Calculate difference between values by group and matched for time

Question

For each individual bird, I would like to calculate the difference between average hourly body temperature (Tb) measurements taken on different days (Tb_Periods). My goal is to be able to compare the change in Tb of BirdX from 0900 PreI to 09:00 DayI, 10:00 PreI to 10:00 PostI etc. The Tb_Period represents the time before manipulation(PreI), day-of-manipulation(DayI), and post-manipulation(PostI). My initial df:

    Date_Time           Bird_ID  Tb   Hour  Treatment  Tb_Period
    2018-04-04 11:01:39   3282   42.2  11    Control     PreI
    2018-04-04 12:31:51   3282   41.2  12    Control     PreI
    ....
    2018-04-05 09:16:54   3282   41.9   9    Control     DayI
    ....
    2018-04-06 08:09:57   3282   41.4   8    Control     PostI

What I have done so far: Each bird has body temperature measurements taken every 10 minutes over a timespan of 48hrs, so I first calculated the average Tb of each bird for each hour using dplyr:

    Tb_Averages <- TbData %>% group_by(Tb_Period, Hour, Bird_ID, Treatment)%>% 
                          summarize(meanHourTb = mean(Tb))

Resulting df:

         Tb_Period  Hour  Bird_ID  Treatment  meanHourTb
         PreI        9      3500       LPS    41.55000
         PreI        10     3500       LPS    41.75000       
         ...
         DayI        9      3500       LPS    40.88182
         DayI        10     3500       LPS    41.24000

Now what I would like is a df that looks like this:

         Bird_ID  Hour  Treatment  Tb_Diff 
          3500     9      LPS        -.67 (40.88-41.55)
          3282     9      LPS         .5 (e.g.)

Based on an answer from Calculate difference between values in consecutive rows by group , I have tried variations (with dplyrs arrange function) of:

           Tb_Averages <- Tb_Averages %>%
           group_by(Tb_Period, Bird_ID, Hour) %>%
           mutate(Tb_Diff = c(NA, diff(meanHourTb))))

but keeping getting NAs for the Tb_Diff column. What is the best approach to solve this problem (ideally using dplyr)?

Answer 1

You're nearly there! The key is to convert Tb_Period to an ordered factor, such that PreI is treated as "less than" DayI , which is in turn less than PostI . Once this is established, we can group by each bird and hour, and sort by Tb_Period to ensure that differences are calculated in the correct order:

df <- read.table(text = 'Tb_Period  Hour  Bird_ID  Treatment  meanHourTb
PreI        9      3500       LPS    41.55000
PreI        10     3500       LPS    41.75000       
DayI        9      3500       LPS    40.88182
DayI        10     3500       LPS    41.24000', header = T, stringsAsFactors = F)

df <- df %>% 
  mutate(Tb_Period = factor(Tb_Period, c('PreI', 'DayI', 'PostI'), ordered = T)) %>% 
  group_by(Bird_ID, Hour) %>% 
  mutate(diff = meanHourTb - lag(meanHourTb, 1))

# A tibble: 4 x 6
# Groups:   Bird_ID, Hour [2]
  Tb_Period  Hour Bird_ID Treatment meanHourTb     diff
      <ord> <int>   <int>     <chr>      <dbl>    <dbl>
1      PreI     9    3500       LPS   41.55000       NA
2      PreI    10    3500       LPS   41.75000       NA
3      DayI     9    3500       LPS   40.88182 -0.66818
4      DayI    10    3500       LPS   41.24000 -0.51000