Computational Ecology: Population Viability Analysis 9/15/2020

Data: Figure 5 (plus 10 years)

Make data vectors, calculate lambda, and put together dataframe with all necessary data.

census

The census period; an index from 1 to 39 of how many years of data have been collected.

census <- 1:39

year t

The year: 1959 to 1997 (Dennis et al use 1959-1987)

year.t   <- 1959:1997

Population size

Population size is recorded as the number of females with …

females.N <- c(44,47,46,44,46,
               45,46,40,39,39,
               42,39,41,40,33,
               36,34,39,35,34,
               38,36,37,41,39,
               51,47,57,48,60,
               65,74,69,65,57,
               70,81,99,99)

Population growth rate: example

Population growth rate is…

Enter the population size for each year

females.N.1959 <- 44
females.N.1960 <- 47

Calculate the ratio of the 2 population sizes

lambda.59_60 <- females.N.1960/females.N.1959

Access the population sizes by using bracket notation rather than hard coding

# Access the data
females.N[2]
#> [1] 47
females.N[1]
#> [1] 44

# store in objects
females.N.1959 <- females.N[1]
females.N.1960 <- females.N[2]

# confirm the output
females.N.1960/females.N.1959
#> [1] 1.068182

Calculate lambda using bracket notation

lambda.59_60 <- females.N[2]/females.N[1]

The first year of data is 1959. What is lambda for 1958 to 1959?

females.N[1]
#> [1] 44
#lambda.58_59 <- females.N[1]/females.N[]

Population growth rate: vectorized

TASK

Briefly describe (1-2 sentence) what this code is doing.

Calculates two population growth rates for years 1959 to 1960 and 1960 to 1961.

females.N[2:3]
#> [1] 47 46
females.N[1:2]
#> [1] 44 47

females.N[2:3]/females.N[1:2]
#> [1] 1.0681818 0.9787234

This is similar t the previous code chunk, just using all of the data (no need to describe)

length(females.N)
#> [1] 39
females.N[2:39]/females.N[1:38]
#>  [1] 1.0681818 0.9787234 0.9565217 1.0454545 0.9782609 1.0222222 0.8695652
#>  [8] 0.9750000 1.0000000 1.0769231 0.9285714 1.0512821 0.9756098 0.8250000
#> [15] 1.0909091 0.9444444 1.1470588 0.8974359 0.9714286 1.1176471 0.9473684
#> [22] 1.0277778 1.1081081 0.9512195 1.3076923 0.9215686 1.2127660 0.8421053
#> [29] 1.2500000 1.0833333 1.1384615 0.9324324 0.9420290 0.8769231 1.2280702
#> [36] 1.1571429 1.2222222 1.0000000

TASK What does this do? Briefly describe in 1 to 2 sentences why I am using length().

It calculates all population growth rates for all the years in the given data.

len <- length(females.N)
females.N[2:len]/females.N[1:len-1]
#>  [1] 1.0681818 0.9787234 0.9565217 1.0454545 0.9782609 1.0222222 0.8695652
#>  [8] 0.9750000 1.0000000 1.0769231 0.9285714 1.0512821 0.9756098 0.8250000
#> [15] 1.0909091 0.9444444 1.1470588 0.8974359 0.9714286 1.1176471 0.9473684
#> [22] 1.0277778 1.1081081 0.9512195 1.3076923 0.9215686 1.2127660 0.8421053
#> [29] 1.2500000 1.0833333 1.1384615 0.9324324 0.9420290 0.8769231 1.2280702
#> [36] 1.1571429 1.2222222 1.0000000

TASK What does this do? Briefly describe in 1 to 2 sentences what is different about this code chunk from the previous one.

You did not assign length to a variable but it will give the same results.


females.N[2:length(females.N)]/females.N[1:length(females.N)-1]
#>  [1] 1.0681818 0.9787234 0.9565217 1.0454545 0.9782609 1.0222222 0.8695652
#>  [8] 0.9750000 1.0000000 1.0769231 0.9285714 1.0512821 0.9756098 0.8250000
#> [15] 1.0909091 0.9444444 1.1470588 0.8974359 0.9714286 1.1176471 0.9473684
#> [22] 1.0277778 1.1081081 0.9512195 1.3076923 0.9215686 1.2127660 0.8421053
#> [29] 1.2500000 1.0833333 1.1384615 0.9324324 0.9420290 0.8769231 1.2280702
#> [36] 1.1571429 1.2222222 1.0000000

Negative indexing

Make a short vector to play with; first 10 years

females.N[1:10]
#>  [1] 44 47 46 44 46 45 46 40 39 39
females.Ntemp <- females.N[1:10]

Check - are there 10 numbers

females.Ntemp
#>  [1] 44 47 46 44 46 45 46 40 39 39

TASK

What does this do? Briefly describe what the [-1] is doing.

Removes the first element of the vector

females.Ntemp[-1]
#> [1] 47 46 44 46 45 46 40 39 39

TASK How many lambdas can I calculate using the first 10 years of data?

9 lambdas.

females.Ntemp[2:10]/females.Ntemp[1:9]
#> [1] 1.0681818 0.9787234 0.9565217 1.0454545 0.9782609 1.0222222 0.8695652
#> [8] 0.9750000 1.0000000

“Negative indexing” allows you to drop a specific element from a vector.

TASK Drop the the first element

females.Ntemp[-1]
#> [1] 47 46 44 46 45 46 40 39 39

TASK Drop the second element

females.Ntemp[-2]
#> [1] 44 46 44 46 45 46 40 39 39

TASK

How do you drop the 10th element? Type in the code below.

females.Ntemp[-10]
#> [1] 44 47 46 44 46 45 46 40 39

TASK How do you access the last element? Do this in a general way without hard-coding.

females.Ntemp[length(females.Ntemp)]
#> [1] 39

TASK How do DROP the last element? Do this in a general way without hard-coding. By general, I mean in a way that if the length of the vector females.Ntemp changed the code would still drop the correct element.

females.Ntemp[-length(females.Ntemp)]
#> [1] 44 47 46 44 46 45 46 40 39

TASK Calculate the first 9 lambdas.

lambda.i <- females.Ntemp[2:10]/females.Ntemp[1:9]

Converting between these 2 code chunks would be a good test question : )

lambda.i <- females.Ntemp[-1]/females.Ntemp[-length(females.Ntemp)]

Calcualte lambdas for all data

TASK

Below each bulleted line describe what the parts of the code do. Run the code to test it.

What does females.N[-1] do?

Removes the first element of the vector.

What does females.N[-length(females.N)?

Removes the last element of the vector.

TASK Calculate lambdas for all of the data

females.N[-1]
#>  [1] 47 46 44 46 45 46 40 39 39 42 39 41 40 33 36 34 39 35 34 38 36 37 41 39 51
#> [26] 47 57 48 60 65 74 69 65 57 70 81 99 99
females.N[-length(females.N)]
#>  [1] 44 47 46 44 46 45 46 40 39 39 42 39 41 40 33 36 34 39 35 34 38 36 37 41 39
#> [26] 51 47 57 48 60 65 74 69 65 57 70 81 99

lambda.i <- females.N[-1]/females.N[-length(females.N) ]

Finish putting together dataframe

Create special columns

TASK

What does this code do? Why do I include NA in the code? (I didn’t cover this in lecture, so just type 1 line - your best guess. “I don’t know” is fine.)

Represents missing values.

lambda.i <- c(lambda.i,NA)

TASK

Check the help file; what type of log does log() calculate (I forgot to put this question on the test!)

Computes the natural log.

lambda_log <- log(lambda.i)

Assemble the dataframe

bear_N <- data.frame(census,
                year.t,
                females.N,
                lambda.i, 
                lambda_log)

TASK

List 3 functions that allow you to examine this dataframe.

dim()
summary()
head()

Examing the population growth rates

Plotting the raw data

TASK

Plot a time series graph of the number of bears (y) versus time (x)
Label the y axis “Population index (females + cubs)”
Label the x axis “Year”
Change the plot to type = “b” so that both points and dots are shown.

plot(year.t, females.N, xlab = "Population index (females + cubs)", ylab = "Year", type = "b")

Bears love to eat trash. Yellowstone closed the last garbage dump in 1970 https://www.yellowstonepark.com/things-to-do/yellowstone-bears-no-longer-get-garbage-treats

Challenge questions

We will cover this all in the next lecture. Feel free to explore this code yourself.

CHALLENGE TASK

Plot a vertical line at 1970. Write a sentence or indicating if you think the population was impacted by this.

plot(year.t, females.N, xlab = "Population index (females + cubs)", ylab = "Year", type = "b")

abline(v = 1970, col = "red")

Plot lambda

CHALLENGE TASK

Make a histogram of population growth rates
Write a brief sentence describing the shape of these data.

The histogram seems a little skewed to the right.

hist(lambda.i)

Plot ln(lambda)

CHALLENGE TASK

Make a histogram of natural log population growth rates
Write a brief sentence describing the shape of these data.

The results look a little skewed to the right.

hist(lambda_log)

Mean of log(lambda)

CHALLENGE TASK

Briefly describe what happens when you delete na.rm = T

For some reason it is having an error saying bear_N is not found.

mean(bear_N$lambda_lo)
#> [1] NA

Histogram with mean

In statistics the mean is often represented as the Greek letter “mu”. This can be represented as “u”.

CHALLENGE TASK Save the mean to an object called u

u = mean(females.N)

CHALLENGE TASK Make a histogram with the mean plotted on it

hist(females.N, mean = u)
#> Warning in plot.window(xlim, ylim, "", ...): "mean" is not a graphical parameter
#> Warning in title(main = main, sub = sub, xlab = xlab, ylab = ylab, ...): "mean"
#> is not a graphical parameter
#> Warning in axis(1, ...): "mean" is not a graphical parameter
#> Warning in axis(2, ...): "mean" is not a graphical parameter
abline(v = u, col = "red")

Density dependence

CHALLENGE TASK Make a graph that indicates if “density dependence” is occurring.

max = max(females.N)
plot(year.t, females.N)
abline( h= max, col = "red")

How do we determine if a population is likely to go extinct?

This will be the core topic of next lecture.

If there is too much variance in population size from year to year.

Computational Ecology: Population Viability Analysis 9/15/2020

Anna Murray

9/15/2020

Data: Figure 5 (plus 10 years)

census

year t

Population size

Population growth rate: example

Population growth rate: vectorized

Negative indexing

Calcualte lambdas for all data

Finish putting together dataframe

Create special columns

Assemble the dataframe

Examing the population growth rates

Plotting the raw data

Challenge questions

Plot lambda

Plot ln(lambda)

Mean of log(lambda)

Histogram with mean

Density dependence

How do we determine if a population is likely to go extinct?