I want to create a custom function to calculate grouped percentages in a large dataset with 100+ columns. Because I have so many columns I want to do a loop or lapply or something to avoid typing the function out 100+ times. The function I wrote works fine when I type it in individually for each column, but I cannot figure out how to do it repeatedly.
Here's a simplified dataframe and function:
# load required libraries:
library(tidyverse)
df<-data.frame(sex=c('M','M','M','F','M','F','M',NA),
school=c('A','A','A','A','B','B','B',NA),
question1=c(NA,1,1,2,2,3,3,3),
question2=c(2,NA,2,4,5,1,2,3))
my_function<-function(dataset,question_number){
question_number_enquo<-enquo(question_number)
dataset%>%
filter(!is.na(!!question_number_enquo)&!is.na(sex))%>%
group_by(school,sex,!!question_number_enquo)%>%
count(!!question_number_enquo)%>%
summarise(number=sum(n))%>%
mutate(percent=number/sum(number)*100)%>%
ungroup()
}
My function works when I type a column name into it:
my_function(df,question1)
A tibble: 5 x 5
school sex question1 number percent
<fct> <fct> <dbl> <int> <dbl>
1 A F 2 1 100
2 A M 1 2 100
3 B F 3 1 100
4 B M 2 1 50
5 B M 3 1 50
Here's what I've tried in terms of reiteration. I want to repeat the function for every column (except for school and sex, because those are my groups).
question_col_names<-(df%>%select(-sex,-school)%>%colnames())
Using lapply with the column names as a quosure:
question_col_names_enquo<-enquo(question_col_names)
lapply(df,my_function(df,!!question_col_names_enquo))
Error: Column `<chr>` must be length 7 (the number of rows) or one, not 2
Trying lapply with unquoted column names:
lapply(df,my_function(df,question_col_names))
Error: Column `question_col_names` is unknown
Trying lapply with quoted column names:
lapply(df,my_function(df,'question_col_names'))
Error: Column `"question_col_names"` can't be modified because it's a grouping variable
I also tried apply, and got the same types of error messages:
apply(df,1,my_function(df,!!question_col_names_enquo))
Error: Column `<chr>` must be length 7 (the number of rows) or one, not 2
apply(df,1,my_function(df,question_col_names))
Error: Column `question_col_names` is unknown
apply(df,1,my_function(df,'question_col_names'))
Error: Column `"question_col_names"` can't be modified because it's a grouping variable
I also tried different variations of a for loop:
for (i in question_col_names){
my_function(df,i)
}
Error: Column `i` is unknown
for (i in question_col_names){
my_function(df,'i')
}
Error: Column `"i"` can't be modified because it's a grouping variable
How can I use iteration to get my function to repeat over all my columns?
I suspect that this has to do with dplyr; I know that it acts funny in custom functions, but I can get it to work in my function, just not in the iteration. I've done a deep dive on Google and Stack Overflow but haven't found anything that answered this.
Thanks in advance!
Your question_col_names are strings. You need sym to convert string to variable inside your function instead
library(tidyverse)
df <- data.frame(
sex = c("M", "M", "M", "F", "M", "F", "M", NA),
school = c("A", "A", "A", "A", "B", "B", "B", NA),
question1 = c(NA, 1, 1, 2, 2, 3, 3, 3),
question2 = c(2, NA, 2, 4, 5, 1, 2, 3)
)
my_function <- function(dataset, question_number) {
question_number_enquo <- sym(question_number)
dataset %>%
filter(!is.na(!!question_number_enquo) & !is.na(sex)) %>%
group_by(school, sex, !!question_number_enquo) %>%
count(!!question_number_enquo) %>%
summarise(number = sum(n)) %>%
mutate(percent = number / sum(number) * 100) %>%
ungroup()
}
my_function(df, "question1")
#> # A tibble: 5 x 5
#> school sex question1 number percent
#> <fct> <fct> <dbl> <int> <dbl>
#> 1 A F 2 1 100
#> 2 A M 1 2 100
#> 3 B F 3 1 100
#> 4 B M 2 1 50
#> 5 B M 3 1 50
question_col_names <- (df %>% select(-sex, -school) %>% colnames())
result <- map_df(question_col_names, ~ my_function(df, .x))
result
#> # A tibble: 10 x 6
#> school sex question1 number percent question2
#> <fct> <fct> <dbl> <int> <dbl> <dbl>
#> 1 A F 2 1 100 NA
#> 2 A M 1 2 100 NA
#> 3 B F 3 1 100 NA
#> 4 B M 2 1 50 NA
#> 5 B M 3 1 50 NA
#> 6 A F NA 1 100 4
#> 7 A M NA 2 100 2
#> 8 B F NA 1 100 1
#> 9 B M NA 1 50 2
#> 10 B M NA 1 50 5
Probably better if you convert your function result to long format
my_function2 <- function(dataset, question_number) {
question_number_enquo <- sym(question_number)
res <- dataset %>%
filter(!is.na(!!question_number_enquo) & !is.na(sex)) %>%
group_by(school, sex, !!question_number_enquo) %>%
count(!!question_number_enquo) %>%
summarise(number = sum(n)) %>%
mutate(percent = number / sum(number) * 100) %>%
ungroup() %>%
gather(key = 'question', value, -school, -sex, -number, -percent)
return(res)
}
result2 <- map_df(question_col_names, ~ my_function2(df, .x))
result2
#> # A tibble: 10 x 6
#> school sex number percent question value
#> <fct> <fct> <int> <dbl> <chr> <dbl>
#> 1 A F 1 100 question1 2
#> 2 A M 2 100 question1 1
#> 3 B F 1 100 question1 3
#> 4 B M 1 50 question1 2
#> 5 B M 1 50 question1 3
#> 6 A F 1 100 question2 4
#> 7 A M 2 100 question2 2
#> 8 B F 1 100 question2 1
#> 9 B M 1 50 question2 2
#> 10 B M 1 50 question2 5
Created on 2019-11-25 by the reprex package (v0.3.0)