CODEWITHSUNDEEP
rdataframe
Mohideen Imran Khan
Mohideen Imran Khan

Reputation: 833

Combining columns in R table function

I am currently learning R and I encountered problems with tabulating data.

I have integer scores in a data frame, model, that range from 1 to 10 (inclusive). When I use the table function, i.e.

table(model$score)

I get the following result:

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

However, I want to tabulate the data in the following format:

  1-2  3-4  5-6  7-8  9-10
   11   15    5   10     5

Is it possible to achieve this with the table function or do I have to seek the help of another function/package? How do I do it then? Is there a similar way for the prop.table function?

Thank you for your help.

Upvotes: 1

Views: 144

Answers (3)

Sotos
Sotos

Reputation: 51592

You could also use zoo package,

library(zoo)
rollapply(table(model$score), 2, by = 2, sum)

Using @Zheyuan Li's example, (updated as per @G.Grothendieck's comment)

tt <- rollapply(table(a), 2, by = 2, sum)
names(tt) <- rollapply(names(table(a)), 2, by = 2, paste, collapse = "-")
tt
# 1-2  3-4  5-6  7-8 9-10 
#  29   45   43   47   36

Upvotes: 4

akrun
akrun

Reputation: 887291

Here is a faster option with RcppRoll and tabulate

library(RcppRoll)
nm1 <- do.call(paste, c(as.data.frame(matrix(1:10, ncol=2, byrow=TRUE)), list(sep="-")))
setNames(roll_sum(tabulate(a),2)[c(TRUE, FALSE)], nm1)
# 1-2  3-4  5-6  7-8 9-10 
#  29   45   43   47   36

Upvotes: 4

Zheyuan Li
Zheyuan Li

Reputation: 73345

Why not simply do this?

x <- table(model$score)
x <- x[c(1,3,5,7,9)] + x[c(2,4,6,8,10)]
names(x) <- c("1-2","3-4","5-6","7-8","9-10")

It does not introduce extra complexity at all.

table will of course give you a vector of length-10, because you have 10 unique levels.

Well, if you insist calling table() to get the result you want, you need to use cut() to classify your data into bins:

set.seed(0); a <- sample(1:10, 200, replace = TRUE)
table(cut(a, breaks = c(0,2,4,6,8,10)))

 (0,2]  (2,4]  (4,6]  (6,8] (8,10] 
    29     45     43     47     36

Change the label? Use labels (inside cut()):

table(cut(a, breaks = c(0,2,4,6,8,10), labels = c("1-2","3-4","5-6","7-8","9-10")))

 1-2  3-4  5-6  7-8 9-10 
  29   45   43   47   36

But you must make sure a is numerical. You will get error if:

a <- factor(a)
table(cut(a, breaks = c(0,2,4,6,8,10)))

Error in cut.default(a, breaks = c(0, 2, 4, 6, 8, 10)) : 
  'x' must be numeric

Upvotes: 6

Related Questions