dplyr::count() multiple columns

user_3022126 0
Viewed 13851

I have the following data set:

dat = structure(list(C86_1981 = c("Outer London", "Buckinghamshire", 
NA, "Ross and Cromarty", "Cornwall and Isles of Scilly", NA, 
"Kirkcaldy", "Devon", "Kent", "Renfrew"), C96_1981 = c("Outer London", 
"Buckinghamshire", NA, "Ross and Cromarty", "Not known/missing", 
NA, "Kirkcaldy", NA, NA, NA), C00_1981 = c("Outer London", "Inner London", 
"Lancashire", "Ross and Cromarty", NA, "Humberside", "Kirkcaldy", 
NA, NA, NA), C04_1981 = c("Kent", NA, NA, "Ross and Cromarty", 
NA, "Humberside", "Not known/missing", NA, NA, "Renfrew"), C08_1981 = c("Kent", 
"Oxfordshire", NA, "Ross and Cromarty", "Cornwall and Isles of Scilly", 
"Humberside", "Dunfermline", NA, NA, "Renfrew"), C12_1981 = c("Kent", 
NA, NA, "Ross and Cromarty", "Cornwall and Isles of Scilly", 
"Humberside", "Dunfermline", NA, NA, "Renfrew")), row.names = c(NA, 
-10L), class = c("tbl_df", "tbl", "data.frame"), .Names = c("C86_1981", 
"C96_1981", "C00_1981", "C04_1981", "C08_1981", "C12_1981"))

I want to dplyr::count() each column. Expected output:

# A tibble: 10 x 3
                       C86_1981 dat86_n dat96_n ...
                          <chr>   <int>   <int>
 1              Buckinghamshire       1       1
 2 Cornwall and Isles of Scilly       1      NA
 3                        Devon       1      NA
 4                         Kent       1      NA
 5                    Kirkcaldy       1       1
 6                 Outer London       1       1
 7                      Renfrew       1      NA
 8            Ross and Cromarty       1       1
 9                         <NA>       2       5
10            Not known/missing      NA       1

Currently I'm doing this manually then dplyr::full_join()ing the result:

library("tidyverse")

dat86_n = dat %>%
  count(C86_1981) %>%
  rename(dat86_n = n)
dat96_n = dat %>%
  count(C96_1981) %>%
  rename(dat96_n = n)
# ...

dat_counts = dat86_n %>%
  full_join(dat96_n, by = c("C86_1981" = "C96_1981"))
  # ...

Which works, but is not exactly robust if any of my data changes later. I had hoped to do this programmatically.

I've tried a loop:

lapply(dat, count)
# Error in UseMethod("groups") : 
# no applicable method for 'groups' applied to an object of class "character"

(purrr::map() gives the same error). I think this error is because count() expects a tbl and a variable as separate arguments, so I tried that too:

lapply(dat, function(x) {
  count(dat, x)
})
# Error in grouped_df_impl(data, unname(vars), drop) : 
# Column `x` is unknown

Again, purrr::map() gives the same error. I've also tried variants of summarise_all():

dat %>% 
  summarise_all(count)
  # Error in summarise_impl(.data, dots) : 
  # Evaluation error: no applicable method for 'groups' applied to an object of class "character".

I feel like I'm missing something obvious and the solution should be straightforward. dplyr solutions particularly welcome as this is what I tend to use most.

dplyr r
3 Answers

Using also the tidyr package, the following code will do the trick:

dat %>% tidyr::gather(name, city) %>% dplyr::group_by(name, city) %>% dplyr::count() %>% dplyr::ungroup %>% tidyr::spread(name, n)

Result:

# A tibble: 15 x 7
                           city C00_1981 C04_1981 C08_1981 C12_1981 C86_1981 C96_1981
 *                        <chr>    <int>    <int>    <int>    <int>    <int>    <int>
 1              Buckinghamshire       NA       NA       NA       NA        1        1
 2 Cornwall and Isles of Scilly       NA       NA        1        1        1       NA
 3                        Devon       NA       NA       NA       NA        1       NA
 4                  Dunfermline       NA       NA        1        1       NA       NA
 5                   Humberside        1        1        1        1       NA       NA
 6                 Inner London        1       NA       NA       NA       NA       NA
 7                         Kent       NA        1        1        1        1       NA
 8                    Kirkcaldy        1       NA       NA       NA        1        1
 9                   Lancashire        1       NA       NA       NA       NA       NA
10            Not known/missing       NA        1       NA       NA       NA        1
11                 Outer London        1       NA       NA       NA        1        1
12                  Oxfordshire       NA       NA        1       NA       NA       NA
13                      Renfrew       NA        1        1        1        1       NA
14            Ross and Cromarty        1        1        1        1        1        1
15                         <NA>        4        5        3        4        2        5
user_80859720

@You-leee just beat me to it ;)

Using the tidyverse;

library(tidyverse)

df <- 
  dat %>% 
  gather (year, county) %>% 
  group_by(year, county) %>% 
  summarise(no = n()) %>% 
  spread (year, no)

# A tibble: 15 x 7
                         county C00_1981 C04_1981 C08_1981 C12_1981 C86_1981 C96_1981
 *                        <chr>    <int>    <int>    <int>    <int>    <int>    <int>
 1              Buckinghamshire       NA       NA       NA       NA        1        1
 2 Cornwall and Isles of Scilly       NA       NA        1        1        1       NA
 3                        Devon       NA       NA       NA       NA        1       NA
 4                  Dunfermline       NA       NA        1        1       NA       NA
 5                   Humberside        1        1        1        1       NA       NA  
 6                 Inner London        1       NA       NA       NA       NA       NA
 7                         Kent       NA        1        1        1        1       NA
 8                    Kirkcaldy        1       NA       NA       NA        1        1
 9                   Lancashire        1       NA       NA       NA       NA       NA
10            Not known/missing       NA        1       NA       NA       NA        1
11                 Outer London        1       NA       NA       NA        1        1
12                  Oxfordshire       NA       NA        1       NA       NA       NA
13                      Renfrew       NA        1        1        1        1       NA
14            Ross and Cromarty        1        1        1        1        1        1
15                         <NA>        4        5        3        4        2        5
user_55035480

The previous answera with gather +count+spread work well, yet not for very large datasets (either large groups or many variables). Here is an alternative, using map-count + join, on a very large data, it seems to be 2 times faster:

library(tidyverse)
N <-  1000000
df <- tibble(x1=sample(letters, N, replace = TRUE),
             x2=sample(letters, N, replace = TRUE),
             x3=sample(letters, N, replace = TRUE),
             x4=sample(letters, N, replace = TRUE),
             x5=sample(letters, N, replace = TRUE))


res1 <- map(c("x1", "x2", "x3", "x4", "x5"), function(x) select_at(df, x) %>%  count(!!rlang::sym(x)) %>% 
         rename(value=!!rlang::sym(x),
                !!rlang::sym(x):=n)) %>% 
  reduce(full_join, by = "value")

res2 <- df %>% 
  tidyr::gather(variable, value) %>% 
  dplyr::group_by(variable, value) %>%
  dplyr::count() %>% dplyr::ungroup()%>%
  tidyr::spread(variable, n)

all.equal(res1, res2)
#> [1] TRUE

library(microbenchmark)
microbenchmark(s1=map(c("x1", "x2", "x3", "x4", "x5"), function(x) select_at(df, x) %>%  count(!!rlang::sym(x)) %>% 
                     rename(value=!!rlang::sym(x),
                            !!rlang::sym(x):=n)) %>% 
                 reduce(full_join, by = "value"),
               s2= df %>% 
                 tidyr::gather(variable, value) %>% 
                 dplyr::group_by(variable, value) %>%
                 dplyr::count() %>% dplyr::ungroup()%>%
                 tidyr::spread(variable, n),
               times = 50, check = "equal")
#> Unit: milliseconds
#>  expr      min       lq     mean   median       uq      max neval
#>    s1 214.9027 220.2292 241.8811 229.0913 242.2507 368.5147    50
#>    s2 412.8934 447.5347 515.2612 528.0221 561.7649 692.5999    50

Created on 2020-05-19 by the reprex package (v0.3.0)

user_24738910
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