Exercise

1. calculate the sum of all numbers from 15 to 75 by
   • using the sum() function and
   • a loop
2. Write a function, which divides a number a by a number b.
3. Calculate all length by high values (ore similar numbers) for the fibulae from archdata by
   • using apply(),
   • a simple vector calculation and
   • a loop
4. write a loop converting the long-form data to wide-form
   • df <- data.frame( key = c(“a”, “a”, “b”, “a”, “b”, “b”, “c”, “c”, “c”), value = c(1, 3, 54, 2, 66, 43, 743, 234, 444) )
   • target: df2 <- data.frame( a = c(), b = c(), c = c() )

Exercise 1: Numbers

• calculate the sum of all numbers from 15 to 75 by
  • using the sum() function and
  • a loop

Exercise 1: Numbers without loop

sum(15:75)

## [1] 2745

Exercise 1: Numbers with loop

nsum <- 0
numbers <- 15:75
for (i in 1:length(numbers)){
    nsum <- nsum + numbers[i]
}
nsum

## [1] 2745

Exercise 2: Division function

• Write a function, which divides a number a by a number b.

division1 <- function(a, b){
    a / b
}
division1(4,3)

## [1] 1.333333

Exercise 2: Division function

This version is required for apply.

division2 <- function(a){
    a[1] / a[2]
}
a <- c(4,3)
division2(a)

## [1] 1.333333

Exercise 3: Fibulae

• Calculate all length by high values (ore similar numbers) for the fibulae from archdata by
  • using apply(),
  • a simple vector calculation and
  • a loop

The package archdata (Carlson/Roth 2021) provides some archaeological data.

library(archdata)
data("Fibulae")
head(Fibulae)

Exercise 3: Fibulae

fib_df <- data.frame(length = Fibulae$Length, hight = Fibulae$BH)
head(fib_df)

Exercise 3: apply

apply(fib_df, MARGIN = 1, FUN = division2)

##  [1] 4.750000 5.000000 4.000000 2.846154 2.956522 3.666667 2.812500 2.222222
##  [9] 4.153846 2.294118 3.086957 4.588235 3.133333 3.153846 2.533333 2.750000
## [17] 3.333333 2.769231 2.722222 3.117647 8.533333 3.277778 3.142857 3.133333
## [25] 2.045455 3.533333 2.947368 2.600000 2.153846 6.111111

Exercise 3: vector

fib_df$length / fib_df$hight

##  [1] 4.750000 5.000000 4.000000 2.846154 2.956522 3.666667 2.812500 2.222222
##  [9] 4.153846 2.294118 3.086957 4.588235 3.133333 3.153846 2.533333 2.750000
## [17] 3.333333 2.769231 2.722222 3.117647 8.533333 3.277778 3.142857 3.133333
## [25] 2.045455 3.533333 2.947368 2.600000 2.153846 6.111111

Exercise 3: loop

result <- c()
for (i in 1:nrow(fib_df)){
    result[i] <- fib_df[i,1] / fib_df[i,2]
}
result

##  [1] 4.750000 5.000000 4.000000 2.846154 2.956522 3.666667 2.812500 2.222222
##  [9] 4.153846 2.294118 3.086957 4.588235 3.133333 3.153846 2.533333 2.750000
## [17] 3.333333 2.769231 2.722222 3.117647 8.533333 3.277778 3.142857 3.133333
## [25] 2.045455 3.533333 2.947368 2.600000 2.153846 6.111111

Exercise 4: Long-wide

• write a loop converting the long-form data to wide-form
  • df <- data.frame( key = c(“a”, “a”, “b”, “a”, “b”, “b”, “c”, “c”, “c”), value = c(1, 3, 54, 2, 66, 43, 743, 234, 444) )
  • target: df2 <- data.frame( a = c(), b = c(), c = c() )

Assuming that the order of the a-entries, the b-entries and the c-entries is correct and that the different categories belong to three observations.

df <- data.frame(
                key = c("a", "a", "b", "a", "b", "b", "c", "c", "c"),
                value = c(1, 3, 54, 2, 66, 43, 743, 234, 444)
                )
df2 <- data.frame(
                  a = rep(0,3),
                  b = rep(0,3),
                  c = rep(0,3)
                  )

Exercise 4: Long-wide: Loop

ai <- 1
bi <- 1
ci <- 1
for (i in 1:nrow(df)){
    if (df$key[i] == "a") {
        df2$a[ai] <- df$value[i]
        ai <- ai + 1
    }
    if (df$key[i] == "b") {
        df2$b[bi] <- df$value[i]
        bi <- bi + 1
        }
    if (df$key[i] == "c") {
        df2$c[ci] <- df$value[i]
        ci <- ci + 1
        }
}
df2