Rubén Pizarro Gurrola
Junio 2018
3 + 5
[1] 8
5 * 6
[1] 30
6 - 7
[1] -1
7 * 8
[1] 56
2 ^ 3
[1] 8
6 %% 3
[1] 0
6 %% 4
[1] 2
6 %/% 3
[1] 2
6 %/% 4
[1] 1
a <- b <- c <- 6
a
[1] 6
b
[1] 6
c
[1] 6
2 -> a
a
[1] 2
'Ruben' -> nombre
nombre
[1] "Ruben"
2 -> a -> b -> c
a
[1] 2
b
[1] 2
c
[1] 2
2 + 5 * 3
[1] 17
(2 + 5) * 3
[1] 21
6 ^ 2 - 20 * 2
[1] -4
(6 ^ 2 - 20) * 2
[1] 32
(4 * 5) - (4 * 2) ^ 2
[1] -44
a <- 4
b <- 6
a > b
[1] FALSE
a < b
[1] TRUE
a >= b
[1] FALSE
a <= b
[1] TRUE
a == b
[1] FALSE
a != b
[1] TRUE
(a == 4)
[1] TRUE
b != (a + 2)
[1] FALSE
a <- 4
b <- 6
(a == 4)
[1] TRUE
b != (a + 2)
[1] FALSE
b == (a + 2)
[1] TRUE
str1 <- 'RUBEN'
str2 <- 'PATY'
str1 == str2
[1] FALSE
str1 == 'RUBEN'
[1] TRUE
str2 <- 'ruben'
str1 == str2
[1] FALSE
str2 <- 'RUBEN'
str1 == str2
[1] TRUE
eval(str1 == str2)
[1] TRUE
a <- 5; b <- 7; c<-6; d <- 5; x = 0.5
a
[1] 5
! TRUE
[1] FALSE
!(FALSE)
[1] TRUE
!(a == 5)
[1] FALSE
(a != b) & !(FALSE)
[1] TRUE
# Evaluar la expresión 0 < x < 1
(0 < x) & (x < 1)
[1] TRUE
a <- 5; b <- 7; c<-6; d <- 5; x = 0.5
(a < b) | (d < c)
[1] TRUE
(a < b) || (d < c)
[1] TRUE
(a < b) & (d < c)
[1] TRUE
!(a < b) && (d < c)
[1] FALSE
a <- 5; b <- 7; c<-6; d <- 5; x = 0.5
v1 <- 'JUAN'; v2 <- 'juan'
identical(v1, v2)
[1] FALSE
v2 <- 'JUAN '
identical(v1, v2)
[1] FALSE
v2 <- trimws(v2,'b') # remueve espacios a ambos lados ? identical
identical(v1, v2)
[1] TRUE
a <- 5; b <- 7; c<-6; d <- 5; x = 0.5
v1 <- 'JUAN'; v2 <- 'juan'
identical(v1, v2)
[1] FALSE
v2 <- 'JUAN '
identical(v1, v2)
[1] FALSE
v2 <- trimws(v2,'b') # remueve espacios a ambos lados ? identical
identical(v1, v2)
[1] TRUE
isTRUE(TRUE)
[1] TRUE
isTRUE(FALSE)
[1] FALSE
a <- 5; b <- 7; c<-6; d <- 5;
(a < 6) & (d == 5)
[1] TRUE
xor((a < 6) & (d == 5), FALSE)
[1] TRUE
! xor((a < 6) & (d == 5), FALSE)
[1] FALSE
(xor((a < 6) & (d == 5), FALSE)) | (TRUE)
[1] TRUE