class9.Rmd

---
title: 'Data Analysis 3: Week 9'
author: "Alexey Bessudnov"
date: "14 March 2019"
output: github_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
knitr::opts_chunk$set(message = FALSE)
knitr::opts_chunk$set(warning = FALSE)
knitr::opts_chunk$set(cache = TRUE)

```
Plan for today:

1. Final statistical report.
2. Assignment 4: solutions.
3. Exercises on conditional statements and iteration.


Exercises.

1. Check if a number is even. If it is print "Even". If it isn't print "Odd". If it's 0 print "Zero".

    ```{r}
x <- 3

if (x == 0) {
  print("Zero")
} else {
    if (x %% 2 == 0) {
      print("Even")
    } else {
        print("Odd")
    }
  }

    ```

2. Write a for loop finding the largest element of a numeric vector and print it on the screen.

    ```{r}
x <- c(3, 5, 8, -2)
max(x)

for (i in 1:3) {
  print(i)
}

for(i in 1:length(x)) {
  print(x[i])
}

max_x <- x[1]
for (i in 2:length(x)) {
  if (x[i] > max_x) {max_x <- x[i]}
}
max_x

    ```

3. Write a for loop finding the largest element of a numeric matrix and print it on the screen.

    ```{r}
x <- matrix(c(1, -3, 7, 4, 10, 7, -5, 0), nrow = 4)
x
max_x <- x[1, 1]
for (i in 1:dim(x)[1]) {
  for (j in 1:dim(x)[2]) {
      if (x[i,j] > max_x) {max_x <- x[i,j]}
  }
}
max_x

    ```

4. Write a while loop finding the largest element of a numeric vector and print it on the screen.

    ```{r}
i <- 1 
while (i < 10) {
  print(i)
  i <- i + 1
}

x <- c(3, 5, 8, -2)

max_x <- x[1]
i <- 2
while (i <= length(x)) {
    if (x[i] > max_x) {max_x <- x[i]}
  i <- i + 1
}
max_x

    ```

5. x is a vector with whole numbers (zero and positive integers).  If the largest even element of x is smaller than the largest odd element of x, all even elements of x are replaced by 0s. Otherwise all odd elements of x are replaced by 0s. For example, if x = {7; 1; 3; 2; 14; 5; 9; 6} the output should be [0; 0; 0; 2; 14; 0; 0; 6].

    ```{r}
x <- c(7, 1, 3, 2, 14, 5, 9, 6)
max <- 0
for (i in seq_along(x)) {
                if (x[i] > max) {max <- x[i]}
}
for (i in seq_along(x)) {
        if (max %% 2 == 0) {
                if (x[i] %% 2 != 0) {x[i] <- 0}
        }
        else {
                if (x[i] %% 2 == 0) {x[i] <- 0}
        }
}
x
    ```

6. x is a vector with whole numbers. Write a programme that count the number of pairs of the elements of x where the sum can be divided by 12 without a remainder.

For example, for x = {8; 10; 14; 7; 13; 5; 30; 9; 6} then the answer is 3 ((10, 14); (7, 5); (30, 6)).

    ```{r}
x <- c(8, 10, 14, 7, 13, 5, 30, 9, 6)
k <- 0
for (i in 1:(length(x)-1)) {
        for (j in (i+1):length(x)) {
                if ((x[i] + x[j]) %% 12 == 0) {
                         k <- k + 1
                         print(c(x[i], x[j]))
                }
        }
}
k
    ```