duplicated example I think

delete?

delete?

are all the bracketed parts supposed to be links?

slice_max is not found

Should read:

The fastest flight is the one with the fastest average ground speed,

The sentence as currently worded describes the flight with the average ground speed, not the fastest.

Is there a reason not to use

idxs <- which(sapply(df, is.numeric)) ?

NA and NaN. Note that 0.286 is equal to 2 out of 7.

"an" ?

numeric values not limited to integers?

summary(is.na(flights)) will give NAs of all variables

should be >= (at least meaning 30 or more)

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))

This worked for me arrange(flights, air_time/distance)

I find this hard to follow. Why not phrase it as the next example? NA | TRUE is TRUE because anything or TRUE is always TRUE

If we were to calculate the proportion of flights not delayed or cancelled, we would need to adjust for NA values of the cancelled flights:

summarise(n = n(), on_time = sum(on_time, na.rm=TRUE) / n)

Exercise 10.5.2

Exercise 10.5.1

only

if we need to view/extract later the words that have at least 7 characters we can:

str_view( stringr::words, "........*", match = T)

adding "*" after the eight "." states that the last "." can be repeated zero or more times.

or

str_view(stringr::words, ".{7,}", match = T)

in both cases we match the words that have at least 7 characters, not only up to the seventh letter

An alternative: str_view(words, "[aeiou]", match = FALSE)

Or, using str_subset: str_subset(words, "[aeiou]", negate = TRUE)

If one adds "%>% distinct(dest)" to the expression, one sees that there are four airports that are not in the FAA list: BQN, SJU, STT, and PSE. Three of these are in Puerto Rico, one (STT) is in US Virgin Islands.

Perhaps the association with visibility is even stronger.

Could not generate the same stat_summary( ) plot with that code, did a little research and stack overflow suggested two solutions: use geom_line( )

ggplot(data = diamonds, mapping = aes(x = cut, y = depth)) + geom_line() + stat_summary(fun.y = "median", geom = "point", size = 3)

or, reduce the amount of data by grouping it

data = diamonds %>% group_by(cut) %>% summarise(min = min(depth), max = max(depth), median = median(depth))

ggplot(data, aes(x = cut, y = median, ymin = min, ymax = max)) + geom_linerange() + geom_pointrange()

Source: https://stackoverflow.com/questions/41850568/r-ggplot2-pointrange-example

I think you flipped these labels. width is horizontal and height is vertical

It would also be helpful to emphasize that unless height and/or weight are explicitly defined as zero, there will be jitter. When I first used the geom_jitter, I did not realize this.

is a combination

contained in

the sentence is not complete.

flew

It seems to refer to diverted flights. The original BTS data also has diverted airport information, including a variable DivArrDelay with the following description: "Difference in minutes between scheduled and actual arrival time for a diverted flight reaching scheduled destination. The ArrDelay column remains NULL for all diverted flights." I browsed this data quickly and, indeed, those missing arr_delay observations have values in the DivArrDelay variable.

is a column.

%% 1440 is brilliant

Did you mean 6.5?

What is viridis library?

n_distinct is a faster and more concise equivalent of length(unique(x)).

pretty sure december as 31 days

manufacturer missing (it is a categorical variable as well)

How cylinder is a continuous variable?

Is anyone getting the above plot? even when I copy paste the code the boxplots are horizontal instead of vertical. tidyverse has the same plot as I do https://ggplot2.tidyverse.org/reference/geom_boxplot.html

If I understand the aes correctly, group should make carat the "categorical" variable and plot vertically as seen here.

If I replace y = price with y = depth it plots vertically as expected.

Can anyone help me understand what's happening?

Retired Selection Helpers

one_of() is retired in favour of the more precise any_of() and all_of() selectors.

https://www.rdocumentation.org/packages/tidyselect/versions/1.0.0/topics/one_of

Change "will the difference" to "is the difference" or "will be the difference"

Shouldn't we update this functions to the newer pivor_longer and pivot_wider? (check ?gather, and also https://r4ds.had.co.nz/tidy-data.html#exercises-24 )

Thank you for writing/maintaining r4ds! Great resource.

I delivered this example in a lecture introducing tidy data, and the solution to have three variables doesn't make sense to me. Why is this not only two variables (sex and pregnant)? Sure, I could argue that count is a variable on this data, but it doesn't seem as natural.

This exercise makes more sense AFTER introducing Position adjustments in part 3.8.

from the documentation: If FALSE, the default, missing values are removed with a warning. If TRUE, missing values are silently removed.

Using geom_count() with position ='jitter' may help a bit with overplotting:

ggplot(data = mpg) +

• geom_count(mapping = aes(x = cty, y = hwy, color = class), position = "jitter")

The ggplot2 v3.2.1 R documentation shows geom_violin() paired with stat_ydensity()

The ggplot2 v3.2.1 R documentation states "geom_bar() uses stat_count() by default". Was ggplot2 updated since this answer was published? My understanding is stat_count() is used for discrete x data and stat_bin() for continuous x data.

missing the variable manufacturer, which is also a categorical (<chr>) variable.

Can we the whole code replace with: ifelse(length(x)==1, x, str_c(str_c(x[-length(x)], collapse = ", ")," and ",str_c(x[length(x)], collapse = ",")))

of course in the form of a function

Repeated word

Why we do that? Just to see if each digit appears or what? Thanks

Thank you for the book! Could you provide a hint as to how you might go about with this if 48 hours here meant any contiguous 48 hours? My guess is using a windowing function, but I couldn't figure it out.

Either it is "the" or "an", but not both.

"with" written 2 times.

this should be "stat_bin()"

I think this should be "no points" (i. e. no points on the scatter plot).

Typo.

Sentence syntax unclear. Maybe this:

If a warning is provided often, but not always, there may be a bug in the code.

vectors, R recycles

min_rank是对数据大小进行编号排序，遇到重复值，排序相同，但每个值都占一个位置，缺失值不计入

affect

already use

Missing "of" between "benefits" and "encoding"

I think common notation is dependent variable (Y) versus independent variable (X), so if you're asking for a y = cyl and x = hwy plot, then I'd say that's cyl vs. hwy (cyl as a function of hwy), not vice versa as written.

Question Why would it be unusual that flights with air time on the left side of the distribution would be the fastest?

Is there not something missing here?

Replace by "not ALL tail"?

Should be "of both the origin and..." :)

Exercise 14.4.2.2 Q 2 - this misses words that are at the end of the sentence with a period after them. I use this: str_extract(sentences, "[A-za-z]+ing[ .]")

The way this is written could be somewhat confusing to a reader, in my opinion, although the code makes the order of the functions clearer..

Suggestion:

If we wanted a numeric vector, we could combine the map() followed with the flatten_dbl(),

Edit suggestion.

like shown:

Hi. is "Highway MPG Relative to Class Average" the name of Y-axis? and the name of X-axis is "Engine Displacement" because we put in X-axis with displ. thanks:)

Minor mistake: a not at

"Not" repeated. Delete?

Edit suggestion,

Neither NaN nor Inf is a number

Or

Both NaN and Inf are not even numbers.

If accepted, then "and so they aren’t even numbers" may be deleted as it becomes superfluous.

Could you please check the sentence? There seems to be some ambiguity.

Looks like a New York issue.

> filter(flights, !is.na(dep_delay), is.na(arr_delay)) %>%
+   count(origin)
# A tibble: 3 x 2
  origin     n
  <chr>  <int>
1 EWR      469
2 JFK      337
3 LGA      369

redundant. delete?

Hi,

Please can you explain to me how to read this line of code and what it means. I'm having difficulty understanding it. Individually I do understand what each symbol means but put together as it is, I'm unable to. Moreover, it looks more efficient than my effort, as shown below.

Thank you.

fizzbuzz <- function(n) {

x <- n

if(( x %% 3 == 0 ) && ( x %% 5 == 0 )) {

print("fizzbuzz")

} else {

if(x %% 3 == 0) {

print("fizz")

} else {

if(x %% 5 == 0) {

print("buzz")

} else { print(x)

} } }

}

Hi Jeffrey. First of all, i really appreciate this Exercise solutions! and.. i need your help! why did you code ".data"? Is there any difference in the case of "data"? when i coded "data", The results were the same as ".data". i had tried to find a difference between them, i couldn`t find it....please let me know. does it have no difference?..

Change word order perhaps? that is

which has

Minor mistake. Should read if it does not

This seems to be a tad long so I apologise in advance. This is a whole new field for me and I would really like to understand.

Could it be AA and MQ use different values to represent tailnum?

filter( planes, tailnum == 0 )

A tibble: 0 x 9

length(is.na( planes$tailnum ))

[1] 3322

nrow(planes)

[1] 3322

filter( flights, tailnum == 0 )

A tibble: 0 x 19

length(is.na( flights$tailnum ) )

[1] 336776

nrow( flights )

[1] 336776

Yet , the anti_join () as shown in your code shows clearly that there are some talinum values in flights that are not represented in the planes datasets. How could that be? The one explanation I could come up with is that the two datasets used different talinum values, so I tried to investigate for AA and MQ.

tailnum_flights <- flights %>% filter( carrier == 'AA'| carrier == 'MQ' ) %>% select ( carrier, tailnum )

tailnum_planes <- planes %>% select( tailnum )

tailnum_planes %in% tailnum_flights

[1] FALSE

So, it looks like the tailnum values are not missing for the ten airlines but are represented with values different in the two datasets (flights and planes).

What are your thoughts? Thank you.

a is missing

Erm, full stop omitted after no.

And thanks, for the link. It is an interesting read.

This worked for me as well:

str_view(stringr::words, ".*[^e]ed$", match = TRUE)

Please, could explain why you included this in the code? I replicated the answer without it. Thanks.

Does this work? E.g. "achieve" does not have a matching PAIR of letters.

Please, could you explain to me why if I pipe this directly to ggplot the colour aesthetics is not applied. See code below, it's basically a replication of yours but with ggplot directly piped to avg_dest_delays:

avg_dest_delays <- flights %>% group_by( dest ) %>% summarise( delay = mean( arr_delay, na.rm = TRUE )) %>% inner_join(airports, by = c( dest = "faa" ) ) %>% ggplot( aes(lon, lat, colour = delay ) ) + borders("state" ) + geom_point( ) + coord_quickmap( )

Thanks

repetition?

This also seems to be the case where the by = argument is not used in a code. In that case, it seems, the semi_join() will give outputs only where the rows for both datasets correctly match, for example:

fueleconomy::vehicles %>% semi_join(fueleconomy::common)

produces the same output as:

fueleconomy::vehicles %>% semi_join(fueleconomy::common, by = c("make", "model"))

Or is that a coincident?

But, fueleconomy::vehicles %>% semi_join(fueleconomy::common, by = "make"

will produce a different output as R will match only by "make" in this example.

The code doesn't affect the output but I thought you might mean, sum( !is.na( dep_delay ))

The code in the solution book totally dropped the data for planes age > 25. How might we combine them into a single row? I didn’t think it could be done without first defining a second tibble that contains all the planes age >25, then merge it with the first tibble that contains the data for planes age <= 25, before carrying out the summarise actions on them after applying the group_by = age argument. older_plane_cohorts <- inner_join( flights, select( planes, tailnum, plane_year = year ), by = "tailnum" ) %>% mutate(age = year - plane_year) %>% filter(!is.na(age)) %>% mutate(age = pmin(46, age) - pmin( 25, age ) ) %>% filter( age != 0 )

Then I got stuck. I can’t figure out how to proceed after that. And frankly, I can say with any certainty if my argument is sensible. And I’m not even sure it’s possible to combine the 17 rows into a single row. Your help will be appreciated.

Thank you in advance and for your help so far. Truly appreciated.

I think you used this line of code to limit the selection to planes not older than 25 years. But in the text above, you stated, "There are few planes older than 30 years, so we combine them into a single category." So, I was expecting the selection to be age <= 30, or using your notation pmin(30, age) and not pmin(25, age). Perhaps an edit of the text may be required unless I'm wrong in my supposition?

Edit suggestion: However, this default...

Hi, I hope this doesn't come across as nitpicking:

If we needed a unique identifier for our analysis, we could add a surrogate key, perhaps?

Shouldn't this be "airports$faa" since airports is a data frame and faa is a variable?

"It looks like it is possible for certain variables to missing for (country, years)." Edit suggestion: It looks like it is possible that certain variables are missing for (country, years).

delete?

run:

Using $,

Please, can you explain to me what I am getting wrong in the below code and especially the error message, which I have tried goggling to no success.

I tried to see if instead of count to show, in my practice exercise, the average price per carat group using the code below:

group_by( diamonds, carat ) %>% summarise( avg_price = mean(price ) ) %>% ggplot( ) + mapping = aes( color = cut_width(carat, 5 ), x = avg_price ) + geom_freqpoly( )

The code returns the following error:

Error in group_by(diamonds, carat) %>% summarise(avg_price = mean(price)) %>% : could not find function "+<-"

I've googled the error without success. So my confusion is in two parts:

1. What is the right code to show the average price per carat type
2. What does the above error mean?

Thanks in advance

Does anyone else notice that the R4DS uses British spelling for visualisation, but the Solution textbook uses the American spelling. It's a bit weird when one switches directly from the text book to the solution. I'm sure not many people notice the difference. I probably do because I generally write Brit English. By the way, I am not censuring, just expressing my thought. I am really grateful to the author for making this available.

Omission of are? Perhaps, you meant there are?

Typo. Either the or these, preferably these in my opinion.

Can anyone please explain the n in this code? Thanks in advance.

cos( )

sin( )

why not just use: arrange(flights, desc(distance / air_time))

Class

Recommendation: For new users of R, how one determines the representation of midnight as 2400 should be explained. At this point in the book, a reader would not have been thought how to do the proper exploration to make this determination alone, so it would seem that explaining how the author of the solutions book acquired that knowledge would be most beneficial.

I think you mixed up these two.

(date + 1L) %in% holidays_2013$date ~ "day before holiday",
(date - 1L) %in% holidays_2013$date ~ "day after holiday",

should be $$\beta \log x$$

Based on the output, it seems to me that appending outperform the pre-allocation. I run the code myself, got same results. However, another R notebook gave opposite results.

Omit.

Many thanks for this lovely example. I hadn't understood geom_bar's NA behavior with factor variables until your explanation.

Is this "buyer's psychology?" The opposite of $4.99 being cheaper than $5.00?

Nice! Your intuition seems correct. But how do you account for the large number of 1.01 carat diamonds?

print(n = Inf) will print all rows of a tibble.

see

More precisely: there's a $90 gap: the closed interval [1455,1545].

I wonder if this includes identifying outliers for possible data errors. E.g. x = 0 for 8 diamonds and z = 0 for 20. Also z = 31.8 for one. y = 0 for 7 diamonds and y = 31.8 and y = 58.9 for one each. I believe all are data errors.

I believe the plots can be generated without id:

diamonds %>%
  select(x, y, z) %>%
  gather(variable, value) %>%
  ggplot(aes(x = value)) +
  geom_density() +
  geom_rug() +
  facet_grid(vars(variable))

Is it a deliberate or an error putting "will" and "is" together?

I think this should read:

bottles <- function(i) {
  if (i > 1) {
    bottles <- str_c(i , " bottles")
  } else if (i == 1) {
    bottles <- "1 bottle"
  } else {
    bottles <- "no more bottles"
  }
  bottles
}

Otherwise you get no bottles of beer twice in the final output

typo

repetition?

I think you forget to change mean(X[i, ]) to mean(X[, i]) in calculating column means

I think it should be corrected this way: for (fname in ...) { df[[fname]] <- bind_rows()

I think this is a lovely use of logicals and cumsum. But I believe that it omits planes whose first flight is delayed by more than an hour. There are 234 of these:

(zero  <- flights %>%
   filter(!is.na(dep_delay)) %>%
   arrange(tailnum,month,day) %>%
   group_by(tailnum) %>%
   mutate(delay_gt1hr = dep_delay > 60) %>%
   filter(row_number() == 1,delay_gt1hr) %>%
   select(tailnum) %>%
   mutate(n = 0)
)

Then bind_rows could concatenate these to the tibble produced by the posted solution.

This is a "two-part" solution. Is there a "one-part" solution?

Since the year column contains only 2013, you could arrange chronologically without this argument.

Or, the two parts could be combined:

rank <- flights %>%
   group_by(dest) %>%
   mutate(n_carriers = n_distinct(carrier)) %>%
   filter(n_carriers > 1) %>%
   group_by(carrier) %>%
   summarize(n_dest = n_distinct(dest)) %>%
   arrange(desc(n_dest))

Here width = Inf will print all columns.

I believe that print(width = Inf) will reliably print all selected columns.

below the

observations of each

colour

change from "color" to "colour" to be consistent within the paragraph

drv

drv

drv

'five' is an error.

Why are not these two plots displayed properly? Only empty background is shown.

Diamond Dimensions

1. If you look at this picture, it seems that depth is always smaller than either width or length. Maybe it is easier to plug it into a ring or other jewelry...
2. Also you said "1. length is less than width, otherwise the width would be called the length". actually, it is the other way round.

I do not understand, why geom_bar is used in addition to geom_histogram. It greates two layers which are essentially the same, right?

You might want to check this last row of the tibble that's not displayed. Saturday actually has the shortest delays of any day of the week. You'll see it right away if you plot it:

flights_dt %>% mutate(wday = wday(dep_time, label = TRUE)) %>% group_by(wday) %>% summarize(ave_dep_delay = mean(dep_delay, na.rm = TRUE)) %>% ggplot(aes(x = wday, y = ave_dep_delay)) + geom_bar(stat = "identity")

flights_dt %>% mutate(wday = wday(dep_time, label = TRUE)) %>% group_by(wday) %>% summarize(ave_arr_delay = mean(arr_delay, na.rm = TRUE)) %>% ggplot(aes(x = wday, y = ave_arr_delay)) + geom_bar(stat = "identity")

should not be %/% ?

This function always returns 0. This is because sum(x-m) (performed before raised to the power 3) will always be 0. Resolved with additional parenthesis:

skewness <- function(x, na.rm = FALSE) { n <- length(x) m <- mean(x, na.rm = na.rm) v <- var(x, na.rm = na.rm) (sum((x - m)^3) / (n - 2)) / v^(3 / 2) }

I appreciate there are several possible formulas for skewness which may not match this one.

Omit.

If you wish, this could be eliminated since the only columns that survive are those in the summarise.

flights at departure

omit

?print.tbl instead of ?print.tbl_df (archived)

or

mtcars %>% map_dbl(mean)

You would do You would use

Do we need a descending order to put the fastest flights first? arrange(flights, desc(distance / air_time 60)) See also: flights %>% mutate(speed = distance / air_time 60) %>% select(tailnum, distance, air_time, speed, dep_time) %>% arrange(desc(speed))

This part is missing

beneficital

omit

I believe you don't need ungroup().

"and standard deviation. The following"

Typo. I suspect you mean cospi(), because "sinpi(x)" had just been introduced

I believe the problem asks for "fast flights per destination". If so, this code produces that list, including ties. Sadly, it uses slice.

fast <- flights %>%
  filter(!is.na(air_time)) %>%
  group_by(dest) %>%
  mutate(z = (air_time - mean(air_time)) / sd(air_time)) %>%
  slice(which(z == min(z))) %>%
select(origin,dest,month,day,carrier,flight,air_time,z) %>%
  arrange(z)

This is a nice plot, but the image below is not it. It appears to have been cut-and-pasted from a previous plot.

Somewhat more succinctly:

standardized_flights <- flights %>%
  filter(!is.na(air_time)) %>%
  group_by(dest,origin) %>%
  mutate(air_time_standard = (air_time - mean(air_time)) / sd(air_time)) %>%
select(origin,dest,month,day,carrier,flight,air_time,air_time_standard)

geom_histogram

preprocesses input...

"to" is missing

mode3

This should be filter(n >= 100): "at least 100 flights" implies that 100 should be included as the least possible value of number of flights.

This part does not make sense . Needs to be reviewed.

Omit, since you've already filtered.

English is inherently ambiguous while only math (vectors, subscripts) and algorithms are not. But then you'd lose your reader.

Here's a slight rephrasing where rank refers to a component of, say, min_rank(x) and value a component of x.

For each set of tied values the min_rank() function assigns a rank equal to the number of values less than that tied value plus one. In contrast, the dense_rank() function assigns a rank equal to the number of distinct values less than that tied value plus one.

English is inherently ambiguous while only math (vectors, subscripts) and algorithms are not. But then you'd lose your reader.

Here's a slight rephrasing where rank refers to a component of, say, min_rank(x) and value a component of x.

For each set of tied values the min_rank() function assigns a rank equal to the number of values less than that tied value plus one. The dense_rank() function assigns a rank equal to the number of distinct values less than that tied value plus one.

the ranking functions handle

functions

Is it "rough" or "exact"?

a vector

I might omit this since you give an explanation of row_number berlow which is accurate. Also, tibble adds its own rownumbers and rownumber(x) appears to be "the rownumber after x is sorted", which I don't find confusing.

ties

This probably is intended to be an argument of mean. It's not necessary, however, since neither dep_delay nor dep_delay_lag have any NAs. Also, could you provide an interpretation of delay_diff? For example, what does it mean that JFK's is negative?

Could be eliminated since the dataset is for one specific year.

Another definition of flight could be the quad (carrier,flight,origin,dest) which, say, might fly daily. Here's a solution with this definition.

dest_delay <- flights %>%
  group_by(dest) %>%
  filter(arr_delay > 0) %>%
  summarize(dest_ad = sum(arr_delay,na.rm=T),dest_ct = n())
flight_delay <- flights %>%
  group_by(carrier,flight,origin,dest) %>%
  filter(arr_delay > 0) %>%
  summarize(flight_ad = sum(arr_delay,na.rm=T),flight_ct = n())
prop <- as_tibble(merge(dest_delay,flight_delay,by="dest")) %>%
  mutate(prop = flight_ad / dest_ad) %>%
select(carrier,flight,origin,dest,dest_ct,flight_ct,prop)

In general selecting just a few columns (often the mutate variables) makes the result clearer since "irrelevant" cols are eliminated.

This could be omitted but keeping it in redundantly may add clarity.

Omit

Since this minimum rank group has an on_time of 0.0, I believe that this tibble is all planes which have never flown an on-time flight.

==

I believe min_rank cannot be 0.

(arr_delay > 0)

Of course precedence "does the right thing" but since R will use coercion to give lgl & dbl a value, putting in the redundant parentheses prevents the not-careful reader (me, for example) from "thinking the wrong thing."

In this dataset the presence of arr_delay implies the presence of arr_time so the boolean cancelled could be eliminated. Of course, the reader hasn't verified this relation so keeping it in makes sense. The definition of cancelled as "having an arrival delay" and "not having an arrival time" seems a bit odd.

Do all functions on grouped tibbles operate only per group and not the entire tibble? E.g.

flights %>% 
  group_by(month,day) %>%
  arrange(desc(dep_delay))

Will the arrange function order both by group and by the entire tibble?

Perhaps, but this is a lovely example of the power of R.

". atan(y,x) returns"

My worry is that the reader would assume that atan(x,y) returns that angle.

Hadley defines this as

not_cancelled <- flights %>% 
  filter(!is.na(dep_delay), !is.na(arr_delay))

But in

flights %>% 
  filter(is.na(dep_time) | is.na(arr_time) | is.na(dep_delay) | is.na(arr_delay)) %>% 
select(sched_arr_time,arr_time,arr_delay,sched_dep_time,dep_time,dep_delay)

the first 6 flights have an NA only in arr_delay. Aren't these likely not to have been cancelled and simply have a missing data entry? Why isn't

not_cancelled <- flights %>%
filter(!is.na(dep_time), !is.na(arr_time))

a "better" definition?

Omit.

Another solution. Now that we know we can "trust" dep_delay we could simply

md <- flights %>%
select(carrier,flight,origin,dest,sched_dep_time,dep_time,dep_delay) %>%
arrange(min_rank(desc(dep_delay)))

There are no ties in the top 10.

If you wish, you could change this to "Eastern Daylight" to agree with its previous reference in that sentence.

I understand the point of the paragraph but this sentence is a bit unclear.

Must be na.rm = TRUE inside the sum parenthesis for the below graph to appear otherwise it will just be a blank graph.

By the way thanks so much for these solutions, they are extremely helpful and teach things which are not taught in the book!