Carrying out a simple dataframe subset with dplyr - r

Question

Consider the following dataframe slice:
df = data.frame(locations = c("argentina","brazil","argentina","denmark"),
score = 1:4,
row.names = c("a091", "b231", "a234", "d154"))
df
locations score
a091 argentina 1
b231 brazil 2
a234 argentina 3
d154 denmark 4
sorted = c("a234","d154","a091") #in my real task these strings are provided from an exogenous function
df2 = df[sorted,] #quick and simple subset using rownames
EDIT: Here I'm trying to subset AND order the data according to sorted - sorry that was not clear before. So the output, importantly, is:
locations score
a234 argentina 1
d154 denmark 4
a091 argentina 3
And not as you would get from a simple subset operation:
locations score
a091 argentina 1
a234 argentina 3
d154 denmark 4
I'd like to do the exactly same thing in dplyr. Here is an inelegant hack:
require(dplyr)
dt = as_tibble(df)
rownames(dt) = rownames(df)
Warning message:
Setting row names on a tibble is deprecated.
dt2 = dt[sorted,]
I'd like to do it properly, where the rownames are an index in the data table:
dt_proper = as_tibble(x = df,rownames = "index")
dt_proper2 = dt_proper %>% ?some_function(index, sorted)? #what would this be?
dt_proper2
# A tibble: 3 x 3
index locations score
<chr> <fct> <int>
1 a091 argentina 1
2 d154 denmark 4
3 a234 argentina 3
But I can't for the life of me figure out how to do this using filter or some other dplyr function, and without some convoluted conversion to factor, re-order factor levels, etc.

Answer 1

Hy,
you can simply use mutate and filter to get the row.names of your data frame into a index column and filter to the vector "sorted" and sort the data frame due to the vector "sorted":
df2 <- df %>% mutate(index=row.names(.)) %>% filter(index %in% sorted)
df2 <- df2[order(match(df2[,"index"], sorted))]

Answer 2

I think I've figured it out:
dt_proper2 = dt_proper[match(sorted,dt_proper$index),]
Seems to be shortest implementation of what df[sorted,] will do.

Answer 3

Functions in the tidyverse (dplyr, tibble, etc.) are built around the concept (as far as I know), that rows only contain attributes (columns) and no row names / labels / indexes. So in order to sort columns, you have to introduce a new column containing the ranks of each row.
The way I would do it is to create another tibble containing your "sorting information" (sorting attribute, rank) and inner join it to your original tibble. Then I could order the rows by rank.
library(tidyverse)
# note that I've changed the third column's name to avoid confusion
df = tibble(
locations = c("argentina","brazil","argentina","denmark"),
score = 1:4,
custom_id = c("a091", "b231", "a234", "d154")
)
sorted_ids = c("a234","d154","a091")
sorting_info = tibble(
custom_id = sorted_ids,
rank = 1:length(sorted_ids)
)
ordered_ids = df %>%
inner_join(sorting_info) %>%
arrange(rank) %>%
select(-rank)