我有大约20年的日常数据，我使用month（）函数提取月份。现在我希望这些月份在一年结束时不会改变为1（1月）。我希望他们继续前进13,14,15等等。 R中有功能吗？或者我该怎么做？
例如我有这个：
1993-01-04 1.953 1
1993-01-05 1.978 1
1993-01-06 2.007 1
1993-01-07 1.957 1
1993-01-08 1.955 1
1993-01-11 1.982 1
1993-01-12 1.958 1
1993-01-13 1.924 1
1993-01-14 1.887 1
1993-01-15 1.971 1
1993-01-19 2.098 1
1993-01-20 2.087 1
1993-01-21 2.062 1
1993-01-22 2.056 1
1993-01-25 2.051 1
1993-01-26 2.038 1
1993-01-27 1.990 1
1993-01-28 2.032 1
1993-01-29 2.027 1
1993-02-01 2.083 2
1993-02-02 2.027 2
1993-02-03 2.009 2
1993-02-04 2.058 2
1993-02-05 2.073 2
1993-02-08 2.071 2
1993-02-09 2.022 2
1993-02-10 2.077 2
1993-02-11 2.116 2
1993-02-12 2.049 2
1993-02-15 2.086 2
1993-02-16 2.072 2
1993-02-17 2.061 2
1993-02-18 2.109 2
1993-02-19 2.082 2
1993-02-23 2.027 2
1993-02-24 1.962 2
1993-02-25 1.966 2
1993-02-26 2.001 2
1993-03-01 2.027 3
1993-03-02 2.013 3
1993-03-03 1.989 3
1993-03-04 1.982 3
1993-03-05 1.949 3
1993-03-08 1.966 3
1993-03-09 1.987 3
1993-03-10 1.989 3
1993-03-11 2.009 3
1993-03-12 1.978 3
1993-03-15 1.970 3
1993-03-16 2.009 3
1993-03-17 1.938 3
1993-03-18 1.930 3
1993-03-19 1.914 3
1993-03-22 1.892 3
1993-03-23 1.902 3
1993-03-24 1.910 3
1993-03-25 1.897 3
1993-03-26 1.853 3
1993-03-29 1.879 3
1993-03-30 1.920 3
1993-03-31 1.906 3
1993-04-01 1.894 4
1993-04-02 1.901 4
1993-04-05 1.888 4
1993-04-07 1.905 4
1993-04-08 1.938 4
1993-04-13 1.935 4
1993-04-14 1.945 4
1993-04-15 1.926 4
1993-04-16 1.895 4
1993-04-19 1.885 4
1993-04-20 1.859 4
1993-04-21 1.858 4
1993-04-22 1.890 4
1993-04-23 1.848 4
1993-04-26 1.861 4
1993-04-27 1.837 4
1993-04-28 1.818 4
1993-04-29 1.812 4
1993-04-30 1.835 4
1993-05-03 1.893 5
1993-05-04 1.847 5
1993-05-05 1.841 5
1993-05-06 1.875 5
1993-05-07 1.885 5
1993-05-10 1.893 5
1993-05-11 1.891 5
1993-05-12 1.851 5
1993-05-13 1.889 5
1993-05-14 1.859 5
1993-05-17 1.827 5
1993-05-18 1.827 5
1993-05-19 1.803 5
1993-05-21 1.823 5
1993-05-24 1.812 5
1993-05-25 1.842 5
1993-05-26 1.783 5
1993-05-27 1.797 5
1993-05-28 1.784 5
1993-06-01 1.777 6
1993-06-02 1.774 6
1993-06-03 1.784 6
1993-06-04 1.804 6
1993-06-07 1.858 6
1993-06-08 1.868 6
1993-06-09 1.852 6
1993-06-11 1.888 6
1993-06-14 1.980 6
1993-06-15 1.972 6
1993-06-16 1.972 6
1993-06-17 2.018 6
1993-06-18 2.018 6
1993-06-21 2.075 6
1993-06-22 2.111 6
1993-06-23 2.140 6
1993-06-24 2.175 6
1993-06-25 2.174 6
1993-06-28 2.192 6
1993-06-29 2.175 6
1993-06-30 2.145 6
1993-07-01 2.104 7
1993-07-02 2.069 7
1993-07-05 2.076 7
1993-07-06 2.102 7
1993-07-07 2.174 7
1993-07-08 2.372 7
1993-07-09 2.402 7
1993-07-12 2.392 7
1993-07-13 2.378 7
1993-07-14 2.379 7
1993-07-15 2.381 7
1993-07-16 2.358 7
1993-07-19 2.413 7
1993-07-20 2.403 7
1993-07-21 2.373 7
1993-07-22 2.423 7
1993-07-23 2.393 7
1993-07-26 2.423 7
1993-07-27 2.382 7
1993-07-28 2.354 7
1993-07-29 2.314 7
1993-07-30 2.240 7
1993-08-02 2.272 8
1993-08-03 2.256 8
1993-08-04 2.282 8
1993-08-05 2.262 8
1993-08-06 2.288 8
1993-08-09 2.296 8
1993-08-10 2.264 8
1993-08-11 2.296 8
1993-08-12 2.333 8
1993-08-13 2.353 8
1993-08-16 2.373 8
1993-08-17 2.408 8
1993-08-18 2.398 8
1993-08-19 2.382 8
1993-08-20 2.368 8
1993-08-23 2.332 8
1993-08-24 2.362 8
1993-08-25 2.358 8
1993-08-26 2.286 8
1993-08-27 2.286 8
1993-08-30 2.313 8
1993-08-31 2.359 8
1993-09-01 2.293 9
1993-09-02 2.313 9
1993-09-03 2.283 9
1993-09-06 2.253 9
1993-09-07 2.237 9
1993-09-08 2.255 9
1993-09-09 2.279 9
1993-09-10 2.258 9
1993-09-13 2.253 9
1993-09-14 2.238 9
1993-09-15 2.238 9
1993-09-16 2.223 9
1993-09-17 2.258 9
1993-09-20 2.266 9
1993-09-21 2.243 9
1993-09-22 2.252 9
1993-09-23 2.303 9
1993-09-24 2.260 9
1993-09-27 2.317 9
1993-09-28 2.283 9
1993-09-29 2.298 9
1993-09-30 2.313 9
1993-10-01 2.315 10
1993-10-04 2.307 10
1993-10-05 2.398 10
1993-10-06 2.508 10
1993-10-07 2.386 10
答案 0 :(得分：0)
尝试这样的事情：
test = c(seq(1,12,by=1),seq(1,12,by=1))
[1] 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12
data.table::rleid(test)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
这可能是一个更完整的答案。
library(data.table)
dt = data.table("Date" = seq.Date(as.Date("2014-12-21"),as.Date("2015-01-01"),by = 1), value = runif(12))
dt[, "Month" := month(Date)]
dt[, "ID" := rleid(Month)]
答案 1 :(得分：0)
我们可以使用base R执行此操作（假设日期已订购）
df1$index <- cumsum(c(TRUE, diff(df1$Month) < 0))