与之前的question保持一致,假设我有一个数据集:
Date rain code
2009-04-01 0.0 0
2009-04-02 0.0 0
2009-04-03 0.0 0
2009-04-04 0.7 1
2009-04-05 54.2 1
2009-04-06 0.0 0
2009-04-07 5.0 1
2009-04-08 9.0 0
2009-04-09 0.0 0
2009-04-10 0.0 0
2009-04-11 0.0 0
2009-04-12 5.3 1
2009-04-13 10.1 1
2009-04-14 6.0 1
2009-04-15 8.7 1
2009-04-16 0.0 0
2009-04-17 0.0 0
2009-04-18 0.0 0
2009-04-19 2.0 0
2009-04-20 3.0 0
2009-04-21 0.0 0
2009-04-22 0.0 0
2009-04-23 0.0 0
2009-04-24 0.0 0
2009-04-25 4.3 1
2009-04-26 42.2 1
2009-04-27 45.6 1
2009-04-28 12.6 1
2009-04-29 6.2 1
2009-04-30 1.0 1
DT = structure(list(Date = structure(c(14335, 14336, 14337, 14338,
14339, 14340, 14341, 14342, 14343, 14344, 14345, 14346, 14347,
14348, 14349, 14350, 14351, 14352, 14353, 14354, 14355, 14356,
14357, 14358, 14359, 14360, 14361, 14362, 14363, 14364), class = "Date"),
rain = c(0, 0, 0, 0.7, 54.2, 0, 5, 9, 0, 0, 0, 5.3, 10.1,
6, 8.7, 0, 0, 0, 2, 3, 0, 0, 0, 0, 4.3, 42.2, 45.6, 12.6,
6.2, 1), code = c(0L, 0L, 0L, 1L, 1L, 0L, 1L, 0L, 0L, 0L,
0L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L,
1L, 1L, 1L, 1L, 1L)), .Names = c("Date", "rain", "code"), row.names = c(NA,
-30L), class = "data.frame")
当code为1时,我试图折叠数据集以获得rain连续值的总和。我需要在第二天之前获得它们的总和事件,包容性。例如,我想分别获取2009-04-13到2009-04-06和2009-04-07到2009-04-08的降雨值总和。因此,我试图找到方法来定义 code 何时等于 1 以及第二天(含)。最终产品应该如下所示:
Date rain code
2009-04-01 0.0 0
2009-04-02 0.0 0
2009-04-03 0.0 0
2009-04-06 54.9 1
2009-04-08 14.0 1
2009-04-09 0.0 0
2009-04-10 0.0 0
2009-04-11 0.0 0
2009-04-16 30.1 1
2009-04-17 0.0 0
2009-04-18 0.0 0
2009-04-19 2.0 0
2009-04-20 3.0 0
2009-04-21 0.0 0
2009-04-22 0.0 0
2009-04-23 0.0 0
2009-04-24 0.0 0
2009-04-30 111.9 1 (if last entry of data frame)
任何有关上述问题的帮助将不胜感激。
最佳答案
这是一种方法:
library(data.table)
setDT(DT)
res = DT[, .(
Date = Date[.N],
rain = sum(rain),
code = code[1L]
), by=.(g = cumsum(shift(!code, fill=FALSE)))]
res[, g := NULL]
Date rain code
1: 2009-04-01 0.0 0
2: 2009-04-02 0.0 0
3: 2009-04-03 0.0 0
4: 2009-04-06 54.9 1
5: 2009-04-08 14.0 1
6: 2009-04-09 0.0 0
7: 2009-04-10 0.0 0
8: 2009-04-11 0.0 0
9: 2009-04-16 30.1 1
10: 2009-04-17 0.0 0
11: 2009-04-18 0.0 0
12: 2009-04-19 2.0 0
13: 2009-04-20 3.0 0
14: 2009-04-21 0.0 0
15: 2009-04-22 0.0 0
16: 2009-04-23 0.0 0
17: 2009-04-24 0.0 0
18: 2009-04-30 111.9 1
工作原理:
shift
正在从前一行获取值- 当像
!code
这样的逻辑值相加时,TRUE/FALSE 被视为 1/0 .N
是by=
组中的最后一行
一般语法为 DT[, j, by]
,其中 j
是使用数据的每个 by
子集计算的。
关于r - 连续日值的总和,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38908841/