How to Count Occurrences Combinations in Data.Table in R

How to count occurrences combinations in data.table in R

You just need to add by=list(a,b).

DT1[,count_combination_in_dt2:=nrow(DT2[J(a,b),nomatch=0]), by=list(a,b)]
DT1
## 
##    a b count_combination_in_dt2
## 1: 3 8                        3
## 2: 2 3                        1

EDIT: Some more details: In your original version, you used DT2[DT1, nomatch=0] (because you used all a, b combinations. If you want to use J(a,b) for each a, b combination separately, you need to use the by argument. The data.table is then grouped by a, b and the nrow(...) is evaluated within each group.

How to find all combinations in column and count occurrences in data

If I have understood you correctly, you need to group_by PersonID and paste the all the unique Animals in the group and count the number of occurrence of their combination which can be done counting the number of rows in the group (n()) and dividing it by number of distinct values (n_distinct).

library(dplyr)

df %>%
  group_by(PersonID) %>%
  summarise(AnimalComb = paste(unique(Animal), collapse = ""), 
            CountbyID = n() / n_distinct(Animal)) 

#  PersonID AnimalComb  CountbyID
#     <int> <chr>           <dbl>
#1        1 DogBird             1
#2        2 SnakeSpider         1
#3        3 Cat                 1
#4        4 CatDog              1

R data table unique record count based on all combination of a given list of values from 2 columns

In base R, you can do:

data.frame(table(dt))

        Var1       Var2 Freq
1 Col1Value1 Col2Value1    1
2 Col1Value2 Col2Value1    1
3 Col1Value3 Col2Value1    1
4 Col1Value1 Col2Value2    1
5 Col1Value2 Col2Value2    0
6 Col1Value3 Col2Value2    1
7 Col1Value1 Col2Value3    1
8 Col1Value2 Col2Value3    1
9 Col1Value3 Col2Value3    1

Counting the number of occurrences of a combination of values in r

I found this temporary solution (thanks to iod's solution on the first example using group_by and mutate).

df1 %>% filter(is.na(old_fase) | old_fase == "Finished") %>% # indicates the beginning of a new proces
group_by(id) %>% 
mutate(occurrence = row_number()) %>% 
select(id, time, occurrence) %>% 
left_join(df1, ., by = c("id", "time")) %>% 
fill(occurrence)

count number of combinations by group

Create a "combination" column in summarise, we can count this column afterwards.

An easy way to count the category is to order them at the beginning, then in this case they will all be in the same order.

library(dplyr)

dd %>% 
  group_by(id) %>% 
  arrange(id, cat) %>% 
  summarize(combination = paste0(cat, collapse = "-"), .groups = "drop") %>% 
  count(combination)

# A tibble: 3 x 2
  combination     n
  <chr>       <int>
1 c-d-f           1
2 c-f             2
3 d-f             2

Count occurrences of factors across multiple columns in grouped dataframe

You can stack col1 & col2 together, count the number of each combination, and then transform the table to a wide form.

library(dplyr)
library(tidyr)

df %>%
  pivot_longer(col1:col2) %>%
  count(grp, name, value) %>%
  pivot_wider(grp, names_from = c(name, value), names_sort = TRUE,
              values_from = n, values_fill = 0)

# A tibble: 3 x 6
  grp   col1_A col1_B col2_B col2_C col2_D
  <chr>  <int>  <int>  <int>  <int>  <int>
1 a          1      2      2      0      1
2 b          2      0      0      2      0
3 c          1      2      0      2      1

A base solution (Thank @GKi to refine the code):

table(cbind(df["grp"], col=do.call(paste0, stack(df[-1])[2:1])))

   col
grp col1A col1B col2B col2C col2D
  a     1     2     2     0     1
  b     2     0     0     2     0
  c     1     2     0     2     1

How to count number of occurrences of data combinations and save in a matrix in R?

You could initialize the matrix with zeros instead of NA and then increment the matrix value like this:

pre = c(3,1,3,2,2,4,3,5,3,4,6,5,6,5,4,5,6,6,5,6,5,7,6,7,7,7,4,8,4,8,8,4,4,8,3,9,8,6,9,8)
post = c(4,3,5,3,4,6,5,6,5,4,5,6,6,5,6,5,7,6,7,7,7,4,8,4,8,8,4,4,8,3,9,8,6,9,8,8,9,7,9,9)
df = data.frame(pre,post)

matrix = matrix(0, nrow=20, ncol=20)
colnames(matrix) = seq(1,20,1)
rownames(matrix) = seq(1,20,1)

for (i in 1:40){matrix[df$post[i], df$pre[i]] = matrix[df$post[i], df$pre[i]] + 1}

By the way, the setting of the matrix colnames and rownames is not needed if you don't need it for any other reasons.

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