jrnlod, your answer to this question is particularly great, thank you!!

I added

if (x == 0) { print(x) }

as I don't think 0 should be a fizz/buzz Is there a less manual way to do this?

Exercise 14.5.2 question 2 We have not learned 'unlist' yet, we learn that only in Ch 21

I got a list and ordered it using the following code, but it takes only the first word in each sentence (I guess), and str_extract_all doesn't work at all. Any other ideas?

tibble(word = (str_extract(sentences, boundary("word")))) %>% mutate(word = str_to_lower(word)) %>% count(word, sort = TRUE) %>% head(5)

A tibble: 5 x 2

word n

<chr> <int>

1 the 262

2 a 72

3 he 24

4 we 13

5 it 12

Exercise 12.4.3.2. We learn str_split only in the next section. My rather inelegant solution was: has_apo <- sentences %>% str_subset("\'") has_ap_sep <- has_apo %>% tibble(sentence = has_apo) %>% extract(sentence, c("before", "apo", "after"), "([A-Za-z]*)(\')([A-Za-z]+)", remove = FALSE)

cf https://en.wiktionary.org/wiki/Category:English_contractions There are some contractions where the apostrophe appears as the first character, so use * not + in the first [A-Za-z] expression

Instead of using year, month, day, hour, you can join on only 'origin' and 'time_hour'

The result shows that this is in fact not a primary key, as n > 1 is very high. Why?

One of the variables is named 'n', so first rename that variable, then check:

babynames %>% rename(no = n) %>% count(year, sex, name, prop) %>% filter(n > 1) %>% nrow()

[1] 0

Your solution works, but we have not been taught the mutate command yet. Given the commands we have already been taught, you can get the same results from: arrange(flights, desc(distance / air_time))