dplyr filter: Get rows with minimum of variable, but only the first if multiple minima
RDplyrR Problem Overview
I want to make a grouped filter using dplyr
, in a way that within each group only that row is returned which has the minimum value of variable x
.
My problem is: As expected, in the case of multiple minima all rows with the minimum value are returned. But in my case, I only want the first row if multiple minima are present.
Here's an example:
df <- data.frame(
A=c("A", "A", "A", "B", "B", "B", "C", "C", "C"),
x=c(1, 1, 2, 2, 3, 4, 5, 5, 5),
y=rnorm(9)
)
library(dplyr)
df.g <- group_by(df, A)
filter(df.g, x == min(x))
As expected, all minima are returned:
Source: local data frame [6 x 3]
Groups: A
A x y
1 A 1 -1.04584335
2 A 1 0.97949399
3 B 2 0.79600971
4 C 5 -0.08655151
5 C 5 0.16649962
6 C 5 -0.05948012
With ddply, I would have approach the task that way:
library(plyr)
ddply(df, .(A), function(z) {
z[z$x == min(z$x), ][1, ]
})
... which works:
A x y
1 A 1 -1.04584335
2 B 2 0.79600971
3 C 5 -0.08655151
Q: Is there a way to approach this in dplyr? (For speed reasons)
R Solutions
Solution 1 - R
Update
With dplyr >= 0.3 you can use the slice
function in combination with which.min
, which would be my favorite approach for this task:
df %>% group_by(A) %>% slice(which.min(x))
#Source: local data frame [3 x 3]
#Groups: A
#
# A x y
#1 A 1 0.2979772
#2 B 2 -1.1265265
#3 C 5 -1.1952004
Original answer
For the sample data, it is also possible to use two filter
after each other:
group_by(df, A) %>%
filter(x == min(x)) %>%
filter(1:n() == 1)
Solution 2 - R
Just for completeness: Here's the final dplyr
solution, derived from the comments of @hadley and @Arun:
library(dplyr)
df.g <- group_by(df, A)
filter(df.g, rank(x, ties.method="first")==1)
Solution 3 - R
For what it's worth, here's a data.table
solution, to those who may be interested:
# approach with setting keys
dt <- as.data.table(df)
setkey(dt, A,x)
dt[J(unique(A)), mult="first"]
# without using keys
dt <- as.data.table(df)
dt[dt[, .I[which.min(x)], by=A]$V1]
Solution 4 - R
This can be accomplished by using row_number
combined with group_by
. row_number
handles ties by assigning a rank not only by the value but also by the relative order within the vector. To get the first row of each group with the minimum value of x
:
df.g <- group_by(df, A)
filter(df.g, row_number(x) == 1)
For more information see the dplyr vignette on window functions.
Solution 5 - R
dplyr
offers slice_min
function, wich do the job with the argument with_ties = FALSE
library(dplyr)
df %>%
group_by(A) %>%
slice_min(x, with_ties = FALSE)
Output :
# A tibble: 3 x 3
# Groups: A [3]
A x y
<fct> <dbl> <dbl>
1 A 1 0.273
2 B 2 -0.462
3 C 5 1.08
Solution 6 - R
Another way to do it:
set.seed(1)
x <- data.frame(a = rep(1:2, each = 10), b = rnorm(20))
x <- dplyr::arrange(x, a, b)
dplyr::filter(x, !duplicated(a))
Result:
a b
1 1 -0.8356286
2 2 -2.2146999
Could also be easily adapted for getting the row in each group with maximum value.
Solution 7 - R
I like sqldf for its simplicity..
sqldf("select A,min(X),y from 'df.g' group by A")
Output:
A min(X) y
1 A 1 -1.4836989
2 B 2 0.3755771
3 C 5 0.9284441
Solution 8 - R
In case you are looking to filter the minima of x and then the minima of y. An intuitive way of do it is just using filtering functions:
> df
A x y
1 A 1 1.856368296
2 A 1 -0.298284187
3 A 2 0.800047796
4 B 2 0.107289719
5 B 3 0.641819999
6 B 4 0.650542284
7 C 5 0.422465687
8 C 5 0.009819306
9 C 5 -0.482082635
df %>% group_by(A) %>%
filter(x == min(x), y == min(y))
# A tibble: 3 x 3
# Groups: A [3]
A x y
<chr> <dbl> <dbl>
1 A 1 -0.298
2 B 2 0.107
3 C 5 -0.482
This code will filter the minima of x and y.
Also you can do a double filter that looks even more readable:
df %>% group_by(A) %>%
filter(x == min(x)) %>%
filter(y == min(y))
# A tibble: 3 x 3
# Groups: A [3]
A x y
<chr> <dbl> <dbl>
1 A 1 -0.298
2 B 2 0.107
3 C 5 -0.482
Solution 9 - R
For the sake of completeness, here's the base R
answer:
df[with(df, ave(x, A, FUN = \(x) rank(x, ties.method = "first")) == 1), ]
# A x y
#1 A 1 0.1076158
#4 B 2 -1.3909084
#7 C 5 0.3511618