R group by show count of all factor levels even when zero dplyr

We can convert 'ID' to factor with levels specified and just use table

table(factor(dat$ID, levels = letters))

---

Or using the same with tidyverse

library(tidyverse)
dat %>% 
  mutate(ID=factor(ID, levels = letters)) %>% 
  complete(ID) %>%
  group_by(ID) %>%
  summarise(no_rows = n())

In the accepted answer by akrun, table() works, but the tidyverse answer gives inaccurate counts (see below). Instead use the .drop = FALSE option:

library(tidyverse)
set.seed(1)
dat <- data.frame(ID = sample(letters,50,rep=TRUE))
dat %>%
  mutate(ID = factor(ID, levels = letters)) %>%
  count(ID, name = "no_rows", .drop = F) %>%
  print.data.frame()
#>    ID no_rows
#> 1   a       3
#> 2   b       2
#> 3   c       1
#> 4   d       1
#> 5   e       3
#> 6   f       3
#> 7   g       2
#> 8   h       1
#> 9   i       2
#> 10  j       5
#> 11  k       1
#> 12  l       3
#> 13  m       0
#> 14  n       3
#> 15  o       3
#> 16  p       0
#> 17  q       0
#> 18  r       1
#> 19  s       1
#> 20  t       3
#> 21  u       3
#> 22  v       1
#> 23  w       2
#> 24  x       0
#> 25  y       5
#> 26  z       1

^{Created on 2019-11-22 by the reprex package (v0.3.0)}

Note that we expect nonzero counts for all letters but m, p, q, and x:

set.seed(1)
dat <- data.frame(ID = sample(letters,50,rep=TRUE))
levels(dat$ID)
#>  [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "n" "o" "r" "s" "t"
#> [18] "u" "v" "w" "y" "z"

But if we use complete() we get ones instead:

set.seed(1)
dat <- data.frame(ID = sample(letters,50,rep=TRUE))
dat %>% 
  mutate(ID=factor(ID, levels = letters)) %>% 
  complete(ID) %>%
  group_by(ID) %>%
  summarise(no_rows = n()) %>%
  print.data.frame()
#>    ID no_rows
# ...
#> 12  l       3
#> 13  m       1  # should be 0
#> 14  n       3
#> 15  o       3
#> 16  p       1  # should be 0
#> 17  q       1  # should be 0
#> 18  r       1
#> 19  s       1
#> 20  t       3
#> 21  u       3
#> 22  v       1
#> 23  w       2
#> 24  x       1  # should be 0
#> 25  y       5
#> 26  z       1

That's because complete() actually adds a single m, p, q, and x to ID so it contains at least one of each letter.