我每隔五天收集一次关于植物发育或物候的数据（使用分类变量和编码编码），沿着横切面划分为78个连续区段。每个物种都在每个区段的横断面上进行调查。
我的研究重复了100年前的历史研究，我保留了最初的物候编码方案，而没有考虑如何在夏天之后分析数据！
我在收集数据时没有考虑的问题是代码遵循一个序列，其中一个代码在夏季早晚重复。具体来说，代码是：
b1 = single flower
b2 = sparse flowers (two or three)
b3 = flowers common (more than three)
B4 = flowering ended
根据原始研究的方法，在夏季为任何开花植物收集的代码序列将类似于b1，b2，b3，b2，b1，b4。请注意，我们每隔五天访问样带，并且代码可能在连续几天内重复，例如b1，b1，b2，b2，b2，b2，b3，b3，b3，b2，b2，b1，b4。
我想重新编码＆＃39; b1＆＃39;和＆＃39; b2＆＃39;代码如下（参见示例和示例数据）：
1.如果＆＃39; b1＆＃39;发生在'b2＆＃39;之前或者＆＃39; b3＆＃39;那应该是＆＃39; b1a＆＃39;如果它发生在＆＃39; b2＆＃39;或者＆＃39; b3＆＃39;那应该是＆＃39; b1b＆＃39;。请注意，有时候没有＆＃39; b2＆＃39;或者＆＃39; b3＆＃39;在观察序列中。
2.如果＆＃39; b2＆＃39;发生在＆＃39; b3＆＃39;之前那应该是＆＃39; b2a＆＃39;如果它发生在＆＃39; b3＆＃39;它应该是＆＃39; b2b＆＃39;。 OR 如果没有＆＃39; b3＆＃39;然后＆＃39; b2＆＃39;应该是＆＃39; b2a＆＃39;。请注意，重要的是要记住，在最后一次出现＆＃39; b3＆＃39;可能有多次观察＆＃b;＆＃39;＆nbsp; （参见示例和示例数据）。
3.考虑一下＆＃39; b1＆＃39;和＆＃39; b2＆＃39;可能会在没有和观察到b3＆＃39;的情况下发生，在这种情况下，两者都会被编码为＆＃39; b1a＆＃39;和＆＃39; b2a＆＃39;。
以下是数据的样子：
Date Segment Species Code
01-Jun-17 1 A b1
06-Jun-17 1 A b1
10-Jun-17 1 A b2
14-Jun-17 1 A b2
19-Jun-17 1 A b2
23-Jun-17 1 A b3
28-Jun-17 1 A b3
03-Jul-17 1 A b2
08-Jul-17 1 A b2
14-Jul-17 1 A b1
19-Jul-17 1 A b4
23-Jul-17 1 A b4
这应该是这样的：
Date Segment Species Code
01-Jun-17 1 A b1
06-Jun-17 1 A b1a
10-Jun-17 1 A b2a
14-Jun-17 1 A b2a
19-Jun-17 1 A b2a
23-Jun-17 1 A b3
28-Jun-17 1 A b3
03-Jul-17 1 A b2b
08-Jul-17 1 A b2b
14-Jul-17 1 A b1b
19-Jul-17 1 A b4
23-Jul-17 1 A b4
以下是示例数据：
Test.Data<- structure(list(Date = structure(c(17318, 17323, 17327, 17331,
17336, 17340, 17345, 17350, 17355, 17361, 17366, 17318, 17323,
17327, 17331, 17336, 17340, 17345, 17350, 17355, 17361, 17366,
17370, 17375, 17318, 17323, 17327, 17331, 17336, 17340, 17345,
17350, 17355, 17361, 17366, 17318, 17323, 17327, 17331, 17336,
17340, 17345, 17350, 17355, 17361, 17366, 17370, 17375, 17355,
17361, 17366, 17370, 17375, 17350, 17355, 17361, 17366, 17370
), class = "Date"), Segment = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 1, 1, 1, 1, 1), Species = c("A", "A", "A", "A", "A", "A",
"A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "B", "B",
"B", "B", "B", "B", "B", "B", "B", "B", "B", "B", "B", "B", "B",
"B", "B", "B", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A",
"A", "A", "A", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C"
), Code = c("b1", "b1", "b2", "b2", "b2", "b3", "b3", "b2", "b2",
"b4", "b4", "b1", "b2", "b2", "b2", "b3", "b3", "b3", "b2", "b2",
"b2", "b1", "b4", "b4", "b1", "b1", "b2", "b2", "b2", "b3", "b3",
"b2", "b2", "b4", "b4", "b1", "b2", "b2", "b2", "b3", "b3", "b3",
"b2", "b2", "b2", "b4", "b4", "b4", "b3", "b3", "b2", "b1", "b4",
"b1", "b1", "b2", "b2", "b4")), .Names = c("Date", "Segment",
"Species", "Code"), row.names = c(NA, -58L), class = "data.frame")
答案 0 :(得分：4)
使用data.table：
library(data.table)
setDT(Test.Data)
Test.Data[, temp := rleid(Code), by = .(Segment, Species)] #unique ids for the sequence of codes
Test.Data[Code == "b2", Code := paste0(Code, letters[rleid(temp)]),
by = .(Segment, Species)] #use the unique ids inside subset
Test.Data[, temp := NULL]
# Date Segment Species Code
# 1: 2017-06-01 1 A b1
# 2: 2017-06-06 1 A b1
# 3: 2017-06-10 1 A b2a
# 4: 2017-06-14 1 A b2a
# 5: 2017-06-19 1 A b2a
# 6: 2017-06-23 1 A b3
# 7: 2017-06-28 1 A b3
# 8: 2017-07-03 1 A b2b
# 9: 2017-07-08 1 A b2b
#10: 2017-07-14 1 A b4
#11: 2017-07-19 1 A b4
#12: 2017-06-01 1 B b1
#13: 2017-06-06 1 B b2a
#14: 2017-06-10 1 B b2a
#15: 2017-06-14 1 B b2a
#16: 2017-06-19 1 B b3
#17: 2017-06-23 1 B b3
#18: 2017-06-28 1 B b3
#19: 2017-07-03 1 B b2b
#20: 2017-07-08 1 B b2b
#21: 2017-07-14 1 B b2b
#</cont>
答案 1 :(得分：2)
您可以使用dplyr包
library(dplyr)
Test.Data %>%
group_by(Species) %>%
mutate(hadb3 = cumsum(Code=="b3")>0) %>%
mutate(Code = ifelse(Code=="b2" & !hadb3,"b2a",Code)) %>%
mutate(Code = ifelse(Code=="b2" & hadb3,"b2b",Code))
结果：
# A tibble: 48 x 5
# Groups: Species [2]
Date Segment Species Code hadb3
<date> <dbl> <chr> <chr> <lgl>
1 2017-06-01 1 A b1 FALSE
2 2017-06-06 1 A b1 FALSE
3 2017-06-10 1 A b2a FALSE
4 2017-06-14 1 A b2a FALSE
5 2017-06-19 1 A b2a FALSE
6 2017-06-23 1 A b3 TRUE
7 2017-06-28 1 A b3 TRUE
8 2017-07-03 1 A b2b TRUE
9 2017-07-08 1 A b2b TRUE
10 2017-07-14 1 A b4 TRUE
# ... with 38 more rows
mutate(hadb3 = cumsum(Code=="b3")>0)创建一个逻辑列，用于检查之前是否出现b3，这足以通过ifelse语句获得结果。