dplyr differences between pairs in nested groups

Question

I'd like to use dplyr to calculate differences in value between people nested in pair by session .

dat <- data.frame(person=c(rep(1, 10), 
                           rep(2, 10),
                           rep(3, 10), 
                           rep(4, 10),
                           rep(5, 10),
                           rep(6, 10),
                           rep(7, 10),
                           rep(8, 10)),
                  pair=c(rep(1, 20),
                         rep(2, 20),
                         rep(3, 20),
                         rep(4, 20)),
                  condition=c(rep("NEW", 10), 
                              rep("OLD", 10),
                              rep("NEW", 10), 
                              rep("OLD", 10),
                              rep("NEW", 10),
                              rep("OLD", 10),
                              rep("NEW", 10),
                              rep("OLD", 10)),
                  session=rep(seq(from=1, to=10, by=1), 8),
                  value=c(0, 2, 4, 8, 16, 16, 18, 20, 20, 20,
                          0, 1, 1, 2, 4, 5, 8, 12, 15, 15,
                          0, 2, 8, 10, 15, 16, 18, 20, 20, 20,
                          0, 4, 4, 6, 6, 8, 10, 12, 12, 18,
                          0, 6, 8, 10, 16, 16, 18, 20, 20, 20,
                          0, 2, 2, 3, 4, 8, 8, 8, 10, 12,
                          0, 10, 12, 16, 18, 18, 18, 20, 20, 20,
                          0, 2, 2, 8, 10, 10, 11, 12, 15, 20)
                  )

For instance, person 1 and 2 make a pair ( pair==1 ):

• person==1 & session==2 : 2
• person==2 & session==2 : 1

Difference ( NEW - OLD ) is 2-1=1 .

Here's what I have tried so far. I think I need to group_by() first and then summarise() , but I have not cracked this nut.

dat %>%
    mutate(session = factor(session)) %>%
    group_by(condition, pair, session) %>%
    summarise(pairDiff = value-first(value))

Desired output:

Answer 1

Your output can be obtained by:

dat %>% group_by(pair,session) %>% arrange(condition) %>% summarise(diff = -diff(value))
Source: local data frame [40 x 3]
Groups: pair [?]

# A tibble: 40 x 3
    pair session  diff
   <dbl>   <dbl> <dbl>
 1     1       1     0
 2     1       2     1
 3     1       3     3
 4     1       4     6
 5     1       5    12
 6     1       6    11
 7     1       7    10
 8     1       8     8
 9     1       9     5
10     1      10     5
# ... with 30 more rows

The arrange ensures that NEW and OLD are in the correct positions, but the solution does depend on there being exactly 2 values for each combination of pair and session.

Answer 2

You can spread condition to headers and then do the subtraction NEW - OLD :

library(dplyr); library(tidyr)

dat %>% 
    select(-person) %>% 
    spread(condition, value) %>% 
    mutate(diff = NEW - OLD) %>% 
    select(session, pair, diff)

# A tibble: 40 x 3
#   session  pair  diff
#     <dbl> <dbl> <dbl>
# 1       1     1     0
# 2       2     1     1
# 3       3     1     3
# 4       4     1     6
# 5       5     1    12
# 6       6     1    11
# 7       7     1    10
# 8       8     1     8
# 9       9     1     5
#10      10     1     5
# ... with 30 more rows