Vectors and Functions Solutions

Part a

mean(c(-100, 0, 1, 2, 3, 6, 50, 73), trim = 0.25)

## [1] 3

Part b

trimfraction <- 10/length(rivers)
mean(rivers, trim = trimfraction)

## [1] 505.719

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Part a

attach(women)
cor(height, weight)

## [1] 0.9954948

Part b

cov(height, weight)

## [1] 69

Part c

cor(height, weight, method = "pearson")

## [1] 0.9954948

cor(height, weight, method = "kendall")

## [1] 1

cor(height, weight, method = "spearman")

## [1] 1

The default is method='pearson'.

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Part a

floor and trunc. If you experienced trouble with the exercise, a good way to proceed would be to first write out for each element of x separately, which of the three functions could have been chosen. E.g.:

300.99 -> 300 : trunc or floor

1.6 -> 1: trunc or floor

583 -> 583: trunc or floor or ceiling

42.10 -> 42: trunc or floor

Because only trunc and floor could have been chosen for all 4 values, only these two could have been chosen for the entire vector as well.

x <- c(300.99, 1.6, 583, 42.1)
floor(x)

## [1] 300   1 583  42

trunc(x)

## [1] 300   1 583  42

Part b

trunc. Note that contrary to part (a), floor doesn’t work, because of the negative value (-2.4) in x.

x <- c(152.34, 1940.63, 1.0001, -2.4, sqrt(26))
trunc(x)

## [1]  152 1940    1   -2    5

Part c

trunc and ceiling. Note: all values are negative because of the - sign before c(...).

x <- -c(3.2, 444.35, 1/9, 100)
trunc(x)

## [1]   -3 -444    0 -100

ceiling(x)

## [1]   -3 -444    0 -100

Part d

ceiling.

x <- c(35.6, 670, -5.4, 3^3)
ceiling(x)

## [1]  36 670  -5  27

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Part a

round with d=0.

x <- c(35.63, 300.2, 0.39, -57.8)
round(x, digits = 0)

## [1]  36 300   0 -58

Part b

signif with d=3

x <- c(35.63, 300.2, 0.39, -57.8)
signif(x, digits = 3)

## [1]  35.60 300.00   0.39 -57.80

Part c

Both signif and round with d=2

x <- c(3.8, 0.983, -23, 7.1)
signif(x, 2)

## [1]   3.80   0.98 -23.00   7.10

round(x, 2)

## [1]   3.80   0.98 -23.00   7.10

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