## B.3 Introduction to R

This section provides a collection of self-explainable code snippets for the programming language R . These snippets are not meant to provide an exhaustive introduction to R but just to illustrate the very basic functions and methods.

In the following, # denotes comments to the code and ## outputs of the code.

### Simple computations

# The console can act as a simple calculator
1.0 + 1.1
2 * 2
3/2
2^3
1/0
0/0

# Use ";" for performing several operations in the same line
(1 + 3) * 2 - 1; 3 + 2

# Elemental mathematical functions
sqrt(2); 2^0.5
exp(1)
log(10) # Natural logarithm
log10(10); log2(10) # Logs in base 10 and 2
sin(pi); cos(0); asin(0)
tan(pi/3)
sqrt(-1)

# Remember to close the parenthesis -- errors below
1 +
(1 + 3
## Error: <text>:24:0: unexpected end of input
## 22: 1 +
## 23: (1 + 3
##    ^

Compute:

1. $$\frac{e^{2}+\sin(2)}{\cos^{-1}\left(\tfrac{1}{2}\right)+2}.$$ Answer: 2.723274.
2. $$\sqrt{3^{2.5}+\log(10)}.$$ Answer: 4.22978.
3. $$(2^{0.93}-\log_2(3 + \sqrt{2+\sin(1)}))10^{\tan(1/3))}\sqrt{3^{2.5}+\log(10)}.$$ Answer: -3.032108.

### Variables and assignment

# Any operation that you perform in R can be stored in a variable
# (or object) with the assignment operator "<-"
x <- 1

# To see the value of a variable, simply type it
x
## [1] 1

# A variable can be overwritten
x <- 1 + 1

# Now the value of x is 2 and not 1, as before
x
## [1] 2

# Capitalization matters
X <- 3
x; X
## [1] 2
## [1] 3

# See what are the variables in the workspace
ls()
## [1] "x" "X"

# Remove variables
rm(X)
X
## Error in eval(expr, envir, enclos): object 'X' not found

Do the following:

1. Store $$-123$$ in the variable y.
2. Store the log of the square of y in z.
3. Store $$\frac{y-z}{y+z^2}$$ in y and remove z.
4. Output the value of y. Answer: 4.366734.

### Vectors

# We combine numbers with the function "c"
c(1, 3)
## [1] 1 3
c(1.5, 0, 5, -3.4)
## [1]  1.5  0.0  5.0 -3.4

# A handy way of creating integer sequences is the operator ":"
1:5
## [1] 1 2 3 4 5

# Storing some vectors
myData <- c(1, 2)
myData2 <- c(-4.12, 0, 1.1, 1, 3, 4)
myData
## [1] 1 2
myData2
## [1] -4.12  0.00  1.10  1.00  3.00  4.00

# Entrywise operations
myData + 1
## [1] 2 3
myData^2
## [1] 1 4

# If you want to access a position of a vector, use [position]
myData[1]
## [1] 1
myData2[6]
## [1] 4

# You can also change elements
myData[1] <- 0
myData
## [1] 0 2

# Think on what you want to access...
myData2[7]
## [1] NA
myData2[0]
## numeric(0)

# If you want to access all the elements except a position,
# use [-position]
myData2[-1]
## [1] 0.0 1.1 1.0 3.0 4.0
myData2[-2]
## [1] -4.12  1.10  1.00  3.00  4.00

# Also with vectors as indexes
myData2[1:2]
## [1] -4.12  0.00
myData2[myData]
## [1] 0

# And also
myData2[-c(1, 2)]
## [1] 1.1 1.0 3.0 4.0

# But do not mix positive and negative indexes!
myData2[c(-1, 2)]
## Error in myData2[c(-1, 2)]: only 0's may be mixed with negative subscripts

# Remove the first element
myData2 <- myData2[-1]

Do the following:

1. Create the vector $$x=(1, 7, 3, 4).$$
2. Create the vector $$y=(100, 99, 98, ..., 2, 1).$$
3. Create the vector $$z=(4, 8, 16, 32, 96).$$
4. Compute $$x_2+y_4$$ and $$\cos(x_3) + \sin(x_2) e^{-y_2}.$$ Answers: 104 and -0.9899925.
5. Set $$x_{2}=0$$ and $$y_{2}=-1.$$ Recompute the previous expressions. Answers: 97 and 2.785875.
6. Index $$y$$ by $$x+1$$ and store it as z. What is the output? Answer: z is c(-1, 100, 97, 96).

### Some functions

# Functions take arguments between parenthesis and transform them
# into an output
sum(myData)
## [1] 2
prod(myData)
## [1] 0

# Summary of an object
summary(myData)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##     0.0     0.5     1.0     1.0     1.5     2.0

# Length of the vector
length(myData)
## [1] 2

# Mean, standard deviation, variance, covariance, correlation
mean(myData)
## [1] 1
var(myData)
## [1] 2
cov(myData, myData^2)
## [1] 4
cor(myData, myData * 2)
## [1] 1
quantile(myData)
##   0%  25%  50%  75% 100%
##  0.0  0.5  1.0  1.5  2.0

# Maximum and minimum of vectors
min(myData)
## [1] 0
which.min(myData)
## [1] 1

# Usually the functions have several arguments, which are set
# by "argument = value" In the next case, the second argument is
# a logical flag to indicate the kind of sorting
sort(myData) # If nothing is specified, decreasing = FALSE is
## [1] 0 2
# assumed
sort(myData, decreasing = TRUE)
## [1] 2 0

# Do not know what are the arguments of a function? Use args
# and help!
args(mean)
## function (x, ...)
## NULL
?mean

Do the following:

1. Compute the mean, median and variance of $$y.$$ Answers: 49.5, 49.5, 843.6869.
2. Do the same for $$y+1.$$ What are the differences?
3. What is the maximum of $$y$$? Where is it placed?
4. Sort $$y$$ increasingly and obtain the 5th and 76th positions. Answer: c(4, 75).
5. Compute the covariance between $$y$$ and $$y.$$ Compute the variance of $$y.$$ Why do you get the same result?

### Matrices, data frames, and lists

# A matrix is an array of vectors
A <- matrix(1:4, nrow = 2, ncol = 2)
A
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4

# Another matrix
B <- matrix(1, nrow = 2, ncol = 2, byrow = TRUE)
B
##      [,1] [,2]
## [1,]    1    1
## [2,]    1    1

# Matrix is a vector with dimension attributes
dim(A)
## [1] 2 2

# Binding by rows or columns
rbind(1:3, 4:6)
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    4    5    6
cbind(1:3, 4:6)
##      [,1] [,2]
## [1,]    1    4
## [2,]    2    5
## [3,]    3    6

# Entrywise operations
A + 1
##      [,1] [,2]
## [1,]    2    4
## [2,]    3    5
A * B
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4

# Accessing elements
A[2, 1] # Element (2, 1)
## [1] 2
A[1, ] # First row -- this is a vector
## [1] 1 3
A[, 2] # First column -- this is a vector
## [1] 3 4

# Obtain rows and columns as matrices (and not as vectors)
A[1, , drop = FALSE]
##      [,1] [,2]
## [1,]    1    3
A[, 2, drop = FALSE]
##      [,1]
## [1,]    3
## [2,]    4

# Matrix transpose
t(A)
##      [,1] [,2]
## [1,]    1    2
## [2,]    3    4

# Matrix multiplication
A %*% B
##      [,1] [,2]
## [1,]    4    4
## [2,]    6    6
A %*% B[, 1]
##      [,1]
## [1,]    4
## [2,]    6
A %*% B[1, ]
##      [,1]
## [1,]    4
## [2,]    6

# Care is needed
A %*% B[1, , drop = FALSE] # Incompatible product
## Error in A %*% B[1, , drop = FALSE]: non-conformable arguments

# Compute inverses with "solve"
solve(A) %*% A
##      [,1] [,2]
## [1,]    1    0
## [2,]    0    1

# A data frame is a matrix with column names
# Useful when you have multiple variables
myDf <- data.frame(var1 = 1:2, var2 = 3:4)
myDf
##   var1 var2
## 1    1    3
## 2    2    4

# You can change names
names(myDf) <- c("newname1", "newname2")
myDf
##   newname1 newname2
## 1        1        3
## 2        2        4

# You can access variables by its name with the "$" operator myDf$newname1
## [1] 1 2

# And create new variables also (they have to be of the same
# length as the rest of variables)
myDf$myNewVariable <- c(0, 1) myDf ## newname1 newname2 myNewVariable ## 1 1 3 0 ## 2 2 4 1 # A list is a collection of arbitrary variables myList <- list(myData = myData, A = A, myDf = myDf) # Access elements by names myList$myData
## [1] 0 2
myList$A ## [,1] [,2] ## [1,] 1 3 ## [2,] 2 4 myList$myDf
##   newname1 newname2 myNewVariable
## 1        1        3             0
## 2        2        4             1

# Reveal the structure of an object
str(myList)
## List of 3
##  $myData: num [1:2] 0 2 ##$ A     : int [1:2, 1:2] 1 2 3 4
##  $myDf :'data.frame': 2 obs. of 3 variables: ## ..$ newname1     : int [1:2] 1 2
##   ..$newname2 : int [1:2] 3 4 ## ..$ myNewVariable: num [1:2] 0 1
str(myDf)
## 'data.frame':    2 obs. of  3 variables:
##  $newname1 : int 1 2 ##$ newname2     : int  3 4
##  $myNewVariable: num 0 1 # A less lengthy output names(myList) ## [1] "myData" "A" "myDf" Do the following: 1. Create a matrix called M with rows given by y[3:5], y[3:5]^2, and log(y[3:5]). 2. Create a data frame called myDataFrame with column names “y”, “y2”, and “logy” containing the vectors y[3:5], y[3:5]^2 and log(y[3:5]), respectively. 3. Create a list, called l, with entries for x and M. Access the elements by their names. 4. Compute the squares of myDataFrame and save the result as myDataFrame2. 5. Compute the log of the sum of myDataFrame and myDataFrame2. Answer:  ## y y2 logy ## 1 9.180087 18.33997 3.242862 ## 2 9.159678 18.29895 3.238784 ## 3 9.139059 18.25750 3.234656 ### More on data frames # The iris dataset is already imported in R # (beware: locfit has also an iris dataset, with different names # and shorter) # The beginning of the data head(iris) ## Sepal.Length Sepal.Width Petal.Length Petal.Width Species ## 1 5.1 3.5 1.4 0.2 setosa ## 2 4.9 3.0 1.4 0.2 setosa ## 3 4.7 3.2 1.3 0.2 setosa ## 4 4.6 3.1 1.5 0.2 setosa ## 5 5.0 3.6 1.4 0.2 setosa ## 6 5.4 3.9 1.7 0.4 setosa # "names" gives you the variables in the data frame names(iris) ## [1] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width" "Species" # So we can access variables by "$" or as in a matrix
iris$Sepal.Length[1:10] ## [1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 iris[1:10, 1] ## [1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 iris[3, 1] ## [1] 4.7 # Information on the dimension of the data frame dim(iris) ## [1] 150 5 # "str" gives the structure of any object in R str(iris) ## 'data.frame': 150 obs. of 5 variables: ##$ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
##  $Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... ##$ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
##  $Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... ##$ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...

# Recall the species variable: it is a categorical variable
# (or factor), not a numeric variable
iris$Species[1:10] ## [1] setosa setosa setosa setosa setosa setosa setosa setosa setosa setosa ## Levels: setosa versicolor virginica # Factors can only take certain values levels(iris$Species)
## [1] "setosa"     "versicolor" "virginica"

# If a file contains a variable with character strings as
# observations (either encapsulated by quotation marks or not),
# the variable will become a factor when imported into R

Do the following:

1. Load the faithful dataset into R.
2. Get the dimensions of faithful and show beginning of the data.
3. Retrieve the fifth observation of eruptions in two different ways.
4. Obtain a summary of waiting.

### Logical conditions and subsetting

# Relational operators: x < y, x > y, x <= y, x >= y, x == y, x!= y
# They return TRUE or FALSE

# Smaller than
0 < 1
## [1] TRUE

# Greater than
1 > 1
## [1] FALSE

# Greater or equal to
1 >= 1 # Remember: ">="" and not "=>"" !
## [1] TRUE

# Smaller or equal to
2 <= 1 # Remember: "<="" and not "=<"" !
## [1] FALSE

# Equal
1 == 1 # Tests equality. Remember: "=="" and not "="" !
## [1] TRUE

# Unequal
1 != 0 # Tests inequality
## [1] TRUE

# TRUE is encoded as 1 and FALSE as 0
TRUE + 1
## [1] 2
FALSE + 1
## [1] 1

# In a vector-like fashion
x <- 1:5
y <- c(0, 3, 1, 5, 2)
x < y
## [1] FALSE  TRUE FALSE  TRUE FALSE
x == y
## [1] FALSE FALSE FALSE FALSE FALSE
x != y
## [1] TRUE TRUE TRUE TRUE TRUE

# Subsetting of vectors
x
## [1] 1 2 3 4 5
x[x >= 2]
## [1] 2 3 4 5
x[x < 3]
## [1] 1 2

# Easy way of work with parts of the data
data <- data.frame(x = c(0, 1, 3, 3, 0), y = 1:5)
data
##   x y
## 1 0 1
## 2 1 2
## 3 3 3
## 4 3 4
## 5 0 5

# Data such that x is zero
data0 <- data[data$x == 0, ] data0 ## x y ## 1 0 1 ## 5 0 5 # Data such that x is larger than 2 data2 <- data[data$x > 2, ]
data2
##   x y
## 3 3 3
## 4 3 4

# Problem -- what happened?
data[x > 2, ]
##   x y
## 3 3 3
## 4 3 4
## 5 0 5

# AND operator "&"
TRUE & TRUE
## [1] TRUE
TRUE & FALSE
## [1] FALSE
FALSE & FALSE
## [1] FALSE

# OR operator "|"
TRUE | TRUE
## [1] TRUE
TRUE | FALSE
## [1] TRUE
FALSE | FALSE
## [1] FALSE

# Both operators are useful for checking for ranges of data
y
## [1] 0 3 1 5 2
index1 <- (y <= 3) & (y > 0)
y[index1]
## [1] 3 1 2
index2 <- (y < 2) | (y > 4)
y[index2]
## [1] 0 1 5

Do the following for the iris dataset:

1. Compute the subset corresponding to Petal.Length either smaller than 1.5 or larger than 2. Save this dataset as irisPetal.
2. Compute and summarize a linear regression of Sepal.Width into Petal.Width + Petal.Length for the dataset irisPetal. What is the $$R^2$$? Solution: 0.101.
3. Check that the previous model is the same as regressing Sepal.Width into Petal.Width + Petal.Length for the dataset iris with the appropriate subset expression.
4. Compute the variance for Petal.Width when Petal.Width is smaller or equal that 1.5 and larger than 0.3. Solution: 0.1266541.

### Plotting functions

# "plot" is the main function for plotting in R
# It has a different behavior depending on the kind of object

# How to plot some data
plot(iris$Sepal.Length, iris$Sepal.Width,
main = "Sepal.Length vs. Sepal.Width")


# Change the axis limits
plot(iris$Sepal.Length, iris$Sepal.Width, xlim = c(0, 10),
ylim = c(0, 10))


# How to plot a curve (a parabola)
x <- seq(-1, 1, l = 50)
y <- x^2
plot(x, y, main = "A red thick parabola", type = "l",
col = "red", lwd = 3)


# Plotting a more complicated curve between -pi and pi
x <- seq(-pi, pi, l = 50)
y <- (2 + sin(10 * x)) * x^2
plot(x, y, type = "l") # Kind of rough...


# Remember that we are joining points for creating a curve!
# More detailed plot
x <- seq(-pi, pi, l = 500)
y <- (2 + sin(10 * x)) * x^2
plot(x, y, type = "l")


# For more options in the plot customization see
?plot
## Help on topic 'plot' was found in the following packages:
##
##   Package               Library
##   base                  /Library/Frameworks/R.framework/Resources/library
##   graphics              /Library/Frameworks/R.framework/Versions/4.2-arm64/Resources/library
##
##
## Using the first match ...
?par

# "plot" is a first level plotting function. That means that
# whenever is called, it creates a new plot. If we want to
# add information to an existing plot, we have to use a second
# level plotting function such as "points", "lines" or "abline"

plot(x, y) # Create a plot
lines(x, x^2, col = "red") # Add lines
points(x, y + 10, col = "blue") # Add points
abline(a = 5, b = 1, col = "orange", lwd = 2) # Add a straight

# line y = a + b * x

Do the following:

1. Plot the faithful dataset.
2. Add the straight line $$y=110-15x$$ (red).
3. Make a new plot for the function $$y=\sin(x)$$ (black). Add $$y=\sin(2x)$$ (red), $$y=\sin(3x)$$ (blue), and $$y=\sin(4x)$$ (orange).

### Distributions

# R allows to sample [r], compute density/probability mass
# functions [d], compute distribution function [p], and compute
# quantiles [q] for several continuous and discrete distributions.
# The format employed is [rdpq]name, where name stands for:
# - norm -> Normal
# - unif -> Uniform
# - exp -> Exponential
# - t -> Student's t
# - f -> Snedecor's F
# - chisq -> Chi squared
# - pois -> Poisson
# - binom -> Binomial
# More distributions:
?Distributions

# Sampling from a normal -- 5 random points from a N(0, 1)
rnorm(n = 5, mean = 0, sd = 1)
## [1] -0.5881703  0.3004478 -2.0931383  1.8219223  0.3728798

# If you want to have always the same result, set the seed of
# the random number generator
set.seed(45678)
rnorm(n = 5, mean = 0, sd = 1)
## [1]  1.4404800 -0.7195761  0.6709784 -0.4219485  0.3782196

# Plotting the density of a N(0, 1) -- the Gaussian bell
x <- seq(-4, 4, l = 100)
y <- dnorm(x = x, mean = 0, sd = 1)
plot(x, y, type = "l")


# Plotting the distribution function of a N(0, 1)
x <- seq(-4, 4, l = 100)
y <- pnorm(q = x, mean = 0, sd = 1)
plot(x, y, type = "l")


# Computing the 95% quantile for a N(0, 1)
qnorm(p = 0.95, mean = 0, sd = 1)
## [1] 1.644854

# All distributions have the same syntax: rname(n,...),
# dname(x,...), dname(p,...) and qname(p,...), but the
# parameters in ... change. Look them in ?Distributions
# For example, here is the same for the uniform distribution

# Sampling from a U(0, 1)
set.seed(45678)
runif(n = 10, min = 0, max = 1)
##  [1] 0.9251342 0.3339988 0.2358930 0.3366312 0.7488829 0.9327177 0.3365313 0.2245505 0.6473663 0.0807549

# Plotting the density of a U(0, 1)
x <- seq(-2, 2, l = 100)
y <- dunif(x = x, min = 0, max = 1)
plot(x, y, type = "l")


# Computing the 95% quantile for a U(0, 1)
qunif(p = 0.95, min = 0, max = 1)
## [1] 0.95

# Sampling from a Bi(10, 0.5)
set.seed(45678)
samp <- rbinom(n = 200, size = 10, prob = 0.5)
table(samp) / 200
## samp
##     1     2     3     4     5     6     7     8     9
## 0.010 0.060 0.115 0.220 0.210 0.215 0.115 0.045 0.010

# Plotting the probability mass of a Bi(10, 0.5)
x <- 0:10
y <- dbinom(x = x, size = 10, prob = 0.5)
plot(x, y, type = "h") # Vertical bars


# Plotting the distribution function of a Bi(10, 0.5)
x <- 0:10
y <- pbinom(q = x, size = 10, prob = 0.5)
plot(x, y, type = "h")

Do the following:

1. Compute the $$90\%,$$ $$95\%$$ and $$99\%$$ quantiles of a $$F$$ distribution with df1 = 1 and df2 = 5. Answer: c(4.060420, 6.607891, 16.258177).
2. Plot the distribution function of a $$\mathcal{U}(0,1).$$ Does it make sense with its density function?
3. Sample $$100$$ points from a Poisson with lambda = 5.
4. Sample $$100$$ points from a $$\mathcal{U}(-1,1)$$ and compute its mean.
5. Plot the density of a $$t$$ distribution with df = 1 (use a sequence spanning from -4 to 4). Add lines of different colors with the densities for df = 5, df = 10, df = 50, and df = 100. Do you see any pattern?

### Functions

# A function is a way of encapsulating a block of code so it can
# be reused easily. They are useful for simplifying repetitive

# This is a silly function that takes x and y and returns its sum
# Note the use of "return" to indicate what should be returned
z <- x + y
return(z)
}

# Calling add -- you need to run the definition of the function
# first!
add(x = 1, y = 2)
## [1] 3
add(1, 1) # Arguments names can be omitted
## [1] 2

# A more complex function: computes a linear model and its
# posterior summary. Saves us a few keystrokes when computing a
# lm and a summary
lmSummary <- function(formula, data) {
model <- lm(formula = formula, data = data)
summary(model)
}
# If no return(), the function returns the value of the last
# expression

# Usage
lmSummary(Sepal.Length ~ Petal.Width, iris)
##
## Call:
## lm(formula = formula, data = data)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -1.38822 -0.29358 -0.04393  0.26429  1.34521
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  4.77763    0.07293   65.51   <2e-16 ***
## Petal.Width  0.88858    0.05137   17.30   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.478 on 148 degrees of freedom
## Multiple R-squared:  0.669,  Adjusted R-squared:  0.6668
## F-statistic: 299.2 on 1 and 148 DF,  p-value: < 2.2e-16

# Recall: there is no variable called model in the workspace.
# The function works on its own workspace!
model

# Add a line to a plot
addLine <- function(x, beta0, beta1) {
lines(x, beta0 + beta1 * x, lwd = 2, col = 2)
}

# Usage
plot(x, y)
addLine(x, beta0 = 0.1, beta1 = 0)


# The function "sapply" allows to sequentially apply a function
sapply(1:10, sqrt)
##  [1] 1.000000 1.414214 1.732051 2.000000 2.236068 2.449490 2.645751 2.828427 3.000000 3.162278
sqrt(1:10) # The same
##  [1] 1.000000 1.414214 1.732051 2.000000 2.236068 2.449490 2.645751 2.828427 3.000000 3.162278

# The advantage of "sapply" is that you can use with any function
myFun <- function(x) c(x, x^2)
sapply(1:10, myFun) # Returns a 2 x 10 matrix
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,]    1    2    3    4    5    6    7    8    9    10
## [2,]    1    4    9   16   25   36   49   64   81   100

# "sapply" is useful for plotting non-vectorized functions
sumSeries <- function(n) sum(1:n)
plot(1:10, sapply(1:10, sumSeries), type = "l")


# "apply" applies iteratively a function to rows (1) or columns
# (2) of a matrix or data frame
A <- matrix(1:10, nrow = 5, ncol = 2)
A
##      [,1] [,2]
## [1,]    1    6
## [2,]    2    7
## [3,]    3    8
## [4,]    4    9
## [5,]    5   10
apply(A, 1, sum) # Applies the function by rows
## [1]  7  9 11 13 15
apply(A, 2, sum) # By columns
## [1] 15 40

# With other functions
apply(A, 1, sqrt)
##         [,1]     [,2]     [,3] [,4]     [,5]
## [1,] 1.00000 1.414214 1.732051    2 2.236068
## [2,] 2.44949 2.645751 2.828427    3 3.162278
apply(A, 2, function(x) x^2)
##      [,1] [,2]
## [1,]    1   36
## [2,]    4   49
## [3,]    9   64
## [4,]   16   81
## [5,]   25  100

Do the following:

1. Create a function that takes as argument $$n$$ and returns the value of $$\sum_{i=1}^n i^2.$$
2. Create a function that takes as input the argument $$N$$ and then plots the curve $$(n, \sum_{i=1}^n \sqrt{i})$$ for $$n=1,\ldots,N.$$ Hint: use sapply.

### Control structures

# The "for" statement allows to create loops that run along a
# given vector
# Print 3 times a message (i varies in 1:3)
for (i in 1:3) {
print(i)
}
## [1] 1
## [1] 2
## [1] 3

# Another example
a <- 0
for (i in 1:3) {
a <- i + a
}
a
## [1] 6

# Nested loops are possible
A <- matrix(0, nrow = 2, ncol = 3)
for (i in 1:2) {
for (j in 1:3) {
A[i, j] <- i + j
}
}

# The "if" statement allows to create conditional structures of
# the forms:
# if (condition) {
#  # Something
# } else {
#  # Something else
# }
# These structures are thought to be inside functions

# Is the number positive?
isPositive <- function(x) {
if (x > 0) {
print("Positive")
} else {
print("Not positive")
}
}
isPositive(1)
## [1] "Positive"
isPositive(-1)
## [1] "Not positive"

# A loop can be interrupted with the "break" statement
# Stop when x is above 100
x <- 1
for (i in 1:1000) {
x <- (x + 0.01) * x
print(x)
if (x > 100) {
break
}
}
## [1] 1.01
## [1] 1.0302
## [1] 1.071614
## [1] 1.159073
## [1] 1.35504
## [1] 1.849685
## [1] 3.439832
## [1] 11.86684
## [1] 140.9406

Do the following:

1. Compute $$\mathbf{C}_{n\times k}$$ in $$\mathbf{C}_{n\times k}=\mathbf{A}_{n\times m} \mathbf{B}_{m\times k}$$ from $$\mathbf{A}$$ and $$\mathbf{B}.$$ Use that $$c_{i,j}=\sum_{\ell=1}^ma_{i,l}b_{l,j}.$$ Test the implementation with simple examples.
2. Create a function that samples a $$\mathcal{N}(0,1)$$ and returns the first sampled point that is larger than $$4.$$
3. Create a function that simulates $$N$$ samples from the distribution of $$\max(X_1,\ldots,X_n)$$ where $$X_1,\ldots,X_n$$ are iid $$\mathcal{U}(0,1).$$

### References

R Core Team. 2020. R: A Language and Environment for Statistical Computing. Vienna. https://www.R-project.org/.