删除重复项，但根据另一列决定保留哪些内容

Question

我有一个纵向收集的样本数据集，它进一步分为三个时间间隔。我正在做纵向分析，但我也在单独查看时间间隔。有些 ID 在每个时间间隔捐赠了多个样本，我想删除重复的 ID。但是，我想在每个时间间隔内保留最接近平均时间点的样本。

关于如何解决这个问题有什么建议吗？

我的数据集看起来像这样：

df <- data.frame(ID = c(1, 2, 3, 4, 5, 2, 3, 4, 5, 4, 3), 
                  TimePoint = c(10, 12, 13, 10, 12, 11, 15, 14, 13, 12, 13),
                  TimeInterval = c("T1", "T1", "T1", "T1", "T1", "T1", "T2", "T2", "T2", "T1", "T2"))

我尝试过

unique()

和

distinct()

，但他们保留了他们遇到的第一个

ID

Answer 1

基础R

df |>
  transform(meanTimePoint = ave(TimePoint, TimeInterval, FUN = mean)) |>
  transform(timeDist = abs(TimePoint - meanTimePoint)) |>
  transform(useThis = ave(timeDist, list(ID, TimeInterval), FUN = function(z) seq_along(z) == which.min(z))) |>
  subset(useThis > 0)
#    ID TimePoint TimeInterval meanTimePoint  timeDist useThis
# 1   1        10           T1      11.42857 1.4285714       1
# 3   3        13           T1      11.42857 1.5714286       1
# 5   5        12           T1      11.42857 0.5714286       1
# 6   2        11           T1      11.42857 0.4285714       1
# 8   4        14           T2      13.75000 0.2500000       1
# 9   5        13           T2      13.75000 0.7500000       1
# 10  4        12           T1      11.42857 0.5714286       1
# 11  3        13           T2      13.75000 0.7500000       1

dplyr

library(dplyr)
df |>
  mutate(
    .by = TimeInterval,
    meanTimePoint = mean(TimePoint),
    timeDist = abs(TimePoint - meanTimePoint)
  ) |>
  slice_min(n = 1, order_by = timeDist, by = c(ID, TimeInterval))
#   ID TimePoint TimeInterval meanTimePoint  timeDist
# 1  1        10           T1      11.42857 1.4285714
# 2  2        11           T1      11.42857 0.4285714
# 3  3        13           T1      11.42857 1.5714286
# 4  4        12           T1      11.42857 0.5714286
# 5  5        12           T1      11.42857 0.5714286
# 6  3        13           T2      13.75000 0.7500000
# 7  4        14           T2      13.75000 0.2500000
# 8  5        13           T2      13.75000 0.7500000