按组计算值之间的差异并匹配时间

时间:2018-05-02 19:55:10

标签: r dplyr

对于每只鸟,我想计算不同日期(Tb_Periods)的平均每小时体温(Tb)测量值之间的差异。我的目标是能够比较BirdX的Tb从0900 PreI到09:00 DayI,10:00 PreI到10:00 PostI等的变化.Tb_Period表示操作前的时间(PreI),操作日(DayI)和后期操作(PostI)。我最初的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

到目前为止我做了什么:每只小鸟在48小时的时间内每10分钟测量一次体温,所以我首先使用dplyr计算每只小鸡每小时的平均Tb:

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

结果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

现在我想要的是一个看起来像这样的df:

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

根据Calculate difference between values in consecutive rows by group的回答,我尝试了以下各种变体(使用dplyrs排列函数):

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

但要继续为Tb_Diff列获取NAs。解决此问题的最佳方法是什么(理想情况下使用dplyr)?

1 个答案:

答案 0 :(得分:2)

你快到了!关键是将Tb_Period转换为有序因子,这样PreI被视为&#34;小于&#34; DayI,反过来小于PostI。一旦建立,我们可以按每个鸟和小时分组,并按Tb_Period排序,以确保以正确的顺序计算差异:

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