b1 <- 10:1 # klesajúca postupnosť
b2 <- seq(2, 20, by = 2) # párne čísla od 2 po 20
b3 <- rep(c(1, 0), times = 5) # opakujúci sa vzor
b4 <- runif(6, 5, 15) # 6 čísel z intervalu [5,15]
b5 <- rnorm(6, mean = 0, sd = 1)
b1; b2; b3; b4; b5
## [1] 10 9 8 7 6 5 4 3 2 1
## [1] 2 4 6 8 10 12 14 16 18 20
## [1] 1 0 1 0 1 0 1 0 1 0
## [1] 6.457955 9.754494 12.350799 14.556809 12.954036 6.853422
## [1] -2.5230657 1.1330321 -2.2776428 1.0072966 -0.9658459 -0.4864616
u <- c(3, 6, 9, 12)
u + 5
## [1] 8 11 14 17
u / 3
## [1] 1 2 3 4
u^2
## [1] 9 36 81 144
sum(u)
## [1] 30
c(1,2,3) * c(4,5,6) # Hadamardov súčin
## [1] 4 10 18
Vytvor vektor t s číslami od 100 do 120 a vypočítaj aritmetický priemer všetkých nepárnych čísel.
t <- 100:120
mean(t[t %% 2 == 1])
## [1] 110
N <- matrix(1:20, nrow = 4, ncol = 5)
N
## [,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
X <- matrix(c(2,4,6,8), nrow = 2)
Y <- matrix(c(1,3,5,7), nrow = 2)
X + Y
## [,1] [,2]
## [1,] 3 11
## [2,] 7 15
X * Y
## [,1] [,2]
## [1,] 2 30
## [2,] 12 56
X %*% Y
## [,1] [,2]
## [1,] 20 52
## [2,] 28 76
t(X)
## [,1] [,2]
## [1,] 2 4
## [2,] 6 8
Vytvorime maticu 4x4 z čísel 1..16 a vypočítajme stĺpcové maximá a súčin matíc (N^tN).
N2 <- matrix(1:16, nrow = 4, byrow = TRUE)
colMaxs <- apply(N2, 2, max)
colMaxs
## [1] 13 14 15 16
t(N2) %*% N2
## [,1] [,2] [,3] [,4]
## [1,] 276 304 332 360
## [2,] 304 336 368 400
## [3,] 332 368 404 440
## [4,] 360 400 440 480
y <- c(5, NA, 7, 9, NA, 11)
is.na(y)
## [1] FALSE TRUE FALSE FALSE TRUE FALSE
mean(y, na.rm = TRUE)
## [1] 8
na.omit(y)
## [1] 5 7 9 11
## attr(,"na.action")
## [1] 2 5
## attr(,"class")
## [1] "omit"
Spočítajme, koľko chýbajúcich hodnôt obsahuje vektor y.
sum(is.na(y))
## [1] 2
q <- c(12, 15, 18, 7, 22, 9)
mean(q)
## [1] 13.83333
median(q)
## [1] 13.5
var(q)
## [1] 31.76667
summary(q)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 7.00 9.75 13.50 13.83 17.25 22.00
sort(q)
## [1] 7 9 12 15 18 22
Určime rozdiel medzi maximom a minimom vo vektore q.
max(q) - min(q)
## [1] 15