汇总但保持长度可变（dplyr）

Question

Basic dplyr question... Respondents could select multiple companies that they use. For example:

library(dplyr)
test <- tibble(
 CompanyA = rep(c(0:1),5),
 CompanyB = rep(c(1),10),
 CompanyC = c(1,1,1,1,0,0,1,1,1,1)
)
test

If it were a forced-choice question - ie, respondents could make only one selection - I would do the following for a basic summary table:

test %>% 
  summarise_all(funs(sum), na.rm = TRUE) %>% 
  gather(Response, n) %>% 
  arrange(desc(n)) %>% 
  mutate("%" = round(100*n/sum(n)))

Note, however, that the "%" column is not what I want. I'm instead looking for the proportion of total respondents for each individual response option (since they could make multiple selections).

I've tried adding mutate(totalrows = nrow(.)) %>% prior to the summarise_all command. This would allow me to use that variable as the denominator in a later mutate command. However, summarise_all eliminates the "totalrows" var.

Also, if there's a better way to do this, I'm open to ideas.

Answer 1

To get the proportion of respondents who chose an option when that variable is binary, you can take the mean. To do this with your test data, you can use sapply :

sapply(test, mean)
CompanyA CompanyB CompanyC 
     0.5      1.0      0.8 

If you wanted to do this in a more complicated fashion (say your data is not binary encoded, but is stored as 1 and 2 instead), you could do that with the following:

test %>% 
    gather(key='Company') %>% 
    group_by(Company) %>% 
    summarise(proportion = sum(value == 1) / n())

# A tibble: 3 x 2
  Company  proportion
  <chr>         <dbl>
1 CompanyA        0.5
2 CompanyB        1  
3 CompanyC        0.8

Answer 2

If you put all functions in a list within summarise, then this will work. You'll need to do some quick tidying up after though.

test %>% 
  summarise_all(
    list(
      rows = length,
      n = function(x){sum(x, na.rm = T)},
      perc = function(x){sum(x,na.rm = T)/length(x)}
    )) %>%
  tidyr::gather(Response, n) %>%
  tidyr::separate(Response, c("Company", "Metric"), '_') %>%
  tidyr::spread(Metric, n)

And you'll get this

  Company      n  perc  rows
  <chr>    <dbl> <dbl> <dbl>
1 CompanyA     5   0.5    10
2 CompanyB    10   1      10
3 CompanyC     8   0.8    10

Answer 3

Here is a solution using tidyr::gather :

test %>% 
  gather(Company, response) %>% 
  group_by(Company) %>% 
  summarise(`%` = 100 * sum(response) / n())