exercise_6_solution.Rmd

---
title: 'Exercise Solutions'
output: 
  html_document:
    toc: false
    code_folding: hide
---

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

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## Exercise 6: Basic programming in R

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Read [Chapter 7](https://intro2r.com/prog-r.html) to help you complete the questions in this exercise.

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1\. Create a function to calculate the area of a circle. Test the function by finding the area of a circle with a diameter of 3.4 cm. Can you use it on a vector of data?

```{r Q1, results = 'asis'}

# area of a circle
# the equation to calculate the area of a circle is pi * radius^2

circle.area <- function(d){ 
	pi * (d/2)^2
}

# to use your new function

circle.area(10)
# [1] 78.53982

# to test on a vector of diameters
# first create a vector with diameters ranging from 0 to 50 in steps of 10

cir.diam <- seq(from = 0, to = 50, by = 10)

# test your function

circle.area(cir.diam)
# [1]    0.00000   78.53982  314.15927  706.85835 1256.63706 1963.49541
```

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2\. Write a function to convert farenheit to centegrade (oC = (oF - 32) x 5/9). Get your function to print out your result in the following format: "Farenheit : *value of oF* is equivalent to *value oC* centigrade."

```{r Q2, tidy = TRUE}

far.cent <- function(a){
	val <- (a-32)*5/9
	print(paste("Fahrenheit: ", round(a, digits = 3), "oF",sep = " "), quote = FALSE)# round 3dp
	print(paste("Centigrade: ", round(val, digits = 3), "oC", sep = " "), quote = FALSE)# round 3dp
}

# alternative Fahrenheit to centigrade using cat function

far.cent2 <- function(a){
	val <- (a - 32) * 5/9	#calculation
	cat("Fahrenheit: ", round(a, digits = 3), "oF", "\n")   # use cat function
	cat("Centigrade: ", round(val, digits = 3), "oC", "\n")   
}
```

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3\. Create a vector of normally distributed data, of length 100, mean 35 and standard deviation of 15. Write a function to calculate the mean, median, and range of the vector, print these values out with appropriate labels. Also get the function to plot a histogram (as a proportion) of the values and add a density curve.

```{r Q3}

# Create a vector of normally distributed data
# length 100, mean 35 and standard deviation of 29

vals <- rnorm(100, 35, 15)	# create some norm dist data mean 35, sd = 15

summary.fun <- function(dat){
	mymean <- round(mean(dat), digits = 4)      	# calc mean
	mymedian <- round(median(dat), digits = 4)    # calc median
	mymin <- round(min(dat), digits = 4)          # calc min
	mymax <- round(max(dat), digits = 4)          # calc max
	print(paste("mean:", mymean, sep = " "), quote = FALSE)     # print mean
	print(paste("median:", mymedian, sep = " "), quote = FALSE)     # print median
	print(paste("range:", "from:", mymin, "to", mymax, sep = " "), quote = FALSE) 
	dens <- density(dat)                          # estimate density curve
	hist(dat, main = "",type = "l",freq = FALSE)  # plot histogram
	lines(dens, lty = 1, col = "red")             # plot density curve
}

# use the function
summary.fun(vals)
```

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4\. Write a function to calculate the median value of a vector of numbers (yes I know there's a ```median()``` function already but this is fun!). Be careful with vectors of an even sample size, as you will have to take the average of the two central numbers (hint: use modulo  %%2 to determine whether the vector is an odd or an even size). Test your function on vectors with both odd and even sample sizes. 

```{r Q4}

# calculate a median

ourmedian <- function(x){
	n <- length(x)
	if (n %% 2 == 1)      # odd numbers
	  sort(x)[(n + 1)/2]  # find the middle number by adding 1 to length and div 2
	else {                # even numbers
	  middletwo <- sort(x)[(n/2) + 0:1]   #find the two middle numbers
	  mean(middletwo)
	  }
}

# use the function
mydat <- sample(1:50, size = 10, replace = TRUE )

# our function
ourmedian(mydat)

# R median function
median(mydat)
```

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5\. You are a population ecologist for the day and wish to investigate the properties of the [Ricker model](https://en.wikipedia.org/wiki/Ricker_model). The Ricker model is defined as:

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$$ N_{t+1} = N_{t} exp\left[r\left(1- \frac{N_{t}}{K} \right) \right] $$

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5\. (cont) Where *N~t~* is the population size at time *t*, *r* is the population growth rate and *K* is the carrying capacity. Write a function to simulate this model so you can conveniently determine the effect of changing *r* and the initial population size N0. *K* is often set to 100 by default, but you want the option of being able to change this with your function. So, you will need a function with the following arguments; nzero which sets the initial population size, *r* which will determine the population growth rate, time which sets how long the simulation will run for and *K* which we will initially set to 100 by default.

```{r Q5, tidy = TRUE}

# function to simulate Ricker model

Ricker.model <- function(nzero, r, time, K=1){     # sets initial parameters
  N <- numeric(time + 1)    # creates a real vector of length time+1 to store values of Nt+1
  N[1] <- nzero             # sets initial population size in first element of N
  for (i in 1:time) {				# loops over time
      N[i+1] <- N[i]*exp(r*(1 - N[i]/K))    
  }
  Time <- 0:time   					 # creates vector for x axis
  plot(Time, N, type = "o", xlim = c(0, time), xlab = "Time", ylab = "Population size (N)", main = paste("r =", r, sep=" "))   # plots     output
}

# To run
# play around with the different parameters, especially r
Ricker.model(nzero=0.1,r=1,time=100)
```

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End of Exercise 6