The Logistic Map: Visualizing Chaos in R

In the 1970s, professor Robert May became interested in the relationship between complexity and stability in animal populations. He noted that even simple equations used to model populations over time can lead to chaotic outcomes. The most famous of these equations is as follows:

xn+1 = rxn(1 – xn)

xn is a number between 0 and 1 that refers to the ratio of the existing population to the maximum possible population. Additionally, r refers to a value between 0 and 4 which indicates the growth rate over time. xn is multiplied by the r value to simulate growth where (1 – xn) represents death in the population.

Lets assume a population of animals is at 50% of the maximum population for a given area. We would allow xn to be .5. Lets also assume a growth rate of 75% allowing r to be .75. After the value xn+1 is computed, we use that new value as the xn in the next iteration and continue to use an r value of .75. We can visualize how xn+1 changes over time.

Visualizing the population with an r value of 50% and a starting population of 50%.

Within 20 iterations, the population dies off. Lets rerun the simulation with an r value greater than 1.

Visualizing the population with an r value of 1.25 and a starting population of 50%.

Notice how the population stabilizes at 20% of the area capacity. When the r value is higher than 3, the population with begin oscillating between multiple values.

Visualizing the population with an r value of 3 and a starting population of 50%.

Expanding beyond an r value of 3.54409 yields rapid changes in oscillation and reveals chaotic behavior.

Visualizing the population with an r value of 3.7 and a starting population of 50%.

Extremely minor changes in the r value yield vastly different distributions of population oscillations. Rather than experiment with different r values, we can visualize the distribution of xn+1 values for a range of r values using the R programming language.

Lets start by building a function in R that returns the first 1000 iterations of xn+1 for a given r value.

logistic_sim <- function(lamda, starting_x = 0.5) {
  # Simulate logistic function
  vals <- c(starting_x)
  iter <- seq(1, 1000, 1)
  for (i in iter) { 
    vals[(i + 1)] <- vals[i] * lamda * (1 - vals[i])
  vals <- vals[-length(vals)]
  tibble::tibble(vals, lamda, iter)

This function returns a dataframe with three columns: the iteration number, the r used for each iteration, and the xn+1 value computed for that iteration.

Now we need to iterate this function over a range of r values. Using purrr::map_dfr we can row bind each iteration of r together into a final dataframe.

build_data <- function(min, max) {
  # Build data for logistic map
  step <- (max - min) / 400
    seq(min, max, step),
  ) |>

Min refers to the lower limit of r while the max refers to the upper limit. The function will return a dataframe of approximately 400,000 values referring to each of the 1000 iterations for the 400 r values between the lower and upper bound. The function returns all 400,000 values in less than a quarter of a second.

With the dataframe of values assembled, we can visualize the distribution of values using ggplot.

build_data(1, 4) |>
  dplyr::filter(iter > 50) |>
  dplyr::slice_sample(prop = 0.1) |>
    x = lamda,
    y = vals,
    color = lamda
  )) +
  ggplot2::geom_point(size = 0.5) +
    x = "Growth Rate",
    y = "Population Capacity",
    title = "Testing Logistic Growth in Chaos"
  ) +
    labels = scales::percent
  ) +
    labels = scales::percent
  ) +
  ggplot2::theme_minimal() +
    legend.position = "none",
    text = element_text(size = 25)
Visualizing the distribution of 400 r values between 0 to 4 for 1000 iterations.

Notice how r values of less than 1 indicate the population dies out. Between 1 and just under three, the population remains relatively stable. At around 3, the populations being oscillating between two points. Beyond an r of 3.54409, chaos ensues. It becomes extremely difficult to predict the value of xn+1 for a given iteration with an r value above 3.54409. So difficult, in fact, that this simple deterministic equation was used as an early random number generator.

So what are the practical applications for this? Representations of chaos (or systems that yield unpredictable results and are sensitive to starting conditions) can be seen across many industries and fields of study. In finance, for example, intra-day security prices have been described as a random walk – extremely difficult to predict. While long term outlooks may show seasonality, chaos theory can help model the extremely chaotic and unpredictable nature of stock prices.

Implementing Fuzzy Matching in Python

Text is all around us; essays, articles, legal documents, text messages, and news headlines are consistently present in our daily lives. This abundance of text provides ample opportunities to analyze unstructured data.

Imagine you are playing a game where someone hands you a index card with the misspelled name of a popular musician. In addition, you have a book containing correctly spelled names of popular musicians. The goal is for you to return the correct spelling of the misspelled name.

In this example, suppose someone hands you a card with “Billie Jole” written on it. You quickly open the book of musicians, find names beginning with the letter B, and find the name “Billy Joel.”

As a human, this was easy for you to complete, but what if you wanted to automate this task? This can be done using Fuzzy Logic, or more specifically, the Levenshtein distance.

The Levenshtein distance considers two pieces of text and determines the minimum number of changes required to convert one string into another. You can utilize this logic to find the closest match to any given piece of text.

I am going to focus on implementing the mechanics of finding a Levenshtein distance in Python rather than the math that makes it possible. There are many resources on YouTube which explain how the Levenshtein distance is calculated.

First, import numpy and define a function. I called the function lv as shorthand for Levenshtein distance. Our function requires two input strings which are used to create a 2D matrix that is one greater than the length of each string.

def ld(s1, s2):    
    rows = len(s1)+1
    cols = len(s2)+1
    dist = np.zeros([rows,cols])

If you were to use the strings “pear” and “peach” in this instance, the function should create a 5 by 6 matrix filled with zeros.

A matrix of zeros.

Next, the first row and column need to count up from zero. Using for loops, we can iterate over the selected values. Our Python function now creates the following matrix.

def ld(s1, s2):
    rows = len(s1)+1
    cols = len(s2)+1
    dist = np.zeros([rows,cols])
    for i in range(1, rows):
        dist[i][0] = i
    for i in range(1, cols):
        dist[0][i] = i
A matrix set up for finding the Levenshtein distance.

Finally, we need to iterate over every column and row combination. By doing this, we can find the minimum value of the cells directly above, to the left, and above to the left of each cell. After the minimum is found, our Python script adds one to this value to the location in question.

def ld(s1, s2):
    rows = len(s1)+1
    cols = len(s2)+1
    dist = np.zeros([rows,cols])
    for i in range(1, rows):
        dist[i][0] = i
    for i in range(1, cols):
        dist[0][i] = i
    for col in range(1, cols):
        for row in range(1, rows):
            if s1[row-1] == s2[col-1]:
                cost = 0
                cost = 1
            dist[row][col] = min(dist[row-1][col] + 1,
                                 dist[row][col-1] + 1,      
                                 dist[row-1][col-1] + cost)
            return dist[-1][-1]

Our matrix should now look like the following with the far bottom right cell representing the number of changes required to convert one string into another. In this instance, it requires 2 changes to convert “peach” into “pear”; deleting the letter “c” in “peach” and replacing the letter “h” with the letter “r”.

A completed Levenshtein distance matrix. The bottom right number (in gold) represents the number of changes required.

What is so great about this function is that it is adaptable and will accept a string of any length to compute the number of changes required. While the mechanics behind this function are relatively simple, its use cases are vast.

Hi! I’m Chris Walker.

I am a Senior Data Scientist at Fannie Mae and a student in Georgia Tech’s MS in Analytics program. I graduated with honors from Texas A&M University with a BS in Urban and Regional Planning with a minor in Economics. I am excited about housing, risk modeling, and data science.

Feel free to contact me on LinkedIn or via email at