dplyr pass NULL to group_by - r

Question

This has probably been answered somewhere, but I cannot find the answer...Mark as a duplicate and downvote as you like, but someone please help me :)
Short question
How can I pass NULL to dplyr::group_by inside a function?
library(dplyr)
dt <- data.frame(a = sample(LETTERS[1:2], 100, replace = TRUE), b = sample(LETTERS[3:4], 100, replace = TRUE), value = rnorm(100,5,1))
f1 <- function(dt, a, b, c) {
dt %>% group_by(a, b, c) %>% summarise(mean = mean(value))
}
f1(dt, a = "a", b = "b", c = NULL)
# Error in grouped_df_impl(data, unname(vars), drop) :
# Column `c` is unknown
Long explanation
I am writing a function where "b" column can be given as NULL meaning that the function should ignore this column. If the "b" column is given as a character the function should use the column to summarize data. Like this:
f2 <- function(dt, a, b) {
if(is.null(b)) {
dt %>% group_by(a) %>% summarize(mean = mean(value))
} else {
dt %>% group_by(a, b) %>% summarize(mean = mean(value))
}
}
The actual function is quite long and complicated, and uses dplyr pipes to make all the summarizing code shorter. I have multiple conditions leading to different outputs and summarizing alternatives, and therefore I have shortened the if else statements by grouping first and summarizing in a separate step:
f3 <- function(dt, a, b, type = "mean") {
if(is.null(b)) {
tmp <- dt %>% group_by(a)
} else {
tmp <- dt %>% group_by(a, b)
}
if(type == "mean") {
tmp %>% summarize(mean = mean(value))
} else {
tmp %>% summarise(sum = sum(value))
}
}
If it was possible to pass NULL to the group_by function, I could considerably shorten my code (NULL is supposed to be empty anyway and such passing works with many functions such as reshape2::melt from the same author).

Answer 1

I'm not sure if this covers all of your use cases, but a function using tidy evaluation (see the programming with dplyr vignette) would be more flexible in that you wouldn't have to worry about how many grouping variables there are and you could pass an arbitrary vector of functions to summarize by. Hopefully, this avoids the need to keep track of NULL columns or use ifelse to choose the summary function.
For example, in the code below, ... is any number of grouping columns, including no grouping columns at all. The type argument allows you to summarize by one or more arbitrary functions:
library(tidyverse)
library(rlang)
set.seed(2)
dt <- data.frame(a = sample(LETTERS[1:2], 100, replace = TRUE),
b = sample(LETTERS[3:4], 100, replace = TRUE),
value = rnorm(100,5,1))
f1 = function(data, value.var, ..., type="mean") {
groups = enquos(...)
value.var = enquo(value.var)
names(type) = paste0(type, "_", quo_text(value.var))
type = syms(type)
data %>% group_by(!!!groups) %>%
summarise_at(vars(!!value.var), funs(!!!type))
}
f1(dt, value, a, b)
a b mean_value
<fct> <fct> <dbl>
1 A C 5.01
2 A D 5.05
3 B C 4.95
4 B D 5.13
f1(dt, value)
mean_value
<dbl>
1 5.03
weird_func = function(x) {
paste(round(cos(x),1)[1:3], collapse="/")
}
f1(dt, value, a, b, type=c("mean", "min", "median", "max", "weird_func"))
a b mean_value min_value median_value max_value weird_func_value
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <chr>
1 A C 5.01 3.26 5.07 7.08 1/-0.1/1
2 A D 5.05 2.90 5.33 6.36 -0.4/0.9/0
3 B C 4.95 3.66 4.73 7.11 0.5/-0.5/0.7
4 B D 5.13 2.98 5.46 7.05 0/0.7/0.7
f1(mtcars, mpg, cyl, type=c("mean", "median"))
cyl mean_mpg median_mpg
<dbl> <dbl> <dbl>
1 4 26.7 26
2 6 19.7 19.7
3 8 15.1 15.2

Answer 2

I think you need to first convert it from NULL to NA, like this (as from your answers you just need to pass the value through without involving it in the calculations)
library(dplyr)
dt <- data.frame(a = sample(LETTERS[1:2], 100, replace = TRUE), b = sample(LETTERS[3:4], 100, replace = TRUE), value = rnorm(100,5,1))
f1 <- function(dt, a, b, c) {
dt %>%
mutate(c = ifelse(is_empty(c)==TRUE,NA,c)) %>%
group_by(a, b,c) %>%
summarise(mean = mean(value))
}
f1(dt, a = "a", b = "b",c=NULL)
Results:
# A tibble: 4 x 4
# Groups: a, b [?]
a b c mean
<fct> <fct> <lgl> <dbl>
1 A C NA 5.27
2 A D NA 5.18
3 B C NA 5.27
4 B D NA 5.49