R指定一个月中的哪一周
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我有一些股票返回每日数据需要变成每周格式。如您所知，仅在周一至周五进行股票交易，我需要将每天的回报加起来以获得累积的每周回报。
我曾考虑使用lubridate的周函数，但是lubridate如何知道本周的开始时间？如何让lubridate使用工作日功能识别周，即“星期一”到“星期五”是一周？
我已经考虑过编写一个循环，例如：如果“星期一”到“星期五”在数据中，那么我会打电话给这一周。但是对于第二周，我应该用R来知道我们进入第二周了吗？然后当我们到达年底并且我们有52周时，如何重置周计数以便我们进入下一年？
这是输入：
dat = structure(list(date = structure(c(4019, 4022, 4023, 4024, 4025,
4026, 4029, 4030, 4031, 4032, 4033, 4036, 4037, 4038, 4039, 4040,
4043, 4044, 4045, 4046, 4047, 4050, 4051, 4052, 4053, 4054, 4057,
4058, 4059, 4060, 4061, 4065, 4066, 4067, 4068, 4071, 4072, 4073,
4074, 4075), class = "Date"), weekday = c("Friday", "Monday",
"Tuesday", "Wednesday", "Thursday", "Friday", "Monday", "Tuesday",
"Wednesday", "Thursday", "Friday", "Monday", "Tuesday", "Wednesday",
"Thursday", "Friday", "Monday", "Tuesday", "Wednesday", "Thursday",
"Friday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday",
"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Tuesday",
"Wednesday", "Thursday", "Friday", "Monday", "Tuesday", "Wednesday",
"Thursday", "Friday"), COMP = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), week = c(1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8,
9, 9), RET = c(-0.005435, 0.040984, -0.015748, -0.021333, 0.002725,
0.01087, 0.024194, -0.002625, 0.013158, 0.033766, 0, -0.007538,
-0.005063, 0, -0.002545, 0.015306, 0.017588, -0.007407, 0.024876,
-0.009709, 0, -0.029412, 0.010101, 0.0075, -0.004963, 0.027431,
-0.002427, 0.007299, -0.009662, -0.004878, 0.014706, -0.004831,
0.004854, -0.009662, -0.021951, -0.014963, 0.005063, -0.005038,
0.010127, 0)), .Names = c("date", "weekday", "COMP", "week",
"RET"), row.names = c(NA, -40L), class = c("data.table", "data.frame"
))
library(data.table)
setDT(dat)
这是公司1每日回报的1981-01-02至1981-02-27两个月的数据。让我们忽略计算当前的回报并首先关注时间。
week列由week（）函数生成。你可以看到周不是我想要的，它从周三开始到周三结束。
工作日由weekdays（）函数生成。
我想制作，例如1981-01-02作为第1周（因为我们这里只有星期五），1981-01-05到1981-01-09作为第2周反之亦然。
答案
使用lubridate，您可以使用isoweek定义周列。
library(lubridate)
df[, wk := isoweek(date)]
哪个给你
# date weekday COMP week RET wk
# 1: 1981-01-02 Friday 1 1 -0.005435 1
# 2: 1981-01-05 Monday 1 1 0.040984 2
# 3: 1981-01-06 Tuesday 1 1 -0.015748 2
# 4: 1981-01-07 Wednesday 1 1 -0.021333 2
# 5: 1981-01-08 Thursday 1 2 0.002725 2
# 6: 1981-01-09 Friday 1 2 0.010870 2
# 7: 1981-01-12 Monday 1 2 0.024194 3
# 8: 1981-01-13 Tuesday 1 2 -0.002625 3
# 9: 1981-01-14 Wednesday 1 2 0.013158 3
# 10: 1981-01-15 Thursday 1 3 0.033766 3
# 11: 1981-01-16 Friday 1 3 0.000000 3
# 12: 1981-01-19 Monday 1 3 -0.007538 4
# 13: 1981-01-20 Tuesday 1 3 -0.005063 4
# 14: 1981-01-21 Wednesday 1 3 0.000000 4
# 15: 1981-01-22 Thursday 1 4 -0.002545 4
# 16: 1981-01-23 Friday 1 4 0.015306 4
# 17: 1981-01-26 Monday 1 4 0.017588 5
# 18: 1981-01-27 Tuesday 1 4 -0.007407 5
# 19: 1981-01-28 Wednesday 1 4 0.024876 5
# 20: 1981-01-29 Thursday 1 5 -0.009709 5
使用dplyr，您可以添加周列
library(dplyr)
df %>%
mutate(wk = isoweek(date))
另一答案
如果您想从数据集开始以来计算星期一......
DT[, wk := {
w = DT[weekday == "Monday"][DT, on=.(date), roll=TRUE, which = TRUE]
if (anyNA(w))
1L + replace(w, is.na(w), 0L)
else
w
}]
这个怎么运作
我们正在对DT的每一行进行滚动连接到DT的子集，其中weekday == "Monday"在子集（on = .(date), roll = TRUE）中滚动到最近的日期并且识别我们所在的子集中的哪些行号（which = TRUE）。
如果第一天不是星期一，我们将丢失值（对于第一个星期一之前的所有天），并且希望将它们替换为1并将所有其他行号递增1。
哦，我猜也有
DT[, wk := (first(weekday) != "Monday") + cumsum(weekday == "Monday")]
...因为如果为FALSE，逻辑条件first(weekday) != "Monday"为0，如果为TRUE，则为1。
另一答案
dat[, wk := .GRP, cut(date, 'week')]
head(dat, 20)
# date weekday COMP week RET wk
# 1: 1981-01-02 Friday 1 1 -0.005435 1
# 2: 1981-01-05 Monday 1 1 0.040984 2
# 3: 1981-01-06 Tuesday 1 1 -0.015748 2
# 4: 1981-01-07 Wednesday 1 1 -0.021333 2
# 5: 1981-01-08 Thursday 1 2 0.002725 2
# 6: 1981-01-09 Friday 1 2 0.010870 2
# 7: 1981-01-12 Monday 1 2 0.024194 3
# 8: 1981-01-13 Tuesday 1 2 -0.002625 3
# 9: 1981-01-14 Wednesday 1 2 0.013158 3
# 10: 1981-01-15 Thursday 1 3 0.033766 3
# 11: 1981-01-16 Friday 1 3 0.000000 3
# 12: 1981-01-19 Monday 1 3 -0.007538 4
# 13: 1981-01-20 Tuesday 1 3 -0.005063 4
# 14: 1981-01-21 Wednesday 1 3 0.000000 4
# 15: 1981-01-22 Thursday 1 4 -0.002545 4
# 16: 1981-01-23 Friday 1 4 0.015306 4
# 17: 1981-01-26 Monday 1 4 0.017588 5
# 18: 1981-01-27 Tuesday 1 4 -0.007407 5
# 19: 1981-01-28 Wednesday 1 4 0.024876 5
# 20: 1981-01-29 Thursday 1 5 -0.009709 5
注意：这与dt[, wk := lubridate::isoweek(date)]的结果相同，除非数据未按日期排序。在这种情况下，我的解决方案以相同的方式对周数进行分组，但wk不会按升序排列。第一周可能会给6等。
另一答案
这是一个更简单的方法（我想更容易理解）来解决这个问题：
# if its a monday, mark as 1, 2, 3 and so on
dt[weekday == 'Monday', is_week := seq(.N)]
# forward fill the missing values
library(zoo)
dt[, is_week := na.locf(is_week,na.rm = F, fromLast = F)]
dt[is.na(is_week), is_week := 0]
# find weekly average return
dt[, mean(RET), is_week]
is_week V1
1: 0 -0.005435000
2: 1 0.003499600
3: 2 0.013698600
4: 3 0.000032000
5: 4 0.005069600
6: 5 0.002131400
7: 6 -0.002950222
8: 7 -0.000962200
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