Types of relationships

Information about solutions

Solutions for these exercises are available immediately below each question.
We would like to emphasise that much evidence suggests that testing enhances learning, and we strongly encourage you to make a concerted attempt at answering each question before looking at the solutions. Immediately looking at the solutions and then copying the code into your work will lead to poorer learning.
We would also like to note that there are always many different ways to achieve the same thing in R, and the solutions provided are simply one approach.

Be sure to check the solutions to last week’s exercises.
You can still ask any questions about previous weeks’ materials if things aren’t clear!

LEARNING OBJECTIVES

• LO1: Understand the basic principles of functions.
• LO2: Understand concept of data transformations.
• LO3: Understand the calculation of z-scores.

Functions

You have seen by now how to visualise the distribution of a variable, and how to visualise a relationship between two variables. Relationships between two variables can look very different, and can follow different patterns. These patterns can be expressed mathematically in the form of functions.

Functions

A function is a mapping between two sets of numbers (\(x\) and \(y\)) - associating every element of \(x\) with an element in \(y\).
We often denote functions using the letter \(f\), in that we state that \(y = f(x)\) (“y equals f of x”).

For example, there is a mapping between these two sets:

\[x=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \\ \end{bmatrix}, \qquad y=\begin{bmatrix} 3 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \\ 9 \\ 10 \\ \end{bmatrix}\]

And we can write this mapping as:
\[f(x) = x + 2\]

And we could visualise this relationship between \(x\) and \(y\):

Linear functions

In statistics, we often attempt to summarise the pattern that is present in the data using linear functions.

Imagine that we plant 10 trees, and measure their heights each year for 10 years. We could visualise this data (the relationship between time and tree height) on a scatterplot (we have added some lines to the plot to show which tree is which):

We might sensibly choose to describe this pattern as a line:

And in order to describe a line like this, we require two things:

1. The starting point (i.e., where it crosses the y-axis)
2. The amount it goes up every year.

When we planted the trees (at year 0), they were on average about 5cm tall. So this is where our line starts.
For every year, the trees grew by about 10cm on average. So we can now describe tree height as a function of time:
\[\textrm{Tree height} = 5 + (10 \times \textrm{Years})\]

We can write this in terms of \(x\) and \(y\):

1. \(y = f(x)\)         “\(y\) is some function \(f\) of \(x\)”
2. \(f(x) = 5 + 10x\)         “the function \(f\) maps each value \(x_i\) to \(5 + (10 \times x_i)\)”

Non-linear functions

Functions don’t have to be linear. Often, we might want to describe relationships which appear to be more complex than a straight line.
For example, it is often suggested that for difficult tasks, some amount of stress may improve performance (but not too little or too much). We might think of the relationship between performance and stress as a curve (Figure 1).

Figure 1: Yerkes Dodson Law

One way to describe curves is to use polynomials (\(x^2\), \(x^3\), etc.).
For example, in the following two sets, \(y\) can be described as \(f(x)\) where \(f(x)=x^2\):

\[x=\begin{bmatrix} -5 \\ -4 \\ -3 \\ -2 \\ -1 \\ 0 \\ 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}, \qquad y=\begin{bmatrix} 25 \\ 16 \\ 9 \\ 4 \\ 1 \\ 0 \\ 1 \\ 4 \\ 9 \\ 16 \\ 25 \end{bmatrix}\]

and when we plot each value of \(x\) against the corresponding value of \(y\):

Optional - the code to create the above plot. Click to expand

# the tibble() function can be used to create a dataframe
# here, we create one with two variables, x and y.
# x is the sequence from -5 to 5,
# y is equal to x squared. 
# we save this data as "poly"
poly <- 
    tibble(
        x = c(-5,-4,-3,-2,-1,0,1,2,3,4,5),
        y = x^2
    )
# create a plot with "poly", with variable "x" on the x-axis,
# and variable "y" on the y-axis. Add geom_points for each entry
ggplot(data = poly, aes(x = x, y = y)) + 
    geom_point()

Transformations

We have seen previously how we might change all the values in a variable, for instance if we want to turn heights from centimetres to metres:

# read in the starwars dataset and assign it the name "starwars2"
starwars2 <- read_csv("https://uoepsy.github.io/data/starwars2.csv")

# take the starwars2 dataframe %>%
# mutate it such that there is a new variable called "height_m",
# the values of which are equal to the "height" variable divided by 100.
# then, select only the "height" and "height_m" columns (this is just 
# to make it easier to see without all the other variables)
starwars2 %>%
    mutate(
        height_m = height/100
    ) %>% 
    select(height, height_m)

## # A tibble: 75 × 2
##    height height_m
##     <dbl>    <dbl>
##  1    172     1.72
##  2    167     1.67
##  3     96     0.96
##  4    202     2.02
##  5    150     1.5 
##  6    178     1.78
##  7    165     1.65
##  8     97     0.97
##  9    183     1.83
## 10    182     1.82
## # … with 65 more rows

What we have done here, can be described as a transformation, in that we have applied a mathematical function to the values in the height variable.

Transformation

Data transformation is when we apply a deterministic function to map each value of a variable to a transformed value.
We transform for various reasons. For instance, we can use it to change the units we are interpreting (e.g., cm to m), or to change the shape of a distribution (e.g., make it less skewed).

We could even plot the heights in cm and heights in m against one another (note what units are on each axis):
The relationship between a variable and a transformed variable need be linear, for example, log transformation:

Optional: Logarithms and Natural Logarithms. Click to expand

Logarithm

A logarithm (log) is the power to which a number must be raised in order to get some other number.
Take as examples \(10^2\) and \(10^3\):

\[ 10^2 = 100 \qquad \Longleftrightarrow \qquad Log_{10} (100) = 2 \\ 10^3 = 1000 \qquad \Longleftrightarrow \qquad Log_{10} (1000) = 3 \\ \]

We refer to \(Log_{10}\) as “Log base 10”.

---

Natural log

A special case of the logarithm is referred to as the natural logarithm, and is denoted by \(ln\) or \(log_e\), where \(e = 2.718281828459...\)

\(e\) is a special number for which \(log_e(e) = 1\).

# Natural logarithm of e is 1:
log(2.718281828459, base = 2.718281828459)

## [1] 1

Centering and Standardisation

Recall our dataset from our introduction to handling numerical data, in which we had data on 120 participants’ IQ scores (measured on the Wechsler Adult Intelligence Scale, WAIS), their ages, and their scores on 2 other tests. We know how to calculate the mean and standard deviation of the IQ scores:

# read in the data
wechsler <- read_csv("https://uoepsy.github.io/data/wechsler.csv")

# calculate the mean and sd of IQs
wechsler %>% 
  summarise(
    mean_iq = mean(iq),
    sd_iq = sd(iq)
  )

## # A tibble: 1 × 2
##   mean_iq sd_iq
##     <dbl> <dbl>
## 1    99.3  15.4

Two very useful transformations we can apply to a variable are centering and standardisation.

1. Centering A transformation which re-expresses each value as the distance from a given number (e.g., the mean).
   
2. Standardising A transformation which re-expresses each value as the distance from the mean in units of standard deviations.

Mean-centering

To Mean-center a variable, we simply subtract the mean from each value, \(x_i - \bar{x}\): \[ \textrm{raw IQ} = \begin{bmatrix} 71 \\ 103 \\ 74 \\ 108 \\ 118 \\ 129 \\ ... \end{bmatrix}, \qquad \textrm{mean centered IQ} = \begin{bmatrix} 71-99.3 \\ 103-99.3 \\ 74-99.3 \\ 108-99.3 \\ 118-99.3 \\ 129-99.3 \\ ... \end{bmatrix} = \begin{bmatrix} -28.3 \\ 3.7 \\ -25.3 \\ 8.7 \\ 18.7 \\ 29.7 \\ ... \end{bmatrix} \]

To mean-center in R, we can simply add a new variable using mutate() and subtract the mean IQ from the IQ variable:

# Take the "wechsler" dataframe, and mutate it,
# such that there is a variable called "iq_meancenter" for which
# the entries are equal to the "iq" variable entries minus the 
# mean of the "iq" variable
wechsler %>%
  mutate(
    iq_meancenter = iq - mean(iq)
  )

## # A tibble: 120 × 6
##    participant    iq   age test1 test2 iq_meancenter
##    <chr>       <dbl> <dbl> <dbl> <dbl>         <dbl>
##  1 ppt_1          71    27    46    50        -28.3 
##  2 ppt_2         103    38    42    29          3.67
##  3 ppt_3          74    20    50    77        -25.3 
##  4 ppt_4         108    46    50    62          8.67
##  5 ppt_5         118    45    60    29         18.7 
##  6 ppt_6         129    33    45    45         29.7 
##  7 ppt_7         103    49    42    41          3.67
##  8 ppt_8         120    27    63    33         20.7 
##  9 ppt_9          96    37    53    44         -3.33
## 10 ppt_10         80    26    53    21        -19.3 
## # … with 110 more rows

Standardisation

When we standardise a variable, we call the transformed values z-scores. To transform a given value \(x_i\) into a z-score, we simply calculate the distance from \(x_i\) to the mean, \(\bar{x}\), and divide this by the standard deviation, \(s\)

\[ z_i = \frac{x_i - \bar{x}}{s} \]

So for each of the raw IQ scores, we can transform them to z-scores by subtracting the mean and then dividing by the standard deviation. The resulting values tell us how low/high each participant’s IQ score is compared to observed distribution of scores:
\[ \textrm{raw IQ} = \begin{bmatrix} 71 \\ 103 \\ 74 \\ 108 \\ 118 \\ 129 \\ ... \end{bmatrix}, \qquad \textrm{standardised IQ} = \begin{bmatrix} \frac{71-99.3}{15.43} \\ \frac{103-99.3}{15.43} \\ \frac{74-99.3}{15.43} \\ \frac{108-99.3}{15.43} \\ \frac{118-99.3}{15.43} \\ \frac{129-99.3}{15.43} \\ ... \end{bmatrix} = \begin{bmatrix} -1.84 \\ 0.238 \\ -1.64 \\ 0.562 \\ 1.21 \\ 1.92 \\ ... \end{bmatrix} \]

We can achieve this in R either by manually performing the calculation:

# Take the "wechsler" dataframe, and mutate it,
# such that there is a variable called "iq_z" for which
# the entries are equal to the "iq" variable entries minus the mean of the "iq"
# variable, divided by the standard deviation of the "iq" variable.  
wechsler %>% 
  mutate(
    iq_z = (iq - mean(iq)) / sd(iq)
  )

## # A tibble: 120 × 6
##    participant    iq   age test1 test2   iq_z
##    <chr>       <dbl> <dbl> <dbl> <dbl>  <dbl>
##  1 ppt_1          71    27    46    50 -1.84 
##  2 ppt_2         103    38    42    29  0.238
##  3 ppt_3          74    20    50    77 -1.64 
##  4 ppt_4         108    46    50    62  0.562
##  5 ppt_5         118    45    60    29  1.21 
##  6 ppt_6         129    33    45    45  1.92 
##  7 ppt_7         103    49    42    41  0.238
##  8 ppt_8         120    27    63    33  1.34 
##  9 ppt_9          96    37    53    44 -0.216
## 10 ppt_10         80    26    53    21 -1.25 
## # … with 110 more rows

Or we can use the scale() function:

# Take the "wechsler" dataframe, and mutate it,
# such that there is a variable called "iq_std" for which
# the entries are equal to the scaled values of the "iq" variable.  
wechsler %>%
  mutate(
    iq_std = scale(iq)
  )

## # A tibble: 120 × 6
##    participant    iq   age test1 test2 iq_std[,1]
##    <chr>       <dbl> <dbl> <dbl> <dbl>      <dbl>
##  1 ppt_1          71    27    46    50     -1.84 
##  2 ppt_2         103    38    42    29      0.238
##  3 ppt_3          74    20    50    77     -1.64 
##  4 ppt_4         108    46    50    62      0.562
##  5 ppt_5         118    45    60    29      1.21 
##  6 ppt_6         129    33    45    45      1.92 
##  7 ppt_7         103    49    42    41      0.238
##  8 ppt_8         120    27    63    33      1.34 
##  9 ppt_9          96    37    53    44     -0.216
## 10 ppt_10         80    26    53    21     -1.25 
## # … with 110 more rows

We can also use the scale() function to mean-center a variable, by setting scale(variable, center = TRUE, scale = FALSE):

# create two new variables in the "wechsler" dataframe, one which is 
# mean centered iq, and one which is standardised iq:
wechsler %>%
  mutate(
    iq_mc = scale(iq, center = TRUE, scale = FALSE),
    iq_std = scale(iq, center = TRUE, scale = TRUE) # these are the default settings
  )

## # A tibble: 120 × 7
##    participant    iq   age test1 test2 iq_mc[,1] iq_std[,1]
##    <chr>       <dbl> <dbl> <dbl> <dbl>     <dbl>      <dbl>
##  1 ppt_1          71    27    46    50    -28.3      -1.84 
##  2 ppt_2         103    38    42    29      3.67      0.238
##  3 ppt_3          74    20    50    77    -25.3      -1.64 
##  4 ppt_4         108    46    50    62      8.67      0.562
##  5 ppt_5         118    45    60    29     18.7       1.21 
##  6 ppt_6         129    33    45    45     29.7       1.92 
##  7 ppt_7         103    49    42    41      3.67      0.238
##  8 ppt_8         120    27    63    33     20.7       1.34 
##  9 ppt_9          96    37    53    44     -3.33     -0.216
## 10 ppt_10         80    26    53    21    -19.3      -1.25 
## # … with 110 more rows

Test norms. Click to expand

Many neuropsychological tests will have norms - parameters which describe the distribution of scores on the test in a normal population. For instance, in a normal adult population, scores on the WAIS have a mean of 100 and a standard deviation of 15.
What this means is that rather than calculating a standardised score against the observed mean of our sample, we might calculate a standardised score against the test norms. In the formula for \(z\), we replace our sample mean \(\bar{x}\) with the population mean \(\mu\), and the sample standard deviation \(s\) with the population standard deviation \(\sigma\):
\[ z = \frac{x - \mu}{\sigma} \]

The resulting values tell us how low/high each participant’s IQ score is compared to the distribution of IQ scores in the population.

---

Glossary

• Function: A mapping between two sets of numbers, associating every element of the first set with an elemet in the second.
  
• Transformation: Applying a function to a variable to map each value to a transformed value.
• Logarithm: The power to which a number must be raised in order to get some other number.

1. Centering: Transformation which re-expresses each value as the distance from a given number (e.g., the mean).
   
2. Standardisation: Transformation which re-expresses each value as the distance from the mean in units of standard deviations.
   
   

• scale() To mean center or standardise a variable (depending upon whether center=TRUE/FALSE and scale=TRUE/FALSE).

---

Exercises

The exercises below are a little different. As opposed to starting with some data, we’re going to work a bit more abstractly, and create some data ourselves which maps one variable to another via some function which we define.

As always, you should begin by opening a new Rmarkdown file, and giving it an appropriate title for this set of exercises.
We’re going to need the tidyverse packages, so load that too.

Question 1

We know how to create a vector in R. For instance, we can create an object with called “x” which has the entries from 1 to 100:

x <- 1:100

For \(x = [1,2,...,99,100]\), give the values resulting from \(f(x) = 3x+3\).

Solution

3*x + 3

##   [1]   6   9  12  15  18  21  24  27  30  33  36  39  42  45  48  51  54  57
##  [19]  60  63  66  69  72  75  78  81  84  87  90  93  96  99 102 105 108 111
##  [37] 114 117 120 123 126 129 132 135 138 141 144 147 150 153 156 159 162 165
##  [55] 168 171 174 177 180 183 186 189 192 195 198 201 204 207 210 213 216 219
##  [73] 222 225 228 231 234 237 240 243 246 249 252 255 258 261 264 267 270 273
##  [91] 276 279 282 285 288 291 294 297 300 303

Question 2

For \(x = [1,2,...,19,20]\), give the values resulting from \(f(x) = x^2+2\).
Plot these against \(x\).

Tip: You don’t have to use ggplot. For a really quick plot in R, you can use plot(). If you want to plot two vectors against one another, you can simply use plot(x = <vector1name>, y = <vector2name>)

Solution

Remember - the object named “x” is currently the numbers 1 to 100, but we only want the numbers 1 to 20. We can reassign the name “x” to these numbers

x <- 1:20
y <- (x^2)+2

plot(x = x, y = y)

Question 3

We can create an object in R called a tibble (basically a dataframe in tidyverse language). We can, for instance, create an object which has a variable called “x” and a variable called “y”, which we define:

# create a new tibble/dataframe, and assign it the name "newdata".
# in it, there is a variable called "x" which takes the numeric
# values from -10 to 10, and the variable called "y" which takes a set of 
# numeric values: 102,83,.... 
newdata <-
  tibble(
    x = -10:10,
    y = c(102,83,66,51,38,27,18,11,6,3,2,3,6,11,18,27,38,51,66,83,102)
  )
newdata

## # A tibble: 21 × 2
##        x     y
##    <int> <dbl>
##  1   -10   102
##  2    -9    83
##  3    -8    66
##  4    -7    51
##  5    -6    38
##  6    -5    27
##  7    -4    18
##  8    -3    11
##  9    -2     6
## 10    -1     3
## # … with 11 more rows

Plot the variables against one another, and write out the function which defines their relationship.
Confirm that you are correct by creating a new column called “fx”, which is calculated from the values in x.

Tip: For variables in a dataframe against one another, you could combine plot() with the $, and use plot(x = data$variable1, y = data$variable2).

Solution

plot(x = newdata$x, y = newdata$y)

Take the first pairing, \(x = -10\) and \(y = 102\).
We could write the translation between these two numbers in many ways.
All of these are compatible:

1. \(y = -10x + 2\)
   
2. \(y = 112 + x\)
   
3. \(y = 1000 + x - 888\)
   
4. \(y = x^2 + 2\)

Now take the second pair of points, \(x = -9\) and \(y = 83\).
From the possible functions a), b), c) and d) above, how many of these satisfy the relationship between this new pair of points? Only one!

1. \(-10x + 2 \quad = \quad 92\)     ✘
   
2. \(112 + x \quad = \quad 103\)     ✘
   
3. \(1000 + x - 888 \quad = \quad 103\)     ✘
   
4. \(x^2 + 2 \quad = \quad 83\)     ✔

The curvature here actually gives us a clue as to what to look for - it suggests that \(y\) is a function of \(x\) which includes the term \(x^2\). Remember that negative numbers squared result in positive square numbers.

\[y = x^2 + 2\]

Let’s confirm this:

newdata %>%
  mutate(
    fx = x^2 + 2
  )

## # A tibble: 21 × 3
##        x     y    fx
##    <int> <dbl> <dbl>
##  1   -10   102   102
##  2    -9    83    83
##  3    -8    66    66
##  4    -7    51    51
##  5    -6    38    38
##  6    -5    27    27
##  7    -4    18    18
##  8    -3    11    11
##  9    -2     6     6
## 10    -1     3     3
## # … with 11 more rows

Question 4

We can also create a plot of a function using stat_function() in a ggplot.
The syntax is a bit odd (and difficult to remember!), but we can give our ggplot the variable from our dataset on the x-axis, and then give the stat_function() function a mathematical function to the fun = argument:

ggplot(data = newdata, aes(x = x)) +
    stat_function(fun = function(.x) .x^2+2)

For \(x in [-10,-9,...,9,10]\), use stat_function() to create three plots, one for each of the functions \(f(x)=x^2\), \(f(x)=x^3\) and \(f(x)=x^4\).

As an extra challenge, colour the lines blue, red and green respectively, and arrange the plots side by side.

You can do this with the package patchwork, which is specifically for arranging ggplots.
You will need to install the package first, by calling install.packages("patchwork") in your console (remember, we install only once so we don’t want to keep this line in our script where we are likely to run code many times). You will then need to load the package using library(package) in your editor.

You can then arrange ggplots using |, +, and /.

To find out how, take a look at the patchwork vignette

Figure 2: Artwork by @allison_horst

Solution

p1 <- ggplot(data = newdata, aes(x = x)) +
  stat_function(fun = function(.x) .x^2, col = "blue") +
  labs(title = "Quadratic, x^2")

p2 <- ggplot(data = newdata, aes(x = x)) +
  stat_function(fun = function(.x) .x^3, col = "red") +
  labs(title="Cubic, x^3")
  
p3 <- ggplot(data = newdata, aes(x = x)) +
  stat_function(fun = function(.x) .x^4, col = "green") + 
  labs(title="Quartic, x^4")

library(patchwork)
p1 + p2 + p3

Question 5

I take out a loan of £1000 from a bank with an interest rate of 15% per annum.
Over the course of 20 years, I never pay any of it back.
If you were to create a plot with “years” on the x-axis and “amount owed” on the y-axis. What would the line look like?

As an extra challenge, try to write out the function f(x) for \(x\) in years \([1,2,...,19,20]\) which computes the amount of the loan.

Solution

The amount added each year gets bigger, which means the line is curved!

• Year 1: 15% of £1000 is £150
• Year 2: 15% of £1150 is £172.5
• Year 3: 15% of £1132.5 is £198.375

The function which describes the relationship between year (\(x\)) and total amount owed is \(f(x) = 1000 \times 1.15^x\)

loan <- 
  tibble(
    years = 1:20
  )
ggplot(data = loan, aes(x = years)) +
  stat_function(fun = function(.x) 1000 * 1.15^.x) + 
  labs(x = "years", y = "amount owed", title="Compound Interest")

Question 6

Read in the data at https://uoepsy.github.io/data/rtdata.csv. It contains measurements (in milliseconds) from a simple reaction time task, conducted on 500 participants. Create a histogram of the 500 participants’ reaction times.

Solution

rtdata <- read_csv("https://uoepsy.github.io/data/rtdata.csv")

ggplot(data = rtdata, aes(x = rt)) + 
  geom_histogram(col="white")

Question 7

Notice that the distribution of reaction times is skewed. Why do you think this might be?

Often, a lot of statistical analyses rely on assuming certain characteristics of a distribution - for example, assuming that it is bell-shaped and symmetric (which these reaction times are not).

We can apply certain transformations to an observed distribution in order to have a distribution of transformed values which follows the shape we want.
One of the most common transformations is the log transformation: \(f(x) = log_e(x)\)
Add a new column to the data which contains log transformed reaction times, and create a histogram of the distribution.

Solution

rtdata <- 
  rtdata %>%
  mutate(
    logrt = log(rt)
  )

ggplot(data = rtdata, aes(x = logrt)) + 
  geom_histogram(col="white")

Question 8

Recall that log transformation is not a linear transformation. Let’s explore what that means.

Consider the following people and their reaction times:

• Person A: 500ms
• Person B: 600ms
• Person C: 700ms
• Person D: 800ms

1. How much slower, in milliseconds, is person B compared to person A?
2. How much slower, in milliseconds, is person D compared to person C?
3. How much slower, in log(milliseconds), is person B compared to person A?
4. How much slower, in log(milliseconds), is person D compared to person C?

Are your answers what you expected?

Solution

1. Person B is 100ms slower than person A (600-500)
2. Person D is 100ms slower than person C (800-700)
3. Person B is 0.182 log(milliseconds) slower than person A (log(600)-log(500))
4. Person D is 0.134 log(milliseconds) slower than person C (log(800)-log(700))

Notice that the a difference of 100ms corresponds to different amounts of the log transformed values, depending upon where the difference happens (800ms relative to 700ms, or 600ms relative to 500ms).
You can see the non-linearity of this when we plot the raw values against the log transformed ones:

ggplot(data = rtdata, aes(x = rt, y = logrt)) + 
  geom_point()

Question 9

Also in the dataset are the participants’ heights:

ggplot(data = rtdata, aes(x = height)) + 
  geom_histogram(col="white")

Standardise the participants’ heights (re-express each value as the distance from the mean height in units of standard deviations). Produce a histogram of standardised heights. How is it different from the histogram above?

Solution

The histogram of standardised values will look identical to the raw values, but the numbers on the x-axis are different, now reflecting standard deviations from the mean.

rtdata <- 
  rtdata %>%
  mutate(
    height_z = scale(height)
  )

ggplot(data = rtdata, aes(x = height_z)) + 
  geom_histogram(col="white")

Question 10

How many standard deviations above the mean height is participant number 45? How tall are they in cm?

Solution

Participant 45 is 199cm tall, and is 4.1 standard deviations above the sample mean.
We can access this data either by filter()-ing the data to where the “ppt” variable is equal to 45:

rtdata %>% filter(ppt == 45)

## # A tibble: 1 × 5
##     ppt    rt height logrt height_z[,1]
##   <dbl> <dbl>  <dbl> <dbl>        <dbl>
## 1    45  477.    199  6.17         4.14

Or by using [] and $:

rtdata[rtdata$ppt == 45, ]

## # A tibble: 1 × 5
##     ppt    rt height logrt height_z[,1]
##   <dbl> <dbl>  <dbl> <dbl>        <dbl>
## 1    45  477.    199  6.17         4.14