O A B AB
dim(e2)
## NULL
length(e2)
## [1] 4
names(e2)
## [1] "vect" "mat" "array" "fac"
a17 <-11:15
a17
## [1] 11 12 13 14 15
d5 <-factor(LETTERS[6:10])
d5
## [1] F G H I J
## Levels: F G H I J
f1 <-data.frame(d5,a17)
f1
## d5 a17
## 1 F 11
## 2 G 12
## 3 H 13
## 4 I 14
## 5 J 15
f1[1,2]
## [1] 11
f1[3,]
## d5 a17
## 3 H 13
f1$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
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
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
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"
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
##
## $vectors
## [,1] [,2]
## [1,] 0.6196295 -0.7848945
## [2,] 0.7848945 0.6196295
•Eksekusi bersyarat • if (kondisi) ekspresi else ekspresi • ifelse(kondisi,ekspresi benar,ekspresi salah) • switch(“kondisi”=ekspresi,…) •Pengulangan (loops) • for (objek in sekuens) ekspresi • while (kondisi) ekspresi • repeat ekspresi (untuk menghentikan gunakan perintah break) •Tanpa pengulangan • apply(array, margin, function, function args)
for (i in 1:5) print(i^2)
## [1] 1
## [1] 4
## [1] 9
## [1] 16
## [1] 25
i<-1
while (i<=5) {
print(i^2)
i=i+1
}
## [1] 1
## [1] 4
## [1] 9
## [1] 16
## [1] 25
y=runif(20)
for (i in y) {
if(i<0.5){
print(100*i)
} else print(i/100)
}
## [1] 1.020809
## [1] 7.197112
## [1] 27.1151
## [1] 0.005583264
## [1] 0.008218678
## [1] 0.00889217
## [1] 38.25674
## [1] 0.009073083
## [1] 21.49073
## [1] 0.006291529
## [1] 9.388005
## [1] 0.008924986
## [1] 10.42855
## [1] 0.006946286
## [1] 0.005411837
## [1] 28.17348
## [1] 39.66309
## [1] 9.979162
## [1] 12.43514
## [1] 40.46978
z=0
while(z<=10) {
y=runif(20)
z=sum(y)
print(z)
}
## [1] 12.00503
acak <-sample(1:5,1)
switch(EXPR=acak, "1" = "a", "2" = "z",
"3" = "m", "4" = "h", "5" = "t")
## [1] "t"
Z6 <-matrix(1:25,5,5)
apply(Z6,1,sum)
## [1] 55 60 65 70 75
apply(Z6,2,sd)
## [1] 1.581139 1.581139 1.581139 1.581139 1.581139
a<-0
for(i in 1:5) { b<-a+i
print(b)
a<-b
}
## [1] 1
## [1] 3
## [1] 6
## [1] 10
## [1] 15
i<-1
z<-1
while(z<15)
{y<-z+i
z<-y
print(z)
i<-i+1
}
## [1] 2
## [1] 4
## [1] 7
## [1] 11
## [1] 16
i<-1
m<-2
repeat
{m<-m+i
print(m)
i<-i+1
if(m>15)
break
}
## [1] 3
## [1] 5
## [1] 8
## [1] 12
## [1] 17
Perl <- paste("P",rep(1:4,each=3), sep="")
Kel <- factor(rep(1:3,4))
Resp <- seq(1,23,by=2)
data1 <- data.frame(Perl,Kel,Resp)
print(data1)
## Perl Kel Resp
## 1 P1 1 1
## 2 P1 2 3
## 3 P1 3 5
## 4 P2 1 7
## 5 P2 2 9
## 6 P2 3 11
## 7 P3 1 13
## 8 P3 2 15
## 9 P3 3 17
## 10 P4 1 19
## 11 P4 2 21
## 12 P4 3 23
Pada data1, buatlah peubah baru1 yang berisi nilai dari 12 sampai 1 secara berurutan.
data1$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
• 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$Kel == 1
data2 <-data1[indeks1,]
data2
## Perl Kel Resp baru1
## 1 P1 1 1 12
## 4 P2 1 7 9
## 7 P3 1 13 6
## 10 P4 1 19 3
Dari data1 tersebut ambillah yang termasuk kelompok 1 atau perlakuan P2
indeks2 <-data1$Kel == 1 | data1$Perl == "P2"
data3 <-data1[indeks2,]
data3
## Perl Kel Resp baru1
## 1 P1 1 1 12
## 4 P2 1 7 9
## 5 P2 2 9 8
## 6 P2 3 11 7
## 7 P3 1 13 6
## 10 P4 1 19 3
Dari data1 tersebut ambillah amatan yang responnya berupa bilangan prima
indeks3 <-data1$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
•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$Kel)
data5 <-data1[indeks4,]
data5
## Perl Kel Resp baru1
## 1 P1 1 1 12
## 4 P2 1 7 9
## 7 P3 1 13 6
## 10 P4 1 19 3
## 2 P1 2 3 11
## 5 P2 2 9 8
## 8 P3 2 15 5
## 11 P4 2 21 2
## 3 P1 3 5 10
## 6 P2 3 11 7
## 9 P3 3 17 4
## 12 P4 3 23 1
Urutkan data1 tersebut berdasarkan kelompok dan respon secara descending
indeks5 <-order(data1$Kel, data1$Resp, decreasing=TRUE)
data6 <-data1[indeks5,]
data6
## Perl Kel Resp baru1
## 12 P4 3 23 1
## 9 P3 3 17 4
## 6 P2 3 11 7
## 3 P1 3 5 10
## 11 P4 2 21 2
## 8 P3 2 15 5
## 5 P2 2 9 8
## 2 P1 2 3 11
## 10 P4 1 19 3
## 7 P3 1 13 6
## 4 P2 1 7 9
## 1 P1 1 1 12
Urutkan data1 tersebut berdasarkan kelompok secara ascending dan respon secara descending
indeks6 <-order(data1$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
•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
•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
•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