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

census <- 1:39
year.t   <- 1959:1997
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)
lambda.i <- females.N[-1]/females.N[-length(females.N)]
lambda.i <- c(lambda.i,NA)
lambda_log <- log(lambda.i)
bear_N <- data.frame(census,
                year.t,
                females.N,
                lambda.i, 
                lambda_log)

plot(females.N ~ year.t, data = bear_N, 
     type = "b",
     ylab = "Population index (females + cubs)",
     xlab = "Year")
abline(v = 1970)

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

hist(bear_N$lambda.i)

A whole distribution of population growth rates. Determine average and then see how variable it is with different circumstances. To model populations dynamics we randomly pull population growth rates out of a hat.

hat_of_lambdas <- bear_N$lambda.i

NA means: not aplicaple missing data NaN, Inf, -Inf (mathematical equation resulting in a non real number), NULL (removes an attribute)

is.na -> gives when a value is a NA (true/false)

is.na(hat_of_lambdas)
#>  [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
#> [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
#> [25] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
#> [37] FALSE FALSE  TRUE

any -> are there any values that are true and are NA

any(is.na(hat_of_lambdas) == TRUE)
#> [1] TRUE

Drop the NA

length(hat_of_lambdas)
#> [1] 39
hat_of_lambdas[39]
#> [1] NA
hat_of_lambdas[-39]
#>  [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
hat_of_lambdas[-length(hat_of_lambdas)] #writing above in more general form
#>  [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

na.omit(hat_of_lambdas)
#>  [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
#> attr(,"na.action")
#> [1] 39
#> attr(,"class")
#> [1] "omit"

Create vector that lacks null value

hat_of_lambdas <- hat_of_lambdas[-length(hat_of_lambdas)]

hist(hat_of_lambdas)

Random Lambda Sampling

Pulls a random lambda value from the lambdas previously coded for.

# pulled a lambda from the hat
sample(x = hat_of_lambdas, 
       size = 1, 
       replace = TRUE) #recycles values
#> [1] 1.108108
# pull lambda from hat and save to object
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)

Head and Tail

Head displays the first couple values in the variable. Tail displays the last couple values in the variable

head(bear_N)
#>   census year.t females.N  lambda.i  lambda_log
#> 1      1   1959        44 1.0681818  0.06595797
#> 2      2   1960        47 0.9787234 -0.02150621
#> 3      3   1961        46 0.9565217 -0.04445176
#> 4      4   1962        44 1.0454545  0.04445176
#> 5      5   1963        46 0.9782609 -0.02197891
#> 6      6   1964        45 1.0222222  0.02197891
tail(bear_N)
#>    census year.t females.N  lambda.i lambda_log
#> 34     34   1992        65 0.8769231 -0.1313360
#> 35     35   1993        57 1.2280702  0.2054440
#> 36     36   1994        70 1.1571429  0.1459539
#> 37     37   1995        81 1.2222222  0.2006707
#> 38     38   1996        99 1.0000000  0.0000000
#> 39     39   1997        99        NA         NA
summary(bear_N)
#>      census         year.t       females.N        lambda.i     
#>  Min.   : 1.0   Min.   :1959   Min.   :33.00   Min.   :0.8250  
#>  1st Qu.:10.5   1st Qu.:1968   1st Qu.:39.00   1st Qu.:0.9452  
#>  Median :20.0   Median :1978   Median :44.00   Median :1.0000  
#>  Mean   :20.0   Mean   :1978   Mean   :49.79   Mean   :1.0281  
#>  3rd Qu.:29.5   3rd Qu.:1988   3rd Qu.:57.00   3rd Qu.:1.1038  
#>  Max.   :39.0   Max.   :1997   Max.   :99.00   Max.   :1.3077  
#>                                                NA's   :1       
#>    lambda_log      
#>  Min.   :-0.19237  
#>  1st Qu.:-0.05639  
#>  Median : 0.00000  
#>  Mean   : 0.02134  
#>  3rd Qu.: 0.09874  
#>  Max.   : 0.26826  
#>  NA's   :1
dim(bear_N)
#> [1] 39  5
N.1997 <- 99

One round of population simulate

Uses lambda to predict number of bears you have in the next year.

1.22807*99
#> [1] 121.5789
lambda_rand.t*N.1997
#> [1] 93.5
N.1998 <- lambda_rand.t*N.1997

Simulation the hard way

Simulating a random process by using a random previous lambda. Predicts the population that will occur in the next year.

# 1997 to 1998
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.1998 <- lambda_rand.t*N.1997

# 1998 to 1999
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.1999 <- lambda_rand.t*N.1998

# 1999 to 2000
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.2000 <- lambda_rand.t*N.1999

# 2000 to 2001
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.2001 <- lambda_rand.t*N.2000

# 2001 to 2002
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.2002 <- lambda_rand.t*N.2001

# 2002 to 2003
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.2003 <- lambda_rand.t*N.2002

# 2003 to 2004
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.2004 <- lambda_rand.t*N.2003

# 2004 to 2005
lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
N.2005 <- lambda_rand.t*N.2004

Pop size prediction

Creates vector or predicted population sizes.

year <- seq(1997, 2004)
N.rand <- c(N.1998,N.1999,N.2000,N.2001,N.2002,N.2003,N.2004,N.2005)
df.rand <- data.frame(N.rand, year)
plot(N.rand ~ year, data = df.rand, type = "b")

R’s Version of a For loop

xlim sets the size of the x axis.

#Initial Conditions

N.1997 <- 99
N.initial <- 99

# Explore xlim = argument
plot(N.1997 ~ c(1997))

plot(N.1997 ~ c(1997), xlim = c(1997, 1997+50))

plot(N.1997 ~ c(1997), xlim = c(1997, 1997+50), ylim = c(0, 550))

# For loop the hard way
#
N.current <- N.initial

# This is where the for loop would be
t <- 1
  
  # grab a lambda
  lambda_rand.t <- sample(x = hat_of_lambdas, size = 1,replace = TRUE)
  
  # determine population size
  N.t <- N.current*lambda_rand.t
  
  # tell R what year were in
  year.t <- 1997+t
  
  # Add the point (updates existing graph)
  points(N.t ~ year.t)

  
  # update N current
  N.current <- N.t

Running a real for loop.


# starts at 1997 pop size
plot(N.1997 ~ c(1997), xlim = c(1997, 1997+50), ylim = c(0, 550))

N.current <- N.1997

# Run a for loop
for(t in 1:50){
  
  # gets a random lambda
  lambda_rand.t <- sample(x = hat_of_lambdas, 
                          size = 1,
                          replace = TRUE)
  
  # gives new population size
  N.t <- N.current*lambda_rand.t
  
  # updates the year
  year.t <- 1997+t
  
  # adds point to the graph
  points(N.t ~ year.t)
  
  # update current N
  N.current <- N.t
}

Goofy R plotting code/ magic. Declares parameters for R’s plotting

par(mfrow = c(3,3), mar = c(1,1,1,1))

Keep track of 9 simulations. Visually see the variability.

plot(N.1997 ~ c(1997), xlim = c(1997, 1997+50), ylim = c(0, 550))
N.current <- N.1997
for(t in 1:50){
  
  lambda_rand.t <- sample(x = hat_of_lambdas, 
                          size = 1,
                          replace = TRUE)
  
  N.t <- N.current*lambda_rand.t
  
  year.t <- 1997+t
  
  points(N.t ~ year.t)
  
  N.current <- N.t
}

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

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

Random Lambda Sampling

Head and Tail

One round of population simulate

Simulation the hard way

Pop size prediction

R’s Version of a For loop