如何用不相等的觀測數匯總時間序列數據與R

Question

我有一個大的數據框（86000行），其中包括幾位患者，每位患者在其逗留期間均進行了幾次驗血（僅進行了3次測試：T1，T2和T3）。 這些患者中有些住院了3天，有些住院了168天。

這只是count函數輸出的一小部分，它顯示了住院時間的巨大變化：

No  Id     Days
148 29757  111
149 30368   36
150 31062   29
151 31993   24
152 32198   51
153 32438    6
154 32836   74
155 32944   24
156 33467   39
157 36108   90
158 36849    6
159 37136    3

我使用匯總來計算均值等，但是我想總結一下誰在逗留期間確實有所改善或惡化。

我認為，這將涉及至少提取第一個和最后一個測試，並取其差（越低越好）。 但是我找不到辦法。

我認為一個更簡單的解決方案是將整個結果轉換為有序數據（根據測試的正常范圍），並查看其中有多少個值異常低或異常高。 不幸的是，幾乎每個病人都有高潮和低潮。

理想情況下，我希望了解幾位患者（或一組患者）隨時間的進展。 但是，由於他們在不同的時間范圍內住院，（過於簡化）的結果是這樣的：

如您所見，第一位患者（紅點）以中等水平開始，迅速惡化（高水平），然后好轉（較低水平）。 第二名患者的進展尚不清楚，因為他/她的住院時間可能很短。

有人可以建議一個入門者（代碼或想法）嗎？ 我用不相等的觀測值檢查了關於多個時間序列圖的 一些問題 ，但是它們對我的情況沒有幫助。 示例數據集（匿名）在這里：

structure(list(Id = c("10200", "10200", "10200", "10200", "10200", 
"10200", "10700", "10700", "10700", "10700", "10700", "10700", 
"10700", "10700", "10700", "10700", "10700", "10700", "10700", 
"10700", "10700", "10766", "10766", "10766", "10766", "10766", 
"10766", "10766", "10766", "10766", "10766", "10766", "10766", 
"10766", "10766", "10766", "10766", "10766", "10766", "10766"
), Date = structure(c(15068, 15068, 15068, 15069, 15069, 15069, 
15072, 15072, 15072, 15072, 15072, 15072, 15073, 15073, 15073, 
15075, 15075, 15075, 15078, 15078, 15078, 15073, 15074, 15074, 
15075, 15075, 15075, 15075, 15076, 15076, 15076, 15078, 15078, 
15078, 15081, 15082, 15083, 15084, 15085, 15085), class = "Date"), 
    Test = c("T1", "T2", "T3", "T1", "T2", "T3", "T1", "T1", 
    "T2", "T2", "T3", "T3", "T1", "T2", "T3", "T1", "T2", "T3", 
    "T1", "T2", "T3", "T1", "T1", "T2", "T1", "T1", "T2", "T2", 
    "T1", "T2", "T3", "T1", "T2", "T3", "T1", "T1", "T2", "T1", 
    "T1", "T2"), Result = c(131, 4.53, 5.4, 108, 3.19, 3.7, 125, 
    NA, 1.26, NA, NA, 3.8, 125, 0.97, 4.2, 73, 0.84, 6.6, 48, 
    0.52, 4.8, 60, 75, 0.83, 52, 51, 0.62, 0.65, 40, 0.57, 4.1, 
    45, 0.54, 3.7, 96, 77, 1.04, 134, 144, 0.95)), .Names = c("Id", 
"Date", "Test", "Result"), row.names = c(3L, 6L, 4L, 2L, 1L, 
5L, 10L, 14L, 9L, 19L, 8L, 11L, 20L, 18L, 7L, 17L, 13L, 21L, 
12L, 15L, 16L, 22L, 28L, 29L, 24L, 31L, 26L, 33L, 34L, 32L, 37L, 
23L, 35L, 25L, 38L, 36L, 30L, 27L, 39L, 40L), class = "data.frame")

Answer 1

我不知道這是不是您想要的，但是您可以使用dplyr包。 下面的代碼將按“ Id”對數據進行分組，然后在Result中查找第一個和最后一個值，最后在新列中計算“差”

mydata <- structure(list(Id=c ( "10200", "10200", "10200", "10200", "10200", "10200", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10700", "10766", "10766", "10766",
"10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766", "10766" ), Date=s tructure(c(15068, 15068, 15068, 15069, 15069, 15069, 15072, 15072, 15072, 15072, 15072, 15072, 15073, 15073,
15073, 15075, 15075, 15075, 15078, 15078, 15078, 15073, 15074, 15074, 15075, 15075, 15075, 15075, 15076, 15076, 15076, 15078, 15078, 15078, 15081, 15082, 15083, 15084, 15085, 15085), class="Date" ), Test=c ( "T1", "T2", "T3", "T1", "T2", "T3", "T1",
"T1", "T2", "T2", "T3", "T3", "T1", "T2", "T3", "T1", "T2", "T3", "T1", "T2", "T3", "T1", "T1", "T2", "T1", "T1", "T2", "T2", "T1", "T2", "T3", "T1", "T2", "T3", "T1", "T1", "T2", "T1", "T1", "T2"), Result=c (131, 4.53, 5.4, 108, 3.19, 3.7, 125, NA, 1.26,
NA, NA, 3.8, 125, 0.97, 4.2, 73, 0.84, 6.6, 48, 0.52, 4.8, 60, 75, 0.83, 52, 51, 0.62, 0.65, 40, 0.57, 4.1, 45, 0.54, 3.7, 96, 77, 1.04, 134, 144, 0.95)), .Names=c ( "Id", "Date", "Test", "Result"), row.names=c (3L, 6L, 4L, 2L, 1L, 5L, 10L, 14L, 9L, 19L,
8L, 11L, 20L, 18L, 7L, 17L, 13L, 21L, 12L, 15L, 16L, 22L, 28L, 29L, 24L, 31L, 26L, 33L, 34L, 32L, 37L, 23L, 35L, 25L, 38L, 36L, 30L, 27L, 39L, 40L), class="data.frame" )

library(dplyr) 

result <- mydata %>%
  group_by(Id) %>%  
  summarise_each(funs(first, last), Result) %>%
  mutate(difference = first - last)
result