问题:
我有赚取和兑换代金券的时间序列数据。优惠券只限一次 3天固定窗口包括他们获得的那一天，例如。 1 月 1 日获得的代金券将在 1 日、2 日和 3 日有效。
我需要做出一个假设，即当涉及到针对他们赚取的赎回时，它是先到先得的。例如。如果我们有数据
Date VouchersEarned VouchersRedeemed
01/01/2020 10 0
02/01/2020 8 9
03/01/2020 4 4
04/01/2020 2 4
05/01/2020 1 4
Date VouchersEarned VouchersRedeemed RemainingVouchers
01/01/2020 10 0 1
02/01/2020 8 9 8
Date VouchersEarned VouchersRedeemed RemainingVouchers
01/01/2020 10 0 0
02/01/2020 8 9 5
03/01/2020 4 4 0
Date VouchersEarned VouchersRedeemed RemainingVouchers
01/01/2020 10 0 0
02/01/2020 8 9 1
03/01/2020 4 4 4
04/01/2020 2 4 2
Date VouchersEarned VouchersRedeemed RemainingVouchers
01/01/2020 10 0 0
02/01/2020 8 9 1**
03/01/2020 4 4 0
04/01/2020 2 4 2
05/01/2020 1 4 1
我需要在任何时间点使用此逻辑流计算有多少凭证过期。我可以在脑海中弄清楚如何做到这一点，像上面一样逐行进行。但是我正在努力了解如何以矢量化方式进行操作，这也可以在 MSSQL 中使用。如果它只是在 R 中，我可以用循环暴力破解它。我也在 Cross Validated 上发布了这个但还没有反馈，因此将其扩展到编程社区。
数据:
我已经包含了一个
R dput最后但这是它的样子编辑:本摘要末尾有一些新数据，这是一个更棘手的示例
Date VouchersEarned VouchersRedeemed ActualActive ActualExpired CumulativeEarned
1 01/01/2020 10 0 10 0 10
2 02/01/2020 8 9 9 0 18
3 03/01/2020 4 4 9 0 22
4 04/01/2020 2 4 7 0 24
5 05/01/2020 1 4 3 1 25
6 06/01/2020 0 1 2 0 25
7 07/01/2020 0 1 0 1 25
8 08/01/2020 0 0 0 0 25
CumulativeRedeemed CumulativeDiff
1 0 10
2 9 9
3 13 9
4 17 7
5 21 4
6 22 3
7 23 2
8 23 2
ActualActive & ActualExpired是我用笔和纸得出的数字。请注意，我将它们放在它们到期的日期上，而不是根据它们获得的日期。要么对我有用，只是改变报告 View 。我可以通过查看 CumulativeEarned 来获得总代金券。 & CumulativeRedeemed然后取差额。我想如果我能得到过期的，那么计算活跃的就很简单了。如果有人有任何想法，我将不胜感激，因为我今天似乎对此有心理障碍!谢谢! :)
编辑:我的实际问题是一个 28 天的窗口，这是一个简化的 View :)
df <- structure(list(Date = c("01/01/2020", "02/01/2020", "03/01/2020",
"04/01/2020", "05/01/2020", "06/01/2020", "07/01/2020", "08/01/2020"
), VouchersEarned = c(10L, 8L, 4L, 2L, 1L, 0L, 0L, 0L), VouchersRedeemed = c(0L,
9L, 4L, 4L, 4L, 1L, 1L, 0L), ActualActive = c(10L, 9L, 9L, 7L,
3L, 2L, 0L, 0L), ActualExpired = c(0L, 0L, 0L, 0L, 1L, 0L, 1L,
0L), CumulativeEarned = c(10L, 18L, 22L, 24L, 25L, 25L, 25L,
25L), CumulativeRedeemed = c(0L, 9L, 13L, 17L, 21L, 22L, 23L,
23L), CumulativeDiff = c(10L, 9L, 9L, 7L, 4L, 3L, 2L, 2L)), class = "data.frame", row.names = c(NA,
-8L))
library(data.table)
dt <- as.data.table(df)
dt[, Date := lubridate::dmy(Date)]
# functional form
findExpiredVouchers <- function(dt, period=3){
# generation of cumulative data
dt[, CumulativeEarned := cumsum(VouchersEarned)]
dt[, CumulativeRedeemed := cumsum(VouchersRedeemed)]
dt[, CumulativeDiff := CumulativeEarned-CumulativeRedeemed]
# I think if we look at the cumulative earned against the cumulative redeemed,
# plus it's lag period from that point, ie the cumulative redeemed in 2 days,
# then we can see for data prior to last 3 which have expired
dt[, LaggedCumulativeRedeemed := shift(CumulativeRedeemed, period-1, type="lead")]
dt[, ExpiredCumulative := CumulativeEarned - LaggedCumulativeRedeemed]
# Now this creates negative values though for eg the first case, I'm not 100%
# if I need to worry about these
dt[ExpiredCumulative < 0, ExpiredCumulative := 0]
# so now it should be the difference in this series that captures the origin
# of an expiration
dt[, Expired := c(NA, diff(ExpiredCumulative))]
dt[1, Expired := ExpiredCumulative]
# and I can shift this by the lag period to get the end result
dt[, OutputExpired := shift(Expired, period, type="lag")]
dt[is.na(OutputExpired), OutputExpired := 0]
# and active
dt[, CumulativeExpired := cumsum(OutputExpired)]
dt[, OutputActive := CumulativeDiff-CumulativeExpired]
}
dt <- findExpiredVouchers(dt, 3)
dt[] # OutputExpired & OutputActive
set.seed(1)
p = 0.2
new_dt <- data.table(
Date = 1:10,
VouchersEarned = sample(0:15, 10, replace=TRUE)
)
new_dt[, CumulativeEarned := cumsum(VouchersEarned)]
# fake VouchersRedeemed
new_dt[, VouchersRedeemed := as.integer(NA)]
new_dt[, CumulativeDiff := CumulativeEarned]
for(i in 1:nrow(new_dt)){
new_value <- sum(rbinom(new_dt$CumulativeDiff[i], 1, p))
new_dt[i, VouchersRedeemed := new_value]
new_dt[i:.N, CumulativeDiff := CumulativeDiff - new_value]
}
new_dt <- findExpiredVouchers(dt=new_dt, 3)
new_dt[] # OutputExpired < 0
Date VouchersEarned VouchersRedeemed OutputExpired
1: 1 8 1 0
2: 2 3 1 0
3: 3 6 3 0
4: 4 0 1 3
5: 5 1 1 2
6: 6 12 5 5
7: 7 6 5 -5
8: 8 10 3 -4
9: 9 13 7 9
10: 10 1 8 -1
最佳答案
使用 tidyverse 库，特别是 dplyr , 和 magrittr ，你可以写这个代码
df %<>%
mutate(VouchersEarnedCumsum = cumsum(VouchersEarned),
VouchersRedeemedCumsum = cumsum(VouchersRedeemed),
VouchersCumsumDifference = VouchersEarnedCumsum - lead(VouchersRedeemedCumsum, 2, default = max(VouchersRedeemedCumsum)),
VouchersCumsumDifference = as.numeric(VouchersCumsumDifference),
VouchersCumsumDifference = case_when(VouchersCumsumDifference < 0 ~ 0, T ~ VouchersCumsumDifference),
ExpiredVouchers = VouchersCumsumDifference - lag(VouchersCumsumDifference, default = 0),
ExpiredVouchersInDate = lag(ExpiredVouchers, 3, default = 0))
# A tibble: 8 x 8
Date VouchersEarned VouchersRedeemed VouchersEarnedCumsum VouchersRedeemedCumsum VouchersCumsumDifference ExpiredVouchers ExpiredVouchersInDate
<chr> <int> <int> <int> <int> <dbl> <dbl> <dbl>
1 01/01/2020 10 0 10 0 0 0 0
2 02/01/2020 8 9 18 9 1 1 0
3 03/01/2020 4 4 22 13 1 0 0
4 04/01/2020 2 4 24 17 2 1 0
5 05/01/2020 1 4 25 21 2 0 1
6 06/01/2020 0 1 25 22 2 0 0
7 07/01/2020 0 1 25 23 2 0 1
8 08/01/2020 0 0 25 23 2 0 0
我不确定此代码是否适用于您将要面对的每种情况，我建议对其进行广泛的测试。
编辑
我认为这个更正的代码解决了这个问题。
df %<>%
mutate(VouchersEarnedCumsum = cumsum(VouchersEarned),
VouchersRedeemedCumsum = cumsum(VouchersRedeemed),
VouchersCumsumDifference = VouchersEarnedCumsum - lead(VouchersRedeemedCumsum, 2, default = max(VouchersRedeemedCumsum)),
VouchersCumsumDifference = as.numeric(VouchersCumsumDifference),
VouchersCumsumDifference = case_when(VouchersCumsumDifference < 0 ~ 0, T ~ VouchersCumsumDifference),
VouchersCumsumDifference = accumulate(VouchersCumsumDifference, max),
ExpiredVouchers = VouchersCumsumDifference - lag(VouchersCumsumDifference, default = 0),
ExpiredVouchersInDate = lag(ExpiredVouchers, 3, default = 0))
关于sql - 计算过期的凭证，给定一个固定的窗口，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/65718399/