I think this questions has multiple other variations (eg: here, here and perhaps here) - probably even an answer somewhere.
How to provide parameters to the filter function.
library(dplyr)
library(lazyeval)
set.seed(10)
data <- data.frame(a=sample(1:10, 100, T))
If I need to count the number of occurrences of the number 1 through 10 occurs and show the count for, say 1, 2 and 3, I would do this:
data %>%
group_by(a) %>%
summarise(n = n()) %>%
filter(a < 4)
Gives:
# A tibble: 3 × 2
a n
<int> <int>
1 1 11
2 2 8
3 3 16
Now, how can I put this into a function ?
Here grp is the grouping var.
fun <- function(d, grp, no){
d %>%
group_by_(grp) %>%
summarise_(n = interp(~ n() )) %>%
filter_( grp < no)
# This final line instead also does not work:
# filter_(interp( grp < no), grp = as.name(grp))
}
Now,
fun(data, 'a', 4)
Gives:
# A tibble: 0 × 2
# ... with 2 variables: a <int>, n <int>
We can use the quosures approach from the devel version of dplyr (soon to be released 0.6.0)
fun <- function(d, grp, no){
grp <- enquo(grp)
d %>%
group_by(UQ(grp)) %>%
summarise(n = n() )%>%
filter(UQ(grp) < no)
}
fun(data, a, 4)
# A tibble: 3 x 2
# a n
# <int> <int>
#1 1 11
#2 2 8
#3 3 16
We use enquo to take the input argument and convert it to quosure, within the group_by/summarise/mutate, the quosure is evaluated by unquoting (UQ or !!)
The above function can be also modified to take both quoted and unquoted arguments
fun <- function(d, grp, no){
lst <- as.list(match.call())
grp <- if(is.character(lst$grp)) {
rlang::parse_quosure(grp)
} else enquo(grp)
d %>%
group_by(UQ(grp)) %>%
summarise(n = n() )%>%
filter(UQ(grp) < no)
}
fun(data, a, 4)
# A tibble: 3 x 2
# a n
# <int> <int>
#1 1 11
#2 2 8
#3 3 16
fun(data, 'a', 4)
# A tibble: 3 x 2
# a n
# <int> <int>
#1 1 11
#2 2 8
#3 3 16