Answers_notebook.Rmd

---
title: "Answers Notebook"
output: html_notebook
---


# Exercise 1 no exercises lecture 1

# Exercise 2 Objects
Look at the following code, guess what kind of object this will create, then check what is created using typeof, str() or class()

1. c(35, 46, 17, 58,2)
```{r}
quickfx <- function(my_object = c(35, 46, 17, 58,2)){
  res <- lapply(list(identity, class, typeof), function(x) x(my_object))
  names(res) <- c("object", "class", "type")
  print(res)
}
quickfx()
```
2. c( 34, 1, "l", TRUE)
```{r}
quickfx(c( 34, 1, "l", TRUE))
```
3. c(T,F,T,F)
```{r}
quickfx(c(T,F,T,F))
```
4. cbind (c(1,0,0), c(0,1,0), c(0,0,1))
```{r}
quickfx(cbind (c(1,0,0), c(0,1,0), c(0,0,1)))
```

5. cbind (c("a", "b", "c"), c (1,2,3))
```{r}
quickfx(cbind (c("a", "b", "c"), c (1,2,3)))
```
6. data.frame (var1 = 1:10, var2 = letters[1:10], var3 = rep(1:2,5), stringsAsFactors = FALSE )
```{r}
quickfx(data.frame (var1 = 1:10, var2 = letters[1:10], var3 = rep(1:2,5), stringsAsFactors = FALSE ))
```

# Exercise 3 Working with vectors
type set.seed (1234) and hit return
This is so we all get the same answer

1. Create a vector a of 10 realisations from the standard normal distribution, use rnorm (10, mean = 0, sd = 1)
2. Calculate max, min, sum of values greater than 1, sum of values less than 1
3. Create a vector b from a standard normal distribution
4. Take the maximum of a and b for each element
5. Identify the number of elements where a > b. What did R do when you ran this?

# Exercise 4 Subsetting vectors
set.seed (1234)

1. Create a vector, x from a Poisson distribution of length 10, with lambda = 5, use rpois
2. Select the first and last elements using an integer vector
3. select all but the first element
4. Return the vector according to the numerical order in two ways. First using `sort`, then using `order`. Why is `order` more general, ie what could you acheive with order that you could not acheive with `sort`?
5. Use logical subsetting to select all values of x which are >5
6. In a single line, identify the minimum value of x which is greater than 3
  + Create a vector 
```{r}
b <- c(a = 1, b = 2, c = 3, d = 4, e = 5)
b
```
7. Return a vector from these elements repeating the first and second element 3 times and twice respectively, using the names attribute of the vector
8. Assign NA to all values of x >5
9. Add 0.5 to any values less than 5. **Hint, need to use is.na and the "&" operator. Also need to consider the length of the vector to the left and to the right of the assignment operator `<-`.**

# Exercise 5 Subsetting other objects
1. Run the following and answer the following questions
```{r, eval = FALSE}
model1 <- lm(Sepal.Length ~ Species, data = iris)
summary (model1)
str(model1)
length(model1)
length(unlist(model1))
```
  + What type of object does model 1 produce?
  + what are its attributes?
  + how are these items organised?
  + what happens when you unlist?

2. Extract the coefficients for the model using the coef(model1) function and using indexing
3. Extract the Species factor from the model using 3 kinds of indexing, what kind of objects do you get with each
4. Turn the swiss dataset into a matrix `swiss <- as.matrix (swiss)`
5. Select all first and third columns and second and fourth rows, use 2 methods
6. Select a single row, first in a way which produces a vector then in a way which produces another matrix. What attributes persist with each method?
7. Calculate the count, mean and sum of all the values are between 25 and 30 
8. Select all the rows for towns with Rive in the names (use the grepl and rownames functions)
9. remove all objects from workspace and create swiss as a matrix and a dataframe, then index each using the following methods.

```{r}
data(swiss)
swiss.m <- as.matrix(swiss)
swiss.df <- swiss
```
  + [1]
  + [[1]]
  + [1:3, 1:2]
  + ["Rive Droite", "Education"]
  + [swiss > 50] # this is an example of indexing with an array


# Exercise 6 working with factors
1. load in the esoph data , `data (esoph)` (not really necessary but in case modfied it) and look at the help file.
2. what class is each of the variables
3. Repace alcohol consumption "120+" with "120 to 150" using indexing with a logical test
  + You can also do this using ifelse.
```{r, eval = FALSE}
a<- ifelse(esoph$alcgp == "120+", "120 to 150", esoph$alcgp)
b<- ifelse(esoph$alcgp == "120+", "120 to 150", as.character(esoph$alcgp))
table(a,b)
```
    + What has R done differently in each case?
4. Dichotomise alcohol consumption into >=80 or <80
5. Run the regression models
```{r, eval = FALSE}
model1 <- glm(cbind(ncases, ncontrols) ~ tobgp, data = esoph, family = binomial())
model2 <- glm(cbind(ncases, ncontrols) ~ factor(tobgp, ordered = FALSE), data = esoph, 
              family = binomial())
```
what has R done differently?

# Exercise 7 Using functions
1. Don't write this down, just look at the menu and get a feel. For c, summary, mean, glm, and plot
  + Is it a generic, primitive, S3 or S4 function
  + What type of object does the first argument accept
  + Which arguments are optional
  + What type of object is output
  + What does the function return if supplied with missing values
2. What will R do if a function argument is not supplied and there is no default?
3. What values will `print (c(a = a, b = b))` result in. Guess before running.
```{r, eval = FALSE}
a <- 1
b <- 4
MyFunction <- function (a, b) {
  a <- a + 1
  b <<- b + 1
}
MyFunction (a = 2, b = 6)
print (c(a = a, b = b))
```

# Exercise 8 apply family
1. In the iris dataset use tapply to calculate the mean Sepal length according to Species 
2. Use lapply to determine the class of each variable and then the mean of each variable
  + How does the R function lapply handle the character vector? What would `mean` have returned if used without an apply function for a character atomic vector?
3. For the vectors of class "numeric", use apply to calculate row sums and column sums and compare the results with rowSums and colSums. [* Hint: Use subsetting to select these columns before trying to use `apply`*]


# Exercise 9 writing functions
1. Look at the esoph dataset type ?esoph into R
2. Summarise the total number of cases and controls by age group
3. Create a variable where the value is true for heavy smokers, defined as >19 gm per day of tobacco
4. Check that there are heavy smokers in each age group
5. Run a logistic regression model of oesophageal cancer on smoking for the whole dataset (have a look at the bottom of the ?esoph page for how to do this with aggregated data)
6. Write a function which will extract the odds ratio for smoking > 19gm/day for each age group stratum.
Use the following steps, similar to the example in the previous session
  + simplify the problem, keeping the same object type
  + work interactively
  + write function
  + test function on different subsets
  + run function using the **by** function, which belongs to the apply family of functions
7. Why couldn't we use `tapply`? look at the help for the first argument in `tapply` and `by`
8. What kind of object does `by` produce?