Tables in R

Tables in R

In this walk-through, we’ll learn the following key concepts and R functions:

  • Using tables to estimate relatively simple probabilities (including conditional and joint probabilities)
  • Making tables with xtabs
  • Modifying tables with prop.table and addmargins
  • piping (%>%) as a way of chaining together computations

The data

Before you get started, download aclfest.csv, which contains data on some bands that played at several major U.S. music festivals (including our own ACL Fest here in Austin).

Getting started

We’ll first load the tidyverse library, which we’ll need for just about every R walkthrough in this course.

## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.0 ──

## ✓ ggplot2 3.3.2     ✓ purrr   0.3.4
## ✓ tibble  3.0.4     ✓ dplyr   1.0.2
## ✓ tidyr   1.1.2     ✓ stringr 1.4.0
## ✓ readr   1.4.0     ✓ forcats 0.5.0

## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()

If you see a similar set of messages to what’s shown above after the ## symbols, you’re good to go!

But if you haven’t installed the tidyverse library, executing this command will give you an error like this:

Error in library(tidyverse) : there is no package called ‘tidyverse’

To avoid the error, you’ll first need to install tidyverse using the Install button under the Packages tab, which is typically located in the bottom-right panel of RStudio’s four-panel layout. Remember, an R library is like an app on your phone: you have to install it just once, but you have to load it each time you want to use it. On a phone, you load an app by clicking on its icon; in R, you load a library using the library command.

Next, use the Import Dataset button to read in the data you downloaded, which you can find under the Environment tab in the top-right panel. Follow the prompts to import aclfest.csv.

For a reminder on how to accomplish these two key steps (loading a library, importing a data set), see the previous walkthrough on Getting Started in R.

If you’re imported the data correctly, you can run the head function to get the first six lines of the file. You should see the following result:

##                          band acl bonnaroo coachella lollapalooza outsidelands
## 1                         ALO   0        0         0            0            1
## 2                     Battles   0        1         1            1            0
## 3                    Bon Iver   0        0         0            0            1
## 4              Flogging Molly   0        0         1            1            0
## 5 Ivan Neville's Dumpstaphunk   0        0         0            0            1
## 6                   Radiohead   0        0         0            1            1
##   year
## 1 2008
## 2 2008
## 3 2008
## 4 2008
## 5 2008
## 6 2008

Each entry is either a 1 or a 0, meaning “yes” and “no”, respectively. So, for example, on the 6th line we see an entry of 1 for Radiohead under the lollapalooza column, which means that Radiohead played at Lollapalooza that year.

Let’s make a few simple tables with this data. This will allow us to estimate probabilities like:

  1. How likely is a band in this sample to play Lollapalooza?
  2. How like is a band to play either ACL Fest and Lollapalooza?
  3. How likely is a band to play ACL Fest, given that they played Lollapalooza?

Simple probabilities from tables

Let’s address question 1: what is P(played Lollapalooza) for a randomly selected band in this sample? To answer this, we’ll use R’s xtabs function to tabulate (i.e. count) the data according to whether a band played Lollapalooza (1) or not (0).

xtabs(~lollapalooza, data=aclfest)
## lollapalooza
##   0   1 
## 800 438

(You might be curious about the little tilde (~) symbol in front of lollapalooza. Roughly speaking, ~ means “by” or “according to”—as in, “cross-tabulate BY the lollapalooza variable.”)

OK, back to the table. Remember that 1 means yes and 0 means no. So of the 1238 bands (800 + 438) in this sample, 438 of them played Lollapalooza. We can now just use R as a calculator to get this proportion:

438/(800 + 438)
## [1] 0.3537964

Using prop.table

Simple enough, right? But we can also get R to turn those counts into proportions for us, using the prop.table function. In general, the more work we can get R to do for us, the better!

One way to do this is as follows. First we’ll make the same table as before, except now we’ll save the result in an “object” whose name we get to choose. We’ll just call it t1 in the code below, although you could call it something more imaginative, like my_great_lollapalooza_table, if you wanted to.

t1 = xtabs(~lollapalooza, data=aclfest)

Notice that nothing gets printed to the screen when you execute this command. But if you ask R what t1 is, it will show you the same table as before:

## lollapalooza
##   0   1 
## 800 438

OK, so why did we bother to store this table in something called t1? Well, remember one of the core ideas in programming: power comes from linking computations together. That’s exactly what we’re doing here: we’ll take this t1 object we’ve created (the first link our chain) and pass it into the prop.table function (the second link in our chain). This function turns a table of counts (like t1) into a table of proportions, like this:

## lollapalooza
##         0         1 
## 0.6462036 0.3537964

Notice how the answer we get—P(plays Lollapalooza) = 0.354—is the same one we got when we did the division “by hand” (i.e. treating R as a calculator to calculate 438/1238).

Using pipes

The above way of doing things—using xtabs creating an “intermediate” object called t1 and then passing t1 into the prop.table function—works just fine. Lots of people write R code this way.

But it turns out there’s a nicer way to accomplish the same task, using a “pipe” (%>%). Here’s how it works:

xtabs(~lollapalooza, data=aclfest) %>%
## lollapalooza
##         0         1 
## 0.6462036 0.3537964

This code block says: “make a table of counts of the lollapalooza variable in the aclfest data set, and pipe (%>%) the resulting table into prop.table to turn the table of counts into a table of proportions.”

The result is exactly the same as before; using the pipe is totally optional. But using pipes tends to make your code easier to easier to write, easier to read, and easier to modify. For a simple calculation like this involving only two steps, the difference is minimal. But for the more complex kinds of calculations we’ll see later in the course, the difference can be substantial.

Piping makes it especially easy to add steps in your pipeline. For example, suppose we wanted to round all numbers to 3 decimal places, to make these easier to read. This is easily done by adding one more pipe to the code block above:

xtabs(~lollapalooza, data=aclfest) %>%
  prop.table %>%
## lollapalooza
##     0     1 
## 0.646 0.354

The table looks a lot nicer now.

For reasons of readability and ease of modification, I strongly recommend learning to write R code using pipes.

Joint probabilities from tables

Let’s now look at a second question: how like is a band to play either ACL Fest and Lollapalooza?

We’ll start off by tabulating the bands according to both of the relevant variables: whether they played at ACL (0 or 1), and whether they played at Lollapalooza (0 or 1). This works as follows:

xtabs(~acl + lollapalooza, data=aclfest)
##    lollapalooza
## acl   0   1
##   0 719 361
##   1  81  77

Here you should interpret the + sign as meaning “and”, not in terms of arithmetic. So in English, the code is saying: “cross-tabulate the aclfest data by the acl AND lollapalooza variables.”

The result of calling xtabs gives us some numbers that represent joint frequencies. Remember, for the row and column labels, 1 means yes and 0 means no. So to be specific, the output is telling us that, of the 1238 bands in the data set:

  • 719 played at neither ACL nor Lollapalooza.
  • 81 played at ACL but not Lollapalooza.
  • 361 played at Lollapalooza but not ACL.
  • 77 played at both ACL and Lollapalooza.

From here, we could actually stop and use use a calculator (or use R as a calculator) to work out the relevant joint probability. But as you’ll see, it’s less labor-intensive if we get R to do some of that work for us.

So instead, let’s use prop.table again, to turn those counts into proportions. As with the first example above, we’ll do this using a pipe:

xtabs(~acl + lollapalooza, data=aclfest) %>%
  prop.table %>%
##    lollapalooza
## acl     0     1
##   0 0.581 0.292
##   1 0.065 0.062

These numbers now represent proportions, not counts. Therefore, all the entries in the table sum to 1, rather than to 1238. So to be specific, the table is telling us that:

  • P(not ACL, not Lollapalooza) = 0.581.
  • P(not ACL, Lollapalooza) = 0.292
  • P(ACL, not Lollapalooza) = 0.065
  • P(ACL, Lollapalooza) = 0.062

And this last number, of course, is the answer to the question we set out to answer: P(ACL, Lollapalooza) = 0.062.

Conditional probabilities

Let turn to our final question: what is P(played ACL | played Lollapalooza)? Or said in English, what is the conditional probability that a band played ACL, given that they played Lollapalooza?

To answer this using the data, we’ll again tabulate the bands by both the acl and lollapalooza variables:

xtabs(~acl + lollapalooza, data=aclfest)
##    lollapalooza
## acl   0   1
##   0 719 361
##   1  81  77

From this table of counts, it’s fairly straightforward to reason as follows:

  • There were 361 + 77 = 438 bands that played at Lollapalooza.
  • Of those 438 bands, 77 played at ACL.
  • Therefore, P(played ACLplayed Lollapalooza) = 77/438.

Let’s use R as a calculator to express this as a decimal number:

## [1] 0.1757991

So about 0.176.

Using margin

But as with the previous two examples, it’s much more satisfying to get R to do the work for us. We’ll do this using prop.table again—but this time, with a twist. Pay close attention to the middle line, where we call prop.table:

xtabs(~acl + lollapalooza, data=aclfest) %>%
  prop.table(margin=2) %>%
##    lollapalooza
## acl     0     1
##   0 0.899 0.824
##   1 0.101 0.176

Notice how we added margin=2 inside parentheses in the prop.table() step. What margin=2 does is to tell prop.table that it should calculate proportions conditional on the second variable we named, which was lollapalooza. (Hence margin=2; if you wanted to condition on the first variable you named, which here is acl, you’d type margin=1 instead.)

So having specified that we want to condition on the lollapalooza variable, now we can read off the relevant conditional probabilities directly from the table—no calculator required. Let’s focus on the lollapalooza = 1 column, since this is what we want to condition on (i.e. that a band played Lollapalooza). The numbers in that column tell us that:

  • P(didn’t play ACLplayed Lollapalooza) = 0.824
  • P(played ACLplayed Lollapalooza) = 0.176

And that second number is precisely the answer we were going for.