Small snippets of R code written while learning R.

R Notes.r

> data.frame(a=1:5, b=6:10)
  a  b
1 1  6
2 2  7
3 3  8
4 4  9
5 5 10

> data.frame(a=1:5, b=6:9)[["a"]]
Error in data.frame(a = 1:5, b = 6:9) : 
  arguments imply differing number of rows: 5, 4

> list(a=1, b=2)[["a"]]
[1] 1
> list(a=1, b=2)[[1]]
[1] 1
> list(a=1, b=2)[1]
$a
[1] 1

> matrix(0, nrow=2, ncol=3)
     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    0    0    0
 
 
> matrix(c(0, 1), nrow=2, ncol=3)
     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    1    1    1

> matrix(c(0, 1), nrow=2, ncol=3, byrow=TRUE)
     [,1] [,2] [,3]
[1,]    0    1    0
[2,]    1    0    1

> m <- matrix(c(0, 1), nrow=2, ncol=3)
> m * m 
     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    1    1    1

> m %*% t(m)
     [,1] [,2]
[1,]    0    0
[2,]    0    3

> a <- if (TRUE) 1 else 2
> a
[1] 1
> a <- if (FALSE) 1 else 2
> a
[1] 2

> f <- function() {
+   1 
+ }
> f()
[1] 1

> lapply 
function (X, FUN, ...) 
{
    FUN <- match.fun(FUN)
    if (!is.vector(X) || is.object(X)) 
        X <- as.list(X)
    .Internal(lapply(X, FUN))
}
<bytecode: 0x101a432a8>
<environment: namespace:base>

> ?lapply
starting httpd help server ... done

> lapply(1:5, function(x) x*x)
[[1]]
[1] 1

[[2]]
[1] 4

[[3]]
[1] 9

[[4]]
[1] 16

[[5]]
[1] 25

> sapply(1:5, function(x) x*x)
[1]  1  4  9 16 25

> sapply(1:5, function(x) c(x*x, x))
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    4    9   16   25
[2,]    1    2    3    4    5

> apply(m, 1, sum)
[1] 0 3
> m
     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    1    1    1
> apply(m, 2, sum)
[1] 1 1 1

> df <- data.frame(a=1:5, b=6:10)
> df
  a  b
1 1  6
2 2  7
3 3  8
4 4  9
5 5 10

> apply(df, 1, sum)
[1]  7  9 11 13 15

> df <- data.frame(a=paste0(as.character(1:5), "str"), b=6:10)
> df
     a  b
1 1str  6
2 2str  7
3 3str  8
4 4str  9
5 5str 10

> head(df)
     a  b
1 1str  6
2 2str  7
3 3str  8
4 4str  9
5 5str 10

> tail(df)
     a  b
1 1str  6
2 2str  7
3 3str  8
4 4str  9
5 5str 10

> unlist(lapply(1:5, function(x) x*x))
[1]  1  4  9 16 25

> length(1:5)
[1] 5
> length(m)
[1] 6
> dim(m)
[1] 2 3
> dim(df)
[1] 5 2
> length(df)
[1] 2

> mode(df)
[1] "list"

> df$a
[1] 1str 2str 3str 4str 5str
Levels: 1str 2str 3str 4str 5str 

> df[["a"]]
[1] 1str 2str 3str 4str 5str
Levels: 1str 2str 3str 4str 5str

> as.numeric(df$a)
[1] 1 2 3 4 5

#The above as.numeric shows how R stores those strings as ints. It's like ENUM of C - it's on by default - so you can switch it off.

> names(df)
[1] "a" "b"

> table(c(1, 2, 3, 1, 2))

1 2 3 
2 2 1 
# This is like a histogram

> t <- table(c(1, 2, 3, 1, 2))
> t["1"]
1 
2 
> t["2"]
2 
2 

#So here, R is storing each of those numbers as a string ENUM, and counting it.

> names(t)
[1] "1" "2" "3"
> as.vector(t)
[1] 2 2 1

> f <- function() { b <- 5 }
> b <- 2
> b
[1] 2
> f()
> b
[1] 2
> f<- function() { b <<- 5 }
> f()
> b
[1] 5

ls()
#prints all the vars in the working memory

> match(1:5, c(1, 0, 2, 0, 3, 0, 4))
[1]  1  3  5  7 NA

> unique(c(1, 2, 1))
[1] 1 2

> mode(data[[1]])
[1] "list"
> length(data[[1]])
[1] 990

> sort(c(2, 1, 3, 5, 4))
[1] 1 2 3 4 5
> sort(c(2, 1, 3, 5, 4), decreasing=TRUE)
[1] 5 4 3 2 1

#In R, x[-n] returns a copy of x with the nth element removed.

> seq(1,10,3)
[1]  1  4  7 10
#This is similar to range(a, b, n) in Python, except 
#Python uses half-open intervals and so the 10 would 
#not be included in this example. The step size 
#argument n defaults to 1 in both R and Python.

#a:b is an abbreviation for seq(a, b, 1).
> 1:5
[1] 1 2 3 4 5

> seq(1,5,length=7)
[1] 1.000000 1.666667 2.333333 3.000000 3.666667 4.333333 5.000000