基于具有重复行的两列R数据帧计算唯一值

Question

I have an R data frame with the following format:

column1    column2
NA         NA
1          A
1          A
1          A
NA         NA
NA         NA
2          B
2          B
NA         NA
NA         NA
3          A
3          A
3          A

df = structure(list(column1 = c(NA, 1L, 1L, 1L, NA, NA, 2L, 2L, NA, 
NA, 3L, 3L, 3L), column2 = c(NA, "A", "A", "A", NA, NA, "B", 
"B", NA, NA, "A", "A", "A")), .Names = c("column1", "column2"
), row.names = c(NA, -13L), class = "data.frame")

If the row in one column has an NA , the other column has an NA . The numerical value in column1 describes a unique group, eg rows 2-4 have the group 1 . The column column2 describes the identity of this grouping. In this data frame, the identity is either A , B , C , or D .

My goal is to tally the number of identities by group within the entire data frame: how many A groups there are, how many B groups, etc.

The correct output for this file (so far) is there are 2 A groups and 1 B group.

How would I calculate this?

At the moment, I would try something like this:

length(df[df$column2 == "B"]) ## outputs 2 

but this is incorrect. If I combined column1 and column2 , took only unique values 1A, 2B, 3A, I guess I could count how many times each label from column2 occurs?

(If it's easier, I'm happy to use data.table for this task.)

Answer 1

You can use rle for runs and table for tabulation:

table(rle(df$column2)$values)

# A B 
# 2 1 

See ?rle and ?table for details.

Or, if you want to take advantage of column1 (which is derived from column2 ):

table(unique(df)$column2)

Answer 2

The 'dplyr' package has simple functions for this

library(dplyr)

df %>%
  filter(complete.cases(.) & !duplicated(.)) %>% 
  group_by(column2) %>%
  summarize(count = n())

1. Filter out rows with NA
2. Filter out duplicated rows; these represent individuals in the same group
3. Group by the identity variable (column2)
4. Count the number of unique groups (column1)

Answer 3

If you want to use data.table:

library(data.table)
setDT(df)

d <- df[!is.na(column1), list(n=.N), by=list(column2,column1)]
d <- d[, list(n=.N), by=list(column2)]
d
   column2 n
1:       A 2
2:       B 1

Or more concisely as a one-liner:

setDT(df)[!is.na(column1), .N, by = .(column2, column1)][, .N, by = column2]