CODEWITHSUNDEEP
rcountcombinations
giac
giac

Reputation: 4309

R - count all combinations

I would like to count all combinations in a data.frame.

The data look like this 

   9 10 11 12
1  1  1  1  1
2  0  0  0  0
3  0  0  0  0
4  1  1  1  1
5  1  1  1  1
6  0  0  0  0
7  1  0  0  1
8  1  0  0  1
9  1  1  1  1
10 1  1  1  1

The output I want is simply 

comb     n 
1 1 1 1  5
0 0 0 0  3 
1 0 0 1  2

Do you know any simple function to do that ?

Thanks 

dt = structure(list(`9` = c(1, 0, 0, 1, 1, 0, 1, 1, 1, 1), `10` = c(1, 
0, 0, 1, 1, 0, 0, 0, 1, 1), `11` = c(1, 0, 0, 1, 1, 0, 0, 0, 
1, 1), `12` = c(1, 0, 0, 1, 1, 0, 1, 1, 1, 1)), .Names = c("9", 
"10", "11", "12"), class = "data.frame", row.names = c(NA, -10L
))

Upvotes: 11

Views: 6645

Answers (6)

Mike Dolan Fliss
Mike Dolan Fliss

Reputation: 237

The dplyr solution above could have been done easier with group_by_all()...

dt %>% group_by_all %>% count

...which as I understand has been superseded by the across() method. Adding in a bit of sorting, and you get:

dt %>% group_by(across()) %>% count %>% arrange(desc(n))

> dt %>% group_by(across()) %>% count %>% arrange(desc(n))
# A tibble: 3 x 5
# Groups:   9, 10, 11, 12 [3]
    `9`  `10`  `11`  `12`     n
  <dbl> <dbl> <dbl> <dbl> <int>
1     1     1     1     1     5
2     0     0     0     0     3
3     1     0     0     1     2

Which you could cast to a matrix if you wished.

Upvotes: 3

akrun
akrun

Reputation: 887851

We can either use data.table or dplyr. These are very efficient. We convert the 'data.frame' to 'data.table' (setDT(dt)), grouped by all the columns of 'dt' (names(dt)), we get the nrow (.N) as the 'Count'

library(data.table)
setDT(dt)[,list(Count=.N) ,names(dt)]

Or we can use a similar methodology using dplyr.

library(dplyr)
names(dt) <- make.names(names(dt))
dt %>%
   group_by_(.dots=names(dt)) %>%
   summarise(count= n())

Benchmarks

In case somebody wants to look at some metrics (and also to backup my claim earlier (efficient!)),

set.seed(24)
df1 <- as.data.frame(matrix(sample(0:1, 1e6*6, replace=TRUE), ncol=6))

akrunDT <-  function() {
  as.data.table(df1)[,list(Count=.N) ,names(df1)]
 }

akrunDplyr <- function() {
  df1 %>%
    group_by_(.dots=names(df1)) %>%
    summarise(count= n())
}

cathG <- function() {
 aggregate(cbind(n = 1:nrow(df1))~., df1, length)
  }

docendoD <- function() {
  as.data.frame(table(comb = do.call(paste, df1)))
}

deena <- function() {
   table(apply(df1, 1, paste, collapse = ","))
}

Here are the microbenchmark results

library(microbenchmark)
microbenchmark(akrunDT(), akrunDplyr(), cathG(), docendoD(),  deena(),
  unit='relative', times=20L)
#   Unit: relative
#        expr       min        lq      mean   median        uq        max neval  cld
#     akrunDT()  1.000000  1.000000  1.000000  1.00000  1.000000  1.0000000    20     a   
#  akrunDplyr()  1.512354  1.523357  1.307724  1.45907  1.365928  0.7539773    20     a   
#       cathG() 43.893946 43.592062 37.008677 42.10787 38.556726 17.9834245    20    c 
#    docendoD() 18.778534 19.843255 16.560827 18.85707 17.296812  8.2688541    20    b  
#       deena() 90.391417 89.449547 74.607662 85.16295 77.316143 34.6962954    20    d

Upvotes: 15

Cath
Cath

Reputation: 24074

A base R solution with aggregate:

aggregate(seq(nrow(dt))~., data=dt, FUN=length)
#  9 10 11 12 seq(nrow(dt))
#1 0  0  0  0             3
#2 1  0  0  1             2
#3 1  1  1  1             5

edit

To get colnames more conformed to your output, you can do:

`colnames<-`(aggregate(seq(nrow(dt))~., data=dt, FUN=length), c("c", "o", "m", "b", "n"))
#  c o m b n
#1 0 0 0 0 3
#2 1 0 0 1 2
#3 1 1 1 1 5

Or, shorter: 

aggregate(cbind(n = 1:nrow(dt))~., dt, length)
#  9 10 11 12 n
#1 0  0  0  0 3
#2 1  0  0  1 2
#3 1  1  1  1 5

Upvotes: 11

CJB
CJB

Reputation: 1819

Also in base R:

Use unique.matrix to get the list of unique combinations.

uncs <- unique.matrix(as.matrix(df), MARGIN = 1)

Then make comparisons and count:

cnts <- colSums(apply(uncs, 1, function(r) apply(dt, 1, function(r2) all(r == r2))))
cbind(comb = apply(uncs, 1, paste), n = cnts)

Upvotes: 3

user5363218
user5363218

Reputation:

Maybe that too : table(apply(dt, 1, paste, collapse = ","))

Upvotes: 5

talat
talat

Reputation: 70336

You could try the following approach using only base R:

as.data.frame(table(comb = do.call(paste, dt)))
#     comb Freq
#1 0 0 0 0    3
#2 1 0 0 1    2
#3 1 1 1 1    5

Upvotes: 8

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