Anand - 1 month ago 5x
R Question

Group by a column and sort by another column in R

I am examining the imdb movie dataset in kaggle with R.

Here is a minimal repro dataset:

``````> movies <- data.frame(movie = as.factor(c("Movie 1", "Movie 2", "Movie 3", "Movie 4")), director = as.factor(c("Dir 1", "Dir 2", "Dir 1", "Dir 3")), director_rating =  c(1000, 2000, 1000, 3000))

> movies
movie director director_rating
1 Movie 1    Dir 1            1000
2 Movie 2    Dir 2            2000
3 Movie 3    Dir 1            1000
4 Movie 4    Dir 3            3000
``````

Note that every row that has the same director has the same value of rating for the director.

I want to list the directors, sorted by rating, and one row per director. The following code works:

``````> library(dplyr)
> movies %>%
group_by(director) %>%
summarize(director_rating = mean(director_rating)) %>%
arrange(desc(director_rating))

# A tibble: 3 x 2
director director_rating
<fctr>           <dbl>
1    Dir 3            3000
2    Dir 2            2000
3    Dir 1            1000
``````

But it seems wasteful to compute the mean when I know that all the ratings for a single director are identical. What is a more idiomatic/efficient way to do this in R?

There's actually no need to group and summarise, since you are just looking for distinct / unique entries. A dplyr option is therefore:

``````select(movies, -movie) %>%
distinct() %>%
arrange(desc(director_rating))
#  director director_rating
#1    Dir 3            3000
#2    Dir 2            2000
#3    Dir 1            1000
``````

Or in case you like to keep other columns:

``````distinct(movies, director, .keep_all = TRUE) %>%   # for dplyr >= 0.5.0
arrange(desc(director_rating))
#    movie director director_rating
#1 Movie 4    Dir 3            3000
#2 Movie 2    Dir 2            2000
#3 Movie 1    Dir 1            1000
``````