CODEWITHSUNDEEP
rdplyr
Chinwobble
Chinwobble

Reputation: 652

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

set.seed(1)
dat <- data.frame(ID = sample(letters,50,rep=TRUE))
dat %>% 
  group_by(ID) %>%
  summarise(no_rows = length(ID))

I have the above code which creates a random sample of letters. However can I make the summarised output show all count levels even when there is zero.

When I run the above code sometimes I get 20 rows and sometimes I get 25, etc. I want this to return 26 rows every time.

Upvotes: 9

Views: 7425

Answers (5)

slowowl
slowowl

Reputation: 694

The tidyverse solution by akrun can be corrected as follows :

set.seed(1)
library(tidyverse)
#> Warning: package 'tibble' was built under R version 4.0.5
dat <- data.frame(ID = sample(letters,50,rep=TRUE))
dat %>% 
  mutate(ID=factor(ID, levels = letters)) %>% 
  group_by(ID) %>%
  summarise(no_rows = n()) %>%
  complete(ID, fill = list(no_rows = 0)) %>%
  print.data.frame()
#> `summarise()` ungrouping output (override with `.groups` argument)
#>    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

Upvotes: 0

Nick
Nick

Reputation: 506

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.

Upvotes: 11

HOSS_JFL
HOSS_JFL

Reputation: 839

Using the accepted solution by akrun I get a wrong result. I expect a frequency table with all letters with no_rows = 0 except "a" with no_rows = 1. 

library(tidyverse)
set.seed(1)
dat <- data.frame(ID = "a")
dat %>% 
  dplyr::mutate(ID=factor(ID, levels = letters)) %>% 
  tidyr::complete(ID) %>%
  dplyr::group_by(ID) %>%
  dplyr::summarise(no_rows = n())

Instead I get a frequency table with no_rows = 1 for all letters:

ID    no_rows
<fct>   <int>
1 a       1
2 b       1
3 c       1
4 d       1
5 e       1
6 f       1
7 g       1
8 h       1
9 i       1
10 j      1
# ... with 16 more rows

Upvotes: 1

Sandipan Dey
Sandipan Dey

Reputation: 23109

This will also work:

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

Upvotes: 1

akrun
akrun

Reputation: 887541

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())

Upvotes: 4

Related Questions