solutions.qmd

# Solutions to exercises {.unnumbered}

## 1. Intro to R

> **Exercise**
>
> Create your own code block below and run a math operation.

```{r}
pi * 2
```

> **Exercise**
>
> Examine the help file of the `log()` function. How can we compute the the base-10 logarithm of my_object? Your code:

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
my_object <- 10
```

1)  Examine the `log()` function.

```{r}
#| eval: false
?log
```

2)  Compute the base-10 logarithm of `my_object`.

```{r}
log(my_object, base = 10)
```

```{r}
# alternative:
log10(my_object)
```

> **Exercise**
>
> Obtain the maximum value of water content per 100g in the data. Your code:

```{r}
#| code-fold: true
#| message: false
#| results: hide
# setup: these steps were executed before the exercise
my_character_vector <- c("Apple", "Orange", "Watermelon", "Banana")
my_data_frame <- data.frame(fruit = my_character_vector,
                            calories_per_100g = c(52, 47, 30, 89),
                            water_per_100g = c(85.6, 86.8, 91.4, 74.9))
my_data_frame
```

```{r}
max(my_data_frame$water_per_100g)
```

## 2. Tidy data analysis I {.unnumbered}

```{r}
#| code-fold: true
#| message: false
#| results: hide
# setup: these steps were executed before the exercises
library(tidyverse)
trump_scores <- read_csv("data/trump_scores_538.csv")
```

> **Exercise**
>
> Select the variables `last_name`, `party`, `num_votes`, and `agree` from the data frame. Your code:

```{r}
trump_scores |> 
  select(last_name, party, num_votes, agree)
```

```{r}
# alternative
trump_scores |> 
  select(last_name, party:agree)
```


> **Exercise**
>
> 1.  Add a new column to the data frame, called `diff_agree`, which subtracts `agree` and `agree_pred`. How would you create `abs_diff_agree`, defined as the absolute value of `diff_agree`? Your code:
>
> 2.  Filter the data frame to only get senators for which we have information on fewer than (or equal to) five votes. Your code:
>
> 3.  Filter the data frame to only get Democrats who agreed with Trump in at least 30% of votes. Your code:

1)  Add a new column to the data frame, called `diff_agree`, which subtracts `agree` and `agree_pred`. How would you create `abs_diff_agree`, defined as the absolute value of `diff_agree`? Your code:

```{r}
trump_scores |> 
  mutate(diff_agree = agree - agree_pred) |> 
  select(last_name, matches("agree")) # just for clarity
```

```{r}
trump_scores |> 
  mutate(abs_diff_agree = abs(agree - agree_pred)) |> 
  select(last_name, matches("agree")) # just for clarity
```

2)  Filter the data frame to only get senators for which we have information on fewer than (or equal to) five votes. Your code:

```{r}
trump_scores |> 
  filter(num_votes <= 5)
```

3)  Filter the data frame to only get Democrats who agreed with Trump in at least 30% of votes. Your code:

```{r}
trump_scores |> 
  filter(party == "D" & agree >= 0.3)
```

> **Exercise**
>
> Arrange the data by `diff_pred`, the difference between agreement and predicted agreement with Trump. (You should have code on how to create this variable from the last exercise). Your code:

```{r}
trump_scores |> 
  mutate(diff_agree = agree - agree_pred)  |> 
  arrange(diff_agree)
```

> **Exercise**
>
> Obtain the maximum absolute difference in agreement with Trump (the `abs_diff_agree` variable from before) for each party.

```{r}
trump_scores |> 
  mutate(abs_diff_agree = abs(agree - agree_pred)) |> 
  summarize(max_abs_diff = max(abs_diff_agree),
            .by = party)
```

> **Exercise**
>
> Draw a column plot with the agreement with Trump of Bernie Sanders and Ted Cruz. What happens if you use `last_name` as the `y` aesthetic mapping and `agree` in the `x` aesthetic mapping? Your code:

```{r}
#| code-fold: true
#| message: false
#| results: hide
# setup: this step was executed before the exercise
trump_scores_ss <- trump_scores |> 
  filter(num_votes >= 10)
```

```{r}
ggplot(trump_scores_ss |> filter(last_name %in% c("Cruz", "Sanders")),
       aes(y = last_name, x = agree)) +
  geom_col()
```

```{r}
# alternative
ggplot(trump_scores_ss |> filter(last_name == "Cruz" | last_name == "Sanders"),
       aes(y = last_name, x = agree)) +
  geom_col()
```

## 3. Matrices {.unnumbered}

> **Exercise**
>
> Get the product of the first three elements of vector $d$. Write the notation by hand and use R to obtain the number.
>
> $$\overrightarrow d = \begin{bmatrix}
> 12 & 7 & -2 & 3 & 1
> \end{bmatrix}$$

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
vector_d <- c(12, 7, -2, 3, -1)
```

$$\prod_{i=1}^3 d_i = 12 \cdot 7 \cdot (-2) = -168$$

```{r}
prod(vector_d[1:3])
```

> **Exercise**
>
> 1)  Calculate $A + B$ $$A= \begin{bmatrix}
>     1 & 0 \\
>     -2 & -1
>     \end{bmatrix}$$
>
> $$B = \begin{bmatrix}
> 5 & 1 \\
> 2 & -1
> \end{bmatrix}$$\
>
> 2)  Calculate $A - B$ $$A= \begin{bmatrix}
>     6 & -2 & 8 & 12 \\
>     4 & 42 & 8 & -6
>     \end{bmatrix}$$ $$B = \begin{bmatrix}
>     18 & 42 & 3 & 7 \\
>     0 & -42 & 15 & 4
>     \end{bmatrix}$$

```{r}
A1 <- matrix(c(1,-2,0,-1), nrow = 2)
B1 <- matrix(c(5,2,1,-1), nrow = 2)
A1 + B1
```

```{r}
A2 <- matrix(c(6,4,-2,42,8,8,12,-6), nrow = 2)
B2 <- matrix(c(18,0,42,-42,3,15,7,4), nrow = 2)
A2 - B2
```

> **Exercise**
>
> Calculate $2\times A$ and $-3 \times B$. Again, do one by hand and the other one using R. $$A= \begin{bmatrix}
> 1 & 4 & 8 \\
> 0 & -1 & 3
> \end{bmatrix}$$ $$ B = \begin{bmatrix}
> -15 & 1 & 5 \\
> 2 & -42 & 0 \\
> 7 & 1 & 6
> \end{bmatrix}$$

```{r}
A3 <- matrix(c(1,0,4,-1,8,3), nrow = 2)
2 * A3
```

```{r}
B3 <- matrix(c(-15,2,7,1,-42,1,5,0,6), nrow = 3)
-3 * B3
```

## 4. Tidy data analysis II {.unnumbered}

> **Exercise**
>
> 1.  Create a dummy variable, `d_large_pop`, for whether the country-year has a population of more than 1 million. Then compute its mean. Your code:
> 2.  Which countries are recorded as "Never colonized"? Change their values to other reasonable codings and compute a tabulation with `count()`. Your code:

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
library(tidyverse)
qog_csv <- read_csv("data/sample_qog_bas_ts_jan23.csv")
qog <- qog_csv
```

1.  Create the dummy variable `d_large_pop`.

```{r}
qog |> 
  mutate(d_large_pop = if_else(wdi_pop >= 1000000, 1, 0)) |> 
  count(d_large_pop) # to check if it went well
```

2.  Change the coding of "Never colonized" countries to something else, and compute a tabulation with `count()`.

```{r}
qog |> 
  filter(ht_colonial == "Never colonized") |> 
  count(cname)
```

```{r}
qog |> 
  mutate(ht_colonial_recoded = case_when(
    cname == "Canada" ~ "French/British",
    cname == "United States" ~ "British",
    .default = ht_colonial
  )) |> 
  count(ht_colonial_recoded)
```

> **Exercise**
>
> Calculate the median value of the corruption variable for each region (i.e., perform a grouped summary). Your code:

```{r}
qog |> 
  summarize(med_corr = median(vdem_corr, na.rm = T), .by = region)
```

> **Exercise**
>
> Convert back `gdp_long` to a wide format using `pivot_wider()`. Check out the help file using `?pivot_wider`. Your code:

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
library(readxl)
gdp <- read_excel("data/wdi_gdp_ppp.xlsx")
gdp_long <- gdp |> 
  pivot_longer(cols = -c(country_name, country_code), 
               names_to = "year", 
               values_to = "wdi_gdp_ppp", 
               names_transform = as.integer) 
```

```{r}
gdp_long |> 
  pivot_wider(id_cols = c(country_name, country_code), # can omit in this case too
              values_from = wdi_gdp_ppp, 
              names_from = year) 
```

> **Exercise**
>
> There is a dataset on country's CO2 emissions, again from the World Bank ([2023](https://data.worldbank.org/)), in "data/wdi_co2.csv". Load the dataset into R and add a new variable with its information, `wdi_co2`, to our `qog_plus` data frame. Finally, compute the average values of CO2 emissions *per capita*, by country. Tip: this exercise requires you to do many steps---plan ahead before you start coding! Your code:

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
library(tidyverse)
qog <- read_csv("data/sample_qog_bas_ts_jan23.csv")
gdp <- readxl::read_excel("data/wdi_gdp_ppp.xlsx")

gdp_long <- gdp |> 
  pivot_longer(cols = -c(country_name, country_code),
               names_to = "year",
               values_to = "wdi_gdp_ppp", 
               names_transform = as.integer)

qog_plus <- left_join(qog,
                      gdp_long,
                      by = c("ccodealp" = "country_code",
                             "year"))
```

1)  Load data (notice the .csv format):

```{r}
emissions <- read_csv("data/wdi_co2.csv")
```

2)  Pivot data to long, creating the "wdi_co2" variable:

```{r}
emissions_long <- emissions |> 
  pivot_longer(cols = -c(country_name, country_code),
               names_to = "year",
               values_to = "wdi_co2",  
               names_transform = as.integer)
```

3)  Merge-in information to our existing `qog_plus` data frame:

```{r}
qog_plus2 <- left_join(qog_plus,
                       emissions_long,
                       by = c("ccodealp" = "country_code",
                              "year"))
```

4)  Create column for emissions *per capita* (here we do per 1,000 people).

5)  Summarize information to get mean values at the country level (remember that `na.rm = T` is always a conscious decision):

```{r}
qog_plus2 |> 
  mutate(emissions_pc = 1000 * wdi_co2 / wdi_pop) |> 
  summarize(emissions_pc_country = mean(emissions_pc, na.rm = T),
            .by = cname)
```

> **Exercise**
>
> Draw a scatterplot with time in the x-axis and democracy scores in the y-axis. Your code:

```{r}
ggplot(qog_plus2) + aes(year, vdem_polyarchy) + geom_point()
```

> **Exercise**
>
> Using your merged dataset from the previous section, plot the trajectories of C02 per capita emissions for the US and Haiti. Use adequate scales.

```{r}
ggplot(qog_plus2 |> filter(cname %in% c("Haiti", "United States")), 
       aes(x = year, y = 1000 * wdi_co2 / wdi_pop)) +
  geom_line() +
  facet_wrap(~cname, scales = "free_y") +
  labs(x = "Year", y = "CO2 Emissions Per Capita",
       title = "CO2 Emissions Per Capita in Haiti and the United States",
       caption = "Source: World Development Indicators (World Bank, 2023) in QOG dataset.")
```

## 5. Functions {.unnumbered}

> **Exercise** When graphed, vertical lines cannot touch functions at more than one point. Why? Which of the following represent functions?
>
> ![Vertical line test: examples.](images/vertical_line_test.png){#fig-vertical-line-test}

A)  Function ✅

B)  Function ✅

C)  NOT a function 🚫

D)  Function ✅

E)  Function ✅

F)  NOT a function 🚫

G)  Function ✅

H)  NOT a function 🚫

> **Exercise**
>
> Create a function that calculates the area of a circle *from its diameter*. So `your_function(d = 6)` should yield the same result as the example above. Your code:

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
circ_area_r <- function(r){
    pi * r ^ 2
}
circ_area_r(r = 3)
```

```{r}
circ_area_d <- function(d){
    pi * (d/2) ^ 2
}
circ_area_d(d = 6)
```

> **Exercise**
>
> Graph the function $y = x^2 + 2x - 10$, i.e., a quadratic function with $a=1$, $b=2$, and $c=-10$. Next, try switching up these values and the `xlim =` argument. How do they each alter the function (and plot)?

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
library(ggplot2)
```

1)  Graph $y = x^2 + 2x - 10$.

```{r}
ggplot() +
  stat_function(fun = function(x){x^2 + 2*x - 10},
                xlim = c(-5, 5)) 
```

2)  Switch up the values and the `xlim =` argument.

```{r}
ggplot() +
  stat_function(fun = function(x){-3*x^2 - 6*x + 9},
                xlim = c(-10, 10)) 
```

> **Exercise**
>
> We'll briefly introduce [Desmos](https://www.desmos.com/calculator), an online graphing calculator. Use Desmos to graph the following function $y = 1x^3 + 1x^2 + 1x + 1$. What happens when you change the $a$, $b$, $c$, and $d$ parameters?

(we'll show how to do this in R here, but you could use Desmos)

1)  Graph $y = 1x^3 + 1x^2 + 1x + 1$.

```{r}
ggplot() +
  stat_function(fun = function(x){x^3 + x^2 + x + 1},
                xlim = c(-10, 10)) 
```

2)  Switch up the values.

```{r}
ggplot() +
  stat_function(fun = function(x){-2*x^3 + 4*x^2 + 8*x + 16},
                xlim = c(-10, 10)) 
```

> **Exercise**
>
> Solve the problems below, simplifying as much as you can. $$log_{10}(1000)$$ $$log_2(\dfrac{8}{32})$$ $$10^{log_{10}(300)}$$ $$ln(1)$$ $$ln(e^2)$$ $$ln(5e)$$

```{r}
log10(1000)
```

```{r}
log2(8/32)
```

```{r}
10^(log10(300))
```

```{r}
log(1)
```

```{r}
log(exp(2))
```

```{r}
log(5*exp(1))
```

> **Exercise**
>
> Compute `g(f(5))` using the definitions above. First do it manually, and then check your answer with R.

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
f <- function(x){x ^ 2}
g <- function(x){x - 3}
```

$$f(5) = 5^2 = 25$$ $$g(25) = 25 - 3 = 22$$

```{r}
g(f(5)) # no pipeline approach
f(5) |> g() # pipeline approach
```

## 6. Calculus {.unnumbered}

> **Exercise**
>
> 1)  Use the slope formula to calculate the rate of change between 5 and 6.
> 2)  Use the slope formula to calculate the rate of change between 5 and 5.5.
> 3)  Use the slope formula to calculate the rate of change between 5 and 5.1.

```{r}
(6^2 - 5^2) / (6 - 5)
```

```{r}
(5.5^2 - 5^2) / (5.5 - 5)
```

```{r}
(5.1^2 - 5^2) / (5.1 - 5)
```

> **Exercise**
>
> Use the differentiation rules we have covered so far to calculate the derivatives of $y$ with respect to $x$ of the following functions:
>
> 1.  $y = 2x^2 + 10$
> 2.  $y = 5x^4 - \frac{2}{3}x^3$
> 3.  $y = 9 \sqrt x$
> 4.  $y = \frac{4}{x^2}$
> 5.  $y = ax^3 + b$, where $a$ and $b$ are constants.
> 6.  $y = \frac{2w}{5}$

1) $4x$ (sum rule, constant rule, coefficient rule, power rule)

2) $20x^3-2x^2$ (sum rule, coefficient rule, power rule)

3) $\frac{-9}{2\sqrt x}$ (power rule)

4) $-\frac{8}{x^3}$ (coefficient rule, power rule)

5) $3ax^2$ (sum rule, constant rule, coefficient rule, power rule)

6) $0$ (constant rule)

> **Exercise**
>
> Compute the following:
>
> 1.  $\frac{d}{dx}(10e^x)$
> 2.  $\frac{d}{dx}(ln(x) - \frac{e^2}{3})$

1) $10e^x$ (coefficient rule, exponent rule)

2) $\frac{1}{x}$ (difference rule, constant rule, logarithm rule)

> **Exercise**
>
> Use the differentiation rules we have covered so far to calculate the derivatives of $y$ with respect to $x$ of the following functions:
>
> 1.  $x^3 \cdot x$
> 2.  $e^x \cdot x^2$
> 3.  $(3x^4-8)^2$

1) $4x^3$ (power rule)

2) $e^x x^2 + 2xe^x$ (product rule, exponent rule, power rule)

3) $24x^3(3x^4-8)$ (chain rule, difference rule, constant rule, power rule)

> **Exercise**
>
> Take the partial derivative with respect to $x$ and with respect to $z$ of the following functions. What would the notation for each look like?
>
> 1.  $y = 3xz - x$
> 2.  $x^3+z^3+x^4z^4$
> 3.  $e^{xz}$

1) 

$\frac{\delta}{\delta x}(3xz - x) = 3z - 1$ (difference rule, coefficient rule, power rule)

$\frac{\delta}{\delta z}(3xz - x) = 3x$ (difference rule, constant rule, coefficient rule)

2) 

$\frac{\delta}{\delta x}(x^3+z^3+x^4z^4) = 4x^3z^4+3x^2$ (add rule, coefficient rule, power rule)

$\frac{\delta}{\delta z}(x^3+z^3+x^4z^4) = 4x^4z^3+3z^2$ (add rule, coefficient rule, power rule)

3)

$\frac{\delta}{\delta x}(e^{xz}) = e^{xz}z$ (chain rule, exponent rule, coefficient rule)

$\frac{\delta}{\delta z}(e^{xz}) = e^{xz}x$ (chain rule, exponent rule, coefficient rule)

> **Exercise**
>
> Identify the global extrema of the function $\displaystyle \frac{x^3}{3} - \frac{3}{2}x^2 -10x$ in the interval $[-6, 6]$.

1) Take the first derivative

$(\frac{x^3}{3} - \frac{3}{2}x^2 -10x)' = x^2 - 3x - 10$ (sum rule, coefficient rule, power rule)

2) Set the derivative equal to zero and obtain its roots (F.O.C)

$ x^2 - 3x - 10 = (x-5)(x+2)$

$(x-5)(x+2) = 0$

$x^*_1 = 5,\, x^*_2=-2$

3) Calculate the second derivative and substitute the roots (S.O.C.)

$(x^2 - 3x - 10)' = 2x -3$

(i) $2x^*_1-3 = 2\cdot5-3 = 7$ (since it is positive, this is a minimum)

(ii) $2x^*_2-3 = 2\cdot(-2)-3 = -7$ (since it is negative, this is a maximum)

4) Adjudicate between these critical points or the bounds.

Minimum critical point: $f(5) = \frac{(5)^3}{3} - \frac{3}{2}(5)^2 -10 \cdot (5) = −45.8\overline{3}$.

Maximum critical point: $f(-2) = \frac{(-2)^3}{3} - \frac{3}{2}(-2)^2 -10 \cdot (-2) = 11.\overline{3}$.

Lower bound: $f(-6) = \frac{(-6)^3}{3} - \frac{3}{2}(-6)^2 -10 \cdot (-6) = −66$

Upper bound: $f(6) = \frac{(-6)^3}{3} - \frac{3}{2}(-6)^2 -10 \cdot (-6) = −42$

So we conclude that, for the $[-6, 6]$ interval, the global minimum is at the lower bound ($x=-6$) and the global maximum is at the critical point at $x=-2$.

> **Exercise**
>
> Solve the following indefinite integrals:
>
> 1.  $\int x^2 \, dx$
> 2.  $\int 3x^2\, dx$
> 3.  $\int x\, dx$
> 4.  $\int (3x^2 + 2x - 7\,)dx$
> 5.  $\int \dfrac{2}{x}\,dx$

1. $\frac{x^3}{3} + C$ (power rule)

2. $x^3 + C$ (coefficient rule, power rule)

3. $\frac{x^2}{2} + C$ (power rule)

4. $x^3 + x^2 - 7x + C$ (sum/difference rule, coefficient rule, power rule)

5. $2 ln(x) + C$ (coefficient rule, reciprocal rule)

> And solve the following definite integrals:
>
> 1.  $\displaystyle\int_{1}^{7} x^2 \, dx$
> 2.  $\displaystyle\int_{1}^{10} 3x^2 \, dx$
> 3.  $\displaystyle\int_7^7 x\, dx$
> 4.  $\displaystyle\int_{1}^{5} 3x^2 + 2x - 7\,dx$
> 5.  $\int_{1}^{e} \dfrac{2}{x}\,dx$

In the following, FTC stands for the Fundamental Theorem of Calculus

1. $114$ (substitute from previous answer, FTC)
2. $999$ (substitute from previous answer, FTC)
3. $0$ (there is no area between 7 and 7)
4. $120$ (substitute from previous answer, FTC)
5. $2$ (substitute from previous answer, FTC)

## 7. Probability, statistics, and simulations {.unnumbered}

> **Exercise**
>
> Compute the probability of seeing between 1 and 10 voters of the candidate in a sample of 100 people.

```{r}
pbinom(q = 10, size = 100, prob = 0.02) - 
  dbinom(x = 0, size = 100, prob = 0.02)
```

> **Exercise**
>
> Evaluate the CDF of $Y \sim U(-2, 2)$ at point $y = 1$. Use the formula and `punif()`.

$$A = F(1) = P(Y\leq 1) = 3 \cdot(1/4) = 0.75$$

```{r}
punif(q = 1, min = -2, max = 2)
```

> **Exercise**
>
> What is the probability of obtaining a value above 1.96 or below -1.96 in a standard normal probability distribution? Hint: use the `pnorm()` function.

```{r}
pnorm(-1.96) + (1 - pnorm(1.96))
```

> **Exercise**
>
> Compute and plot `my_rnorm`, a vector with one million draws from a Normal distribution $Z$ with mean equal to zero and standard deviation equal to one ($Z\sim N(0,1)$). You can recycle code from what we did for the uniform distribution!

```{r}
set.seed(1) # set a seed
my_rnorm <- rnorm(n = 1000000)

ggplot(data.frame(my_rnorm), aes(x = my_rnorm)) +
  geom_histogram(binwidth = 0.25, boundary = 0, closed = "right") +
  scale_x_continuous(breaks = seq(-5, 5, 1), limits = c(-5, 5))
```

## 8. Text analysis {.unnumbered}

> **Exercise**
>
> What score (out of 10) would you give Barbie or Oppenheimer? Write your score in one sentence (e.g., I would give Barbie seven of ten stars.) If you have not seen either, write a sentence about which you would like to see more.
>
> Store that text as a string (`string3`) and combine it with our existing `cat_string` to produce a new concatenated string called `cat_string2`. Finally, count the total number of characters within `cat_string2`. Your code:

```{r}
#| code-fold: true
#| message: false
# setup: these steps were executed before the exercise
library(stringr)
my_string <- "I know people who have seen the Barbie movie 2, 3, even 4 times!"
my_string2 <- "I wonder if they have seen Oppenheimer, too."
cat_string <- str_c(my_string, my_string2, sep = " ")
```

```{r}
string3 <- "I would give Barbie 7 out of 10 stars."
string3
cat_string2 <- str_c(cat_string, string3, sep = " ")
cat_string2
str_length(cat_string2)
```

> **Exercise**
>
> Look up the lyrics to your favorite song at the moment (no guilty pleasures here!). Then, follow the process described above to count the words: store the text as a string, convert to a tibble, tokenize, and count.
>
> When you are done counting, create a visualization for the chorus using the `ggplot` code above. Your code:

1.  Store the text as a string.

```{r}
library(tidytext)
dummy <- c("I been goin' dummy (Huh)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Yeah)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy",
           "Dumbass, I been goin' dummy")
```

2.  Convert to a tibble.

```{r}
dummy_df <- tibble(line = 1:9, text = dummy)
dummy_df
```

3.  Tokenize.

```{r}
dummy_tok <- unnest_tokens(dummy_df, word, text)
```

4.  Count.

```{r}
dummy_tok |>
  count(word, sort = TRUE)
```

5.  Visualize.

```{r}
dummy_tok |>
  count(word, sort = TRUE) |>
  mutate(word = reorder(word, n)) |>
  ggplot(aes(n, word)) +
  geom_col()
```