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Complete the exercises from today’s two lessons:

Work through as many of the following challenge exercises as you can in the remaining time. Take a few minutes to ponder each challenge on your own, but feel free to engage with fellow participants and SESYNC instructors for guidance or ideas.

Fix each of the following common data frame subsetting errors. The `plots`

variable
can be read in with `read.csv('data/plots.csv')`

.

- Fix
`plots[plots$id = 4, ]`

so it returns the rows with column`id`

equal to 4. - Fix
`plots[-1:4, ]`

so it returns all rows but the first four. - Fix
`plots[plots$id <= 5]`

so it returns the rows with column`id`

less than or equal to 5. - Fix
`plots[plots$id == 4 | 6, ]`

so it returns the rows with column`id`

equal to 4 or 6.

Were you to ever interview for a “data scientist” postition, you may be asked to
complete this common challenge (possibly using multiple approaches). Write a
“for loop” in the R language that calculates the Fibonnacci sequence up through
it’s 12th entry. Start your script with the line `fib <- c(1, 1)`

, and recall
that `c`

combines vectors.

Create a data frame from scratch that will have three columns and 40 rows as
follows. In a column named “id”, put a repeating sequence from 1 to 10. For a
column named “group”, create a factor from an alternating seqence of 10 `a`

s
followed by 10 `b`

s (hint: `?seq`

). In a colume named “var” put 40 random
samples from a uniform distribution (hint: `?runif`

) in order from smallest to
largest (hint: `?sort`

). Plot the data with a smooth line showing the
relationship between the increasing `id`

and `var`

with different colors for
each group.

Try that Fibonnacci challenge again, but use an approach called “recursion”. That is, write a function that calls itself (i.e. a recursion) to calculate any Fibonacci number. Just because this is an “advanced” approach doesn’t mean it’s a good one: don’t try it for any number over your age!

`plots[plots$id == 4, ]`

`plots[-(1:4), ]`

or`plots[-1:-4, ]`

`plots[plots$id <= 5, ]`

`plots[plots$id == 4 | plots$id == 6, ]`

```
fib <- c(1, 1)
for (i in 3:12) {
fib[i] <- fib[i - 1] + fib[i - 2]
}
```

```
df <- data.frame(
id = 1:10,
group = rep(c('a', 'b'), each = 10),
prob = sort(runif(40)))
ggplot(df, aes(x = id, y = prob, color = group)) +
geom_point() +
geom_smooth(method = 'lm')
```

```
fibn <- function(n) {
if (n <= 2) {
fib <- 1
} else if (n > 2) {
fib <- fibn(n - 1) + fibn(n - 2)
}
return(fib)
}
fibn(12)
```

Note that this recursive solution is very inefficient, but could be greatly improved with memoization.