CODEWITHSUNDEEP
rsplitcountcategories
ju.arroyom
ju.arroyom

Reputation: 182

Count the number of consecutive pairs in a vector

Suppose that I have the following vector: 

    V<-c(-1,-1,-1,-1,-1,-1,-1,-1,1,1)

and I want to count the number of CONSECUTIVE pairs in the following categories:

    (1,1), (-1,1), (1,-1), and (-1,-1).

In my example, there are seven consecutive pairs of (-1,-1), one pair of (-1,1), and 1 pair of (1,1).

I am trying to solve this problem using the split function, but I have not been able to figure out the correct factors.

The idea is to pair the 1st observation with the 2nd. The 2nd with the 3rd and so on. The final pair should be the (n-1)th observation with the nth observation.

Upvotes: 4

Views: 752

Answers (4)

David Arenburg
David Arenburg

Reputation: 92282

Maybe something like that

library(zoo)
table(rollapply(V, 2, toString))
# -1, -1  -1, 1   1, 1 
#      7      1      1

Or with base R

table(paste(head(V, -1), tail(V, -1)))
# -1 -1  -1 1   1 1 
#     7     1     1

Or as per @akruns comment, without paste

table(head(V, -1), tail(V, -1))
#    -1 1
# -1  7 1
#  1  0 1

Or

as.data.frame(table(head(V, -1), tail(V, -1)))
#   Var1 Var2 Freq
# 1   -1   -1    7
# 2    1   -1    0
# 3   -1    1    1
# 4    1    1    1

Upvotes: 9

Josh O&#39;Brien
Josh O&#39;Brien

Reputation: 162321

With data.table:

library(data.table)
dt <- rev(data.table(embed(V,2)))
dt[,.N, by=names(dt)]
#    V2 V1 N
# 1: -1 -1 7
# 2: -1  1 1
# 3:  1  1 1

Upvotes: 7

Martin Morgan
Martin Morgan

Reputation: 46856

All consecutive pairs can be represented by two parallel vectors, omitting the last or the first observation

x <- V[-length(V)]
y <- V[-1]

and then cross-tabulating these

> xtabs(~ x + y)
    y
x    -1 1
  -1  7 1
  1   0 1

or in slightly different form

> as.data.frame(xtabs(~x+y))
   x  y Freq
1 -1 -1    7
2  1 -1    0
3 -1  1    1
4  1  1    1

Upvotes: 11

Neal Fultz
Neal Fultz

Reputation: 9687

In base R, add the right half of the pair to the difference to create a score:

V<-c(-1,-1,-1,-1,-1,-1,-1,-1,1,1)
table(diff(V) + V[-1])    
#-1  1  3 
# 7  1  1

This is how each pair is scored:

(-1,-1) => (-1) - (-1) + (-1) = -1
(-1, 1) => ( 1) - (-1) +  (1) =  3
( 1,-1) => (-1) -  (1) + (-1) = -3
( 1, 1) => ( 1) -  (1) +  (1) =  1

Upvotes: 2

Related Questions