R：成对的t.test删除不成对的观测值（长格式）

Question

I have the following data with paired observations in long format. I am trying to do a paired t-test along time variable in R on the long format, but by first detecting obs that are not available in both time 1 and 2 (obs B and E in this case), and then perhaps creating a new dataframe with the observations in order. Is there a way to do this without reshaping the data into wide format first? Help and suggestions would be appreciated, R newbie here.

obs time value
A   1    5.5
B   1    7.1
C   1    4.3
D   1    6.4
E   1    6.6
F   1    5.6
G   1    6.6
A   2    6.5
C   2    6.7
D   2    7.8
F   2    5.7
G   2    8.9   

Answer 1

As an alternative to the use of duplicated in @CPak's long-format answer you can group by the observation and filter for where the count of the observations is not equal to 1:

library(dplyr)

p = 
  group_by(df, obs) %>%
  filter(n() != 1) %>%
  arrange(time, obs) %>%
  ungroup()

Leads to the same result in any event, as when applying the t-test as shown in @CPak's answer:

ans <- with(p, t.test(value ~ time, paired=TRUE))

> ans

    Paired t-test

data:  value by time
t = -3.3699, df = 4, p-value = 0.02805
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.6264228 -0.2535772
sample estimates:
mean of the differences 
                  -1.44 

Answer 2

You can use duplicated both in the forward and reverse fromLast=TRUE direction to filter your data

library(dplyr)
p <- df %>%
       filter(duplicated(obs) | duplicated(obs, fromLast=TRUE)) %>%
       arrange(time, obs)

   # obs time value
# 1    A    1   5.5
# 2    C    1   4.3
# 3    D    1   6.4
# 4    F    1   5.6
# 5    G    1   6.6
# 6    A    2   6.5
# 7    C    2   6.7
# 8    D    2   7.8
# 9    F    2   5.7
# 10   G    2   8.9

Then perform the paired t.test

ans <- with(p, t.test(value ~ time, paired=TRUE))

        # Paired t-test

# data:  value by time
# t = -3.3699, df = 4, p-value = 0.02805
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
 # -2.6264228 -0.2535772
# sample estimates:
# mean of the differences 
                  # -1.44    

Your original data

df <- read.table(text="obs time value
A   1    5.5
B   1    7.1
C   1    4.3
D   1    6.4
E   1    6.6
F   1    5.6
G   1    6.6
A   2    6.5
C   2    6.7
D   2    7.8
F   2    5.7
G   2    8.9", header=TRUE, stringsAsFactors=FALSE)