9.8 Exercise 4. Matrix manipulation

Create the script “exercise4.R” and save it to the “Rcourse/Module1” directory: you will save all the commands of exercise 4 in that script.
Remember you can comment the code using #.

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getwd()
setwd("Rcourse/Module1")
setwd("~/Rcourse/Module1")

1- Create three numeric vectors x, y, z, each of 4 elements of your choice.

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x <- 2:5
y <- 6:9
z <- 7:4

Use rbind() to create a matrix mat (3 rows and 4 columns) out of x, y and z.

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mat <- rbind(x, y, z)

2- Create the same matrix now using the matrix function.

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mat <- matrix(data=c(x, y, z), nrow=3, ncol=4)
# Try with the "byrow=TRUE" parameter: what is different ?
mat <- matrix(data=c(x, y, z), nrow=3, ncol=4, byrow=TRUE)

3- Add names to mat’s columns: “a”, “b”, “c”, “d”, respectively.

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colnames(mat) <- c("a", "b", "c", "d")

4- Calculate the sum of each row, and the sum of each column

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rowSums(mat); colSums(mat)

## [1] 14 30 22

##  a  b  c  d 
## 15 16 17 18

5- Create the matrix mat2 as:

mat2 <- matrix(c(seq(from=1, to=10, by=2), 5:1, rep(x=2017, times=5)), ncol=3)

What does function seq() do?

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seq generate sequences of numbers. Here, it creates a sequences from 1 to 10 with a step of 2 numbers.

6- What are the dimensions of mat2 (number of rows and number of columns)?

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# number of rows
nrow(mat2)

## [1] 5

# number of columns
ncol(mat2)

## [1] 3

# dimensions: number of rows, number of columns
dim(mat2)

## [1] 5 3

7- Add column names to mat2: “day”, “month” and “year”, respectively.

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colnames(mat2) <- c("day", "month", "year")

8- Add row names to mat2: letters “A” to “E”

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rownames(mat2) <- c("A", "B", "C", "D", "E")
rownames(mat2) <- LETTERS[1:5]

9- Shows row(s) of mat2 where the month column is greater than or equal to 3.

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# select column month
mat2[, "month"]

## A B C D E 
## 5 4 3 2 1

# element(s) of column month that is (are) greater than or equal to 3
mat2[,"month"] >= 3

##     A     B     C     D     E 
##  TRUE  TRUE  TRUE FALSE FALSE

# finally select row(s) where the month columns is greater than or equal to 3
mat2[mat2[,"month"] >= 3,]

##   day month year
## A   1     5 2017
## B   3     4 2017
## C   5     3 2017

10- Replace all elements of mat2 that are equal to 2017 with 2018.

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# which elements of mat2 that are exactly equal to 2017
mat2==2017

##     day month year
## A FALSE FALSE TRUE
## B FALSE FALSE TRUE
## C FALSE FALSE TRUE
## D FALSE FALSE TRUE
## E FALSE FALSE TRUE

# retrieve actual elements
mat2[mat2==2017]

## [1] 2017 2017 2017 2017 2017

# replace all 2017 with 2018
mat2[mat2==2017] <- 2018

11- Multiply all elements of the 2nd column of mat2 by 7. Reassign mat2!

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# multiply all elements of the 2nd column of mat2 by 7
mat2[,2] * 7

##  A  B  C  D  E 
## 35 28 21 14  7

# reassign mat2 with the new values of column 2
mat2[,2] <- mat2[,2] * 7

12- Add the column named “time” to mat2, that contains values 8, 12, 11, 10, 8. Save in the new object mat3.

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mat3 <- cbind(mat2, time=c(8, 12, 11, 10, 8))

13- Replace all elements of mat3 that are less than 3 with NA.

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# which elements of mat3 that are less than 3
mat3 < 3

##     day month  year  time
## A  TRUE FALSE FALSE FALSE
## B FALSE FALSE FALSE FALSE
## C FALSE FALSE FALSE FALSE
## D FALSE FALSE FALSE FALSE
## E FALSE FALSE FALSE FALSE

# actually elements of mat3 that are less than 3
mat3[mat3 < 3]

## [1] 1

# reassign elements of mat3 that are less than 3 with NA
mat3[mat3 < 3] <- NA

14- Remove rows from mat3 if a NA is present. Save in the new object mat4.

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mat4 <- na.omit(mat3)

15- Retrieve the smaller value of each column of mat4.

Try different approaches:

Retrieve the minimum for each column one by one.

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min(mat4[,"day"])

## [1] 3

min(mat4[,"month"])

## [1] 7

min(mat4[,"year"])

## [1] 2018

min(mat4[,"time"])

## [1] 8

Retrieve the minimum of all columns simultaneously using the apply() function.

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# mat4: object
# 2: by column
# min: function to apply
apply(mat4, 2, min)

##   day month  year  time 
##     3     7  2018     8