使用 tidyverse 对多个列应用相同的操作汇总

Question

I'm trying to create a summary table that gives me the proportion of yes responses for 17 questions sorted by year. I just don't know how to apply the summarize operation to multiple columns easily without hard-coding it.

Unfortunately, I can't use the summarize_at or summarize_all functions because I'm working with a dataframe. I was thinking of writing a function, looping through the columns, and rbinding the summary columns together, but summarize is a bit weird with column names, which can't be of type character. What do you recommend?

Here's what I currently have:

s2 <- db %>%
  group_by(Year)%>%
  summarize(Q1=round(sum(Q1d, na.rm=TRUE)*100/length(which(!is.na(Q1d))),1),
            Q2=round(sum(Q2d, na.rm=TRUE)*100/length(which(!is.na(Q2d))),1),
            Q3=round(sum(Q3d, na.rm=TRUE)*100/length(which(!is.na(Q3d))),1),
            Q4=round(sum(Q4d, na.rm=TRUE)*100/length(which(!is.na(Q4d))),1),
            Q5=round(sum(Q5d, na.rm=TRUE)*100/length(which(!is.na(Q5d))),1),
            Q6=round(sum(Q6d, na.rm=TRUE)*100/length(which(!is.na(Q6d))),1),
            Q7=round(sum(Q7d, na.rm=TRUE)*100/length(which(!is.na(Q7d))),1),
            Q8=round(sum(Q8d, na.rm=TRUE)*100/length(which(!is.na(Q8d))),1),
            Q9=round(sum(Q9d, na.rm=TRUE)*100/length(which(!is.na(Q9d))),1),
            Q10=round(sum(Q10d, na.rm=TRUE)*100/length(which(!is.na(Q10d))),1),
            Q11=round(sum(Q11d, na.rm=TRUE)*100/length(which(!is.na(Q11d))),1),
            Q12=round(sum(Q12d, na.rm=TRUE)*100/length(which(!is.na(Q12d))),1),
            Q13=round(sum(Q13d, na.rm=TRUE)*100/length(which(!is.na(Q13d))),1),
            Q14=round(sum(Q14d, na.rm=TRUE)*100/length(which(!is.na(Q14d))),1),
            Q15=round(sum(Q15d, na.rm=TRUE)*100/length(which(!is.na(Q15d))),1),
            Q16=round(sum(Q16d, na.rm=TRUE)*100/length(which(!is.na(Q16d))),1),
            Q17=round(sum(Q17d, na.rm=TRUE)*100/length(which(!is.na(Q17d))),1),
            )

Note: Q1d, Q2d... are the names of the columns

Answer 1

We can use across in dplyr

library(dplyr)
library(stringr)
db %>%
    group_by(Year) %>%
    summarise(across(matches('^Q\\d+d$'), ~ 
              sum(., na.rm = TRUE) * 100 /sum(!is.na(.))), 
         .groups = 'drop') %>%
    rename_with(~ str_remove(., 'd$'), -Year)

---

or using collapse

library(collapse)
f1 <- function(x) sum(x, na.rm = TRUE) * 100/sum(!is.na(x))
collap(db, ~ Year, FUN = f1)
#   Year      Q1d Q2d
#1 2010 250.0000 350
#2 2015 293.3333 320

data

db <- data.frame(Year = c(2010, 2010, 2015, 2015, 2015, 2015),
   Q1d = c(2.5, NA, 3, 3.5, NA, 2.3), Q2d = c(NA, 3.5, NA, 2, 4.6, 3))