I am trying to use the group_by function inside of a function but it doesn't seem to work. I found an example in another post as below (this works) :-
dat <- mtcars[c(2:4,11)]
grp <- function(x) {
group_by(dat,!!as.name(x)) %>%
summarise(n=n()) %>%
mutate(pc=scales::percent(n/sum(n))) %>%
arrange(desc(n)) %>% head()
}
lapply(colnames(dat), grp)
What I don't understand is why do I need to data frame name in the group_by function - doesn't group_by function work this way :-
data %>% group_by(lgID) %>% summarise(mean_run = mean(HR))
where the data is piped to the group_by function?
Also, why do I need '!!as.name(x)' - what does this do?
Further, why does the version shown above work and this version shown below doesn't?
grp <- function(x) {
group_by(x) %>%
summarise(n=n()) %>%
mutate(pc=scales::percent(n/sum(n))) %>%
arrange(desc(n)) %>% head()
}
lapply(colnames(dat), grp)
Obviously I am missing something here!
Best regards
Deepak
If we need to pass both index and strings as 'x', wrap it inside across within group_by
library(dplyr) # version >= 1.0.0
f1 <- function(data, x) {
data %>%
group_by(across(all_of(x))) %>%
summarise(n=n(), .groups = 'drop') %>%
mutate(pc=scales::percent(n/sum(n))) %>%
arrange(desc(n)) %>%
head()
}
If we have an older version, use group_by_at(x)
-apply the function
out1 <- lapply(colnames(dat), function(x) f1(dat, x))
Or use index
out2 <- lapply(seq_along(dat), function(i) f1(dat, i))
identical(out1, out2)
#[1] TRUE
-output
out1[[1]]
# A tibble: 3 x 3
# cyl n pc
# <dbl> <int> <chr>
#1 8 14 43.8%
#2 4 11 34.4%
#3 6 7 21.9%
out2[[1]]
# A tibble: 3 x 3
# cyl n pc
# <dbl> <int> <chr>
#1 8 14 43.8%
#2 4 11 34.4%
#3 6 7 21.9%
Related
I would like to write a function that uses dplyr inside and I supply variable names as strings. Unfortunately dplyr-s use of NSE makes it rather complicated. From Programming with dplyr I get the following example
my_summarise <- function(df, var) {
var <- enquo(var)
df %>%
group_by(!!var) %>%
summarise(a = mean(a))
}
my_summarise(df, g1)
However, I would like to write function where instead of g1 I could provide "g1" and I am not able to wrap my head around how to do that.
dplyr >= 1.0
Use combination of double braces and the across function:
my_summarise2 <- function(df, group_var) {
df %>% group_by(across({{ group_var }})) %>%
summarise(mpg = mean(mpg))
}
my_summarise2(mtcars, "cyl")
# A tibble: 3 x 2
# cyl mpg
# <dbl> <dbl>
# 1 4 26.7
# 2 6 19.7
# 3 8 15.1
# same result as above, passing cyl without quotes
my_summarise(mtcars, cyl)
dplyr < 1.0
As far as I know, you could use as.name or sym (from the rlang package - I don't know if dplyr will import it eventually):
library(dplyr)
my_summarise <- function(df, var) {
var <- rlang::sym(var)
df %>%
group_by(!!var) %>%
summarise(mpg = mean(mpg))
}
or
my_summarise <- function(df, var) {
var <- as.name(var)
df %>%
group_by(!!var) %>%
summarise(mpg = mean(mpg))
}
my_summarise(mtcars, "cyl")
# # A tibble: 3 × 2
# cyl mpg
# <dbl> <dbl>
# 1 4 26.66364
# 2 6 19.74286
# 3 8 15.10000
Using the .data pronoun from rlang is another option that works directly with column names stored as strings.
The function with .data would look like
my_summarise <- function(df, var) {
df %>%
group_by(.data[[var]]) %>%
summarise(mpg = mean(mpg))
}
my_summarise(mtcars, "cyl")
# A tibble: 3 x 2
cyl mpg
<dbl> <dbl>
1 4 26.7
2 6 19.7
3 8 15.1
This is how to do it using only dplyr and the very useful as.name function from base R:
my_summarise <- function(df, var) {
varName <- as.name(var)
enquo_varName <- enquo(varName)
df %>%
group_by(!!enquo_varName) %>%
summarise(a = mean(a))
}
my_summarise(df, "g1")
Basically, with as.name() we generate a name object that matches var (here var is a string). Then, following Programming with dplyr, we use enquo() to look at that name and return the associated value as a quosure. This quosure can then be unquoted inside the group_by() call using !!.
I would like to write a function that uses dplyr inside and I supply variable names as strings. Unfortunately dplyr-s use of NSE makes it rather complicated. From Programming with dplyr I get the following example
my_summarise <- function(df, var) {
var <- enquo(var)
df %>%
group_by(!!var) %>%
summarise(a = mean(a))
}
my_summarise(df, g1)
However, I would like to write function where instead of g1 I could provide "g1" and I am not able to wrap my head around how to do that.
dplyr >= 1.0
Use combination of double braces and the across function:
my_summarise2 <- function(df, group_var) {
df %>% group_by(across({{ group_var }})) %>%
summarise(mpg = mean(mpg))
}
my_summarise2(mtcars, "cyl")
# A tibble: 3 x 2
# cyl mpg
# <dbl> <dbl>
# 1 4 26.7
# 2 6 19.7
# 3 8 15.1
# same result as above, passing cyl without quotes
my_summarise(mtcars, cyl)
dplyr < 1.0
As far as I know, you could use as.name or sym (from the rlang package - I don't know if dplyr will import it eventually):
library(dplyr)
my_summarise <- function(df, var) {
var <- rlang::sym(var)
df %>%
group_by(!!var) %>%
summarise(mpg = mean(mpg))
}
or
my_summarise <- function(df, var) {
var <- as.name(var)
df %>%
group_by(!!var) %>%
summarise(mpg = mean(mpg))
}
my_summarise(mtcars, "cyl")
# # A tibble: 3 × 2
# cyl mpg
# <dbl> <dbl>
# 1 4 26.66364
# 2 6 19.74286
# 3 8 15.10000
Using the .data pronoun from rlang is another option that works directly with column names stored as strings.
The function with .data would look like
my_summarise <- function(df, var) {
df %>%
group_by(.data[[var]]) %>%
summarise(mpg = mean(mpg))
}
my_summarise(mtcars, "cyl")
# A tibble: 3 x 2
cyl mpg
<dbl> <dbl>
1 4 26.7
2 6 19.7
3 8 15.1
This is how to do it using only dplyr and the very useful as.name function from base R:
my_summarise <- function(df, var) {
varName <- as.name(var)
enquo_varName <- enquo(varName)
df %>%
group_by(!!enquo_varName) %>%
summarise(a = mean(a))
}
my_summarise(df, "g1")
Basically, with as.name() we generate a name object that matches var (here var is a string). Then, following Programming with dplyr, we use enquo() to look at that name and return the associated value as a quosure. This quosure can then be unquoted inside the group_by() call using !!.
I need to repeat an operation many times for a different combinations of two different variables (trying to create data for stacked barplots showing percentage. Could anyone turn the code below into a function (of dataset, and the two variables x and y) in order to create the new data sets quickly? Or give me some good reference or link for learning about functions and dplyr. Thanks.
dat = df %>%
select(x, y) %>%
group_by(x, y) %>%
summarise(n = n()) %>%
mutate(percentage = round(n/sum(n)*100, 1)) %>%
ungroup() %>%
group_by(x) %>%
mutate(pos = cumsum(percentage) - (0.5 * percentage)) %>%
ungroup()
return(dat)
As suggested in the comments above, step-by-step explanations can be found here: dplyr.tidyverse.org/articles/programming.html
This guide will provide explanation of quo() function and !! symbols.
For your example you can create a function like so:
df1<- data.frame(x1 = c(rep(3,5), rep(7,2)),
y1 = c(rep(2,4), rep(5,3)))
my.summary <- function(df, x, y){
df %>%
select(!!x, !!y) %>%
group_by(!!x, !!y) %>%
summarise(n = n()) %>%
mutate(percentage = round(n/sum(n)*100, 1)) %>%
ungroup() %>%
group_by(!!x) %>%
mutate(pos = cumsum(percentage) - (0.5 * percentage)) %>%
ungroup()
}
my.summary(df1, quo(x1), quo(y1))
# # A tibble: 3 x 5
# x1 y1 n percentage pos
# <dbl> <dbl> <int> <dbl> <dbl>
# 1 3 2 4 80 40
# 2 3 5 1 20 90
# 3 7 5 2 100 50
I would like to write a function that uses dplyr inside and I supply variable names as strings. Unfortunately dplyr-s use of NSE makes it rather complicated. From Programming with dplyr I get the following example
my_summarise <- function(df, var) {
var <- enquo(var)
df %>%
group_by(!!var) %>%
summarise(a = mean(a))
}
my_summarise(df, g1)
However, I would like to write function where instead of g1 I could provide "g1" and I am not able to wrap my head around how to do that.
dplyr >= 1.0
Use combination of double braces and the across function:
my_summarise2 <- function(df, group_var) {
df %>% group_by(across({{ group_var }})) %>%
summarise(mpg = mean(mpg))
}
my_summarise2(mtcars, "cyl")
# A tibble: 3 x 2
# cyl mpg
# <dbl> <dbl>
# 1 4 26.7
# 2 6 19.7
# 3 8 15.1
# same result as above, passing cyl without quotes
my_summarise(mtcars, cyl)
dplyr < 1.0
As far as I know, you could use as.name or sym (from the rlang package - I don't know if dplyr will import it eventually):
library(dplyr)
my_summarise <- function(df, var) {
var <- rlang::sym(var)
df %>%
group_by(!!var) %>%
summarise(mpg = mean(mpg))
}
or
my_summarise <- function(df, var) {
var <- as.name(var)
df %>%
group_by(!!var) %>%
summarise(mpg = mean(mpg))
}
my_summarise(mtcars, "cyl")
# # A tibble: 3 × 2
# cyl mpg
# <dbl> <dbl>
# 1 4 26.66364
# 2 6 19.74286
# 3 8 15.10000
Using the .data pronoun from rlang is another option that works directly with column names stored as strings.
The function with .data would look like
my_summarise <- function(df, var) {
df %>%
group_by(.data[[var]]) %>%
summarise(mpg = mean(mpg))
}
my_summarise(mtcars, "cyl")
# A tibble: 3 x 2
cyl mpg
<dbl> <dbl>
1 4 26.7
2 6 19.7
3 8 15.1
This is how to do it using only dplyr and the very useful as.name function from base R:
my_summarise <- function(df, var) {
varName <- as.name(var)
enquo_varName <- enquo(varName)
df %>%
group_by(!!enquo_varName) %>%
summarise(a = mean(a))
}
my_summarise(df, "g1")
Basically, with as.name() we generate a name object that matches var (here var is a string). Then, following Programming with dplyr, we use enquo() to look at that name and return the associated value as a quosure. This quosure can then be unquoted inside the group_by() call using !!.
Let the data frame be:
set.seed(123)
df<-data.frame(name=sample(LETTERS,260,replace=TRUE),
hobby=rep(c("outdoor","indoor"),260),chess=rnorm(1:10))
and the condition which I will use to extract from df be:
df_cond<-df %>% group_by(name,hobby) %>%
summarize(count=n()) %>%
mutate(sum.var=sum(count),sum.name=length(name)) %>%
filter(sum.name==2) %>%
mutate(min.var=min(count)) %>%
mutate(use=ifelse(min.var==count,"yes","no")) %>%
filter(grepl("yes",use))
I want to randomly extract the rows from df that correspond to the (name,hobby,count) combination in df_cond along with the rest of df. I am having bit of a trouble combining %in% and sample.Thanks for any clue!
Edit: For example:
head(df_cond)
name hobby count sum.var sum.name min.var use
<fctr> <fctr> <int> <int> <int> <int> <chr>
1 A indoor 2 6 2 2 yes
2 B indoor 8 16 2 8 yes
3 B outdoor 8 16 2 8 yes
4 C outdoor 6 14 2 6 yes
5 D indoor 10 24 2 10 yes
6 E outdoor 8 18 2 8 yes
Using the above data frame, I want to randomly extract 2 rows (=count) with the combination A+indoor(row1) from df,
8 rows with the combination B+indoor (row 2) from df ....and so on.
Combining #denrous and #Jacob answers to get what I need. like so:
m2<-df_cond %>%
mutate(data = map2(name, hobby, function(x, y) {df %>% filter(name == x, hobby == y)})) %>%
ungroup() %>%
select(data) %>%
unnest()
test<-m2 %>%
group_by(name,hobby) %>%
summarize(num.levels=length(unique(hobby))) %>%
ungroup() %>%
group_by(name) %>%
summarize(total_levels=sum(num.levels)) %>%
filter(total_levels>1)
fin<-semi_join(m2,test)
If I understand correctly, you could use purrr to achieve what you want:
df_cond %>%
mutate(data = map2(name, hobby, function(x, y) {filter(df, name == x, hobby == y)})) %>%
mutate(data = map2(data, count, function(x, y) sample_n(x, size = y)))
And if you want the same form as df:
df_cond %>%
mutate(data = map2(name, hobby, function(x, y) {df %>% filter(name == x, hobby == y)})) %>%
mutate(data = map2(data, count, function(x, y) sample_n(x, size = y))) %>%
ungroup() %>%
select(data) %>%
unnest()
Edited based on OP clarification.
There has to better way but I'd use a loop:
library(dplyr)
master_df <- data.frame()
for (i in 1:nrow(df_cond)){
name = as.character(df_cond[i, 1])
hobby = as.character(df_cond[i, 2])
n = as.numeric(df_cond[i, 3])
temp_df <- df %>% filter(name == name, hobby == hobby)
temp_df <- sample_n(temp_df, n)
master_df <- rbind(master_df, temp_df)
}
Not clear if this is exactly what you want, but you may be looking for left_join:
df %>%
left_join(df_cond, by = "name")