R Fundamentals I

Vectorisation

Most of R’s functions are vectorised, meaning that the function will operate on all elements of a vector without needing to loop through and act on each element one at a time. This makes writing code more concise, easy to read, and less error prone.

x <- 1:4
x * 2

[1] 2 4 6 8

The multiplication happened to each element of the vector.

We can also add two vectors together:

y <- 6:9
x + y

[1]  7  9 11 13

Each element of x was added to its corresponding element of y:

x:  1  2  3  4
    +  +  +  +
y:  6  7  8  9
---------------
    7  9 11 13

Vectorised operations work element-wise on matrices:

m <- matrix(1:12, nrow = 3, ncol = 4)
m * -1  

     [,1] [,2] [,3] [,4]
[1,]   -1   -4   -7  -10
[2,]   -2   -5   -8  -11
[3,]   -3   -6   -9  -12

To combine a matrix with a vector, keep in mind that the element-wise combination happens by columns:

m

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

x

[1] 1 2 3 4

m * x

     [,1] [,2] [,3] [,4]
[1,]    1   16   21   20
[2,]    4    5   32   33
[3,]    9   12    9   48

To do it by rows, first create a matrix from the vector and then combine the two matrices:

xm <- matrix(rep(x, 3), nrow = 3, byrow = TRUE)
xm

     [,1] [,2] [,3] [,4]
[1,]    1    2    3    4
[2,]    1    2    3    4
[3,]    1    2    3    4

m * xm

     [,1] [,2] [,3] [,4]
[1,]    1    8   21   40
[2,]    2   10   24   44
[3,]    3   12   27   48

m <- matrix(1:12, nrow = 3, ncol = 4)
m

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

Applying functions across rows/columns

What if we need an operation (average, sum etc.) across rows or across columns?

To support this, we can use the apply function.

apply allows us to repeat a function on all of the rows (MARGIN = 1) or columns (MARGIN = 2) of a data frame. For example, using the m matrix, instead of

mean(m[1,])

[1] 5.5

mean(m[2,])

[1] 6.5

mean(m[3,])

[1] 7.5

we do

apply(m, 1, mean) # average across rows

[1] 5.5 6.5 7.5

Also, we can do

apply(m, 2, min) # minimum across columns

[1]  1  4  7 10

apply(m, 2, sum) # sum across columns

[1]  6 15 24 33

Similarly, with our pierce subset dataset created in the previous session, we can get household totals:

apply(pierce[,2:6], 2, sum)

hh2016 hh2020 hh2030 hh2040 hh2050 
320054 335245 386446 438732 488373 

and the all-cities hh dataset:

apply(hh[,2:6], 2, sum)

 hh2016  hh2020  hh2030  hh2040  hh2050 
1581863 1657497 1912422 2172765 2419919 

However, for a dataset with missing values:

df <- data.frame(id = letters[1:5], X = c(NA, rep(100, 4)), Y = c(rep(5, 3), NA, NA))
df

  id   X  Y
1  a  NA  5
2  b 100  5
3  c 100  5
4  d 100 NA
5  e 100 NA

we get:

apply(df[,2:3], 2, sum)

 X  Y 
NA NA 

because

sum(c(3, NA))

[1] NA

but

sum(c(3, NA), na.rm=TRUE)

[1] 3

Therefore

apply(df[,2:3], 2, sum, na.rm=TRUE)

  X   Y 
400  15 

Challenge solutions

m <- matrix(1:12, nrow = 3, ncol = 4)
m

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

          [,1]      [,2]      [,3]       [,4]
[1,] 1.0000000 0.2500000 0.1428571 0.10000000
[2,] 0.5000000 0.2000000 0.1250000 0.09090909
[3,] 0.3333333 0.1666667 0.1111111 0.08333333

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    0    0    0    0
[3,]   -3   -6   -9  -12

      [,1]  [,2]  [,3]  [,4]
[1,]  TRUE FALSE  TRUE FALSE
[2,] FALSE  TRUE FALSE  TRUE
[3,]  TRUE FALSE  TRUE FALSE