How do I subset a column of the current group in dplyr? - r

Question

I am having difficulty parameterizing some code I wrote. It works fine when not in a function like this:
new_df = group_by(groupby1, groupby2) %>%
mutate(new_value=
slider_helper(
slide(cur_data()[, c('string1', 'string2'], ~.x, .before = Inf, .after = -1),
cur_data()$string2,
'string1',
beta
But when I try to create a function where you can pass strings for the variables to group by and the variables to slide over like so:
my_fun <- function(df, groupby1, groupby2, string1, string2, beta) {
return(df %>%
group_by({{groupby1}}, {{groupby2}}) %>%
mutate(new_value=
slider_helper(
slide(cur_data()[, c({{string1}}, {{string2}}], ~.x, .before = Inf, .after = -1),
cur_data()[[{{string2}}]],
{{string1}},
beta)))
}
I get this vague stack trace:
The error occurred in group 1: "groupby1" = "groupby1", "groupby2" = "groupby2".
Caused by error in `.subset()`:
! invalid subscript type 'closure'
What is the proper way to parameterize dplyr functions to work with column names passed as strings?
EDIT:
Here is a reproducible example
slider_helper <- function(left, right, string1, beta) {
cbind_helper <- function(left, right) {
todaysDate = rep(right, nrow(left))
return(cbind(left, todaysDate))
}
date_helper <- function(today, date) {
return(1/as.integer(today - date))
}
df = data.frame(t(mapply(cbind_helper, left, right)))
df$val1= mapply(date_helper, df[,'todaysDate'], df[, date])
df$val1_product= mapply('%*%', df$val1, df[[target]]) / sapply(df$val1, FUN=sum, na.rm=T)
df$val2= 1/seq(nrow(df), 1)
df$val2_product= sapply(mapply('*', df$val2, df[[target]]), FUN=sum, na.rm=T) / sum(df$val2, na.rm=T)
w_sum = beta * df$val2_product+ (1-beta) * df$val2_product
return(w_sum)
}
my_fun <- function(df, groupby1, groupby2, string1, string2, beta) {
return(df %>%
group_by({{groupby1}}, {{groupby2}}) %>%
mutate(new_value=
slider_helper(
slide(cur_data()[, c({{string1}}, {{string2}})], ~.x, .before = Inf, .after = -1),
cur_data()[[{{string2}}]],
{{string1}},
beta)))
}
df= data.frame(sample(1:2, 20, replace=T), sample(1:2, 20, replace=T), seq(from=-1, to=.9, by = .1), seq.Date(from=as.Date('2011-01-01'), to=as.Date('2011-01-20'), by = 1))
colnames(df) = c('groupby1', 'groupby2', 'string1', 'string2')
my_fun(testy, 'groupby1', 'groupby2', 'string1', 'string2', 0.5)

Answer 1

Without sample data and more context this is kind of a guess.
But, if you are using a simple function then you need to remove the {{}}. The function does not need those brackets.
my_fun <- function(df, groupby1, groupby2, string1, string2, beta) {
return(df %>%
group_by(groupby1, groupby2) %>%
mutate(new_value=
slider_helper(
slide(cur_data()[, c(string1, string2)], ~.x, .before = Inf, .after = -1),
cur_data()[[string2]],
string1,
beta)))
}