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这个问题类似于我如何做一个条件和，它只在某些日期标准之间看起来但略有不同，并且答案不适合当前问题。主要区别在于基于每个组的日期列可能不一定完整（即，可能缺少某些日期）

输入：

input <- read.table(text="
2017-04-01     A     1
2017-04-02     B     2
2017-04-02     B     2
2017-04-02     C     2
2017-04-02     A     2
2017-04-03     C     3
2017-04-04     A     4
2017-04-05     B     5
2017-04-06     C     6
2017-04-07     A     7
2017-04-08     B     8
2017-04-09     C     9")
colnames(input) <- c("Date","Group","Score")

规则：对于每个组在每个日期，回顾 3 个日历日期（包括当前日期）。计算总和。

预期输出：

    Date Group 3DaysSumPerGroup
    2017-04-01     A                1 #1  previous two dates are not available. partial is allowed
    2017-04-02     A                3 #2+1 both 4-01 and 4-02 are in the range
    2017-04-04     A                6 #4+2
    2017-04-07     A                7 #7
    2017-04-02     B                4 # 2+2 at the same day
    2017-04-05     B                5
    2017-04-08     B                8
    2017-04-02     C                2
    2017-04-03     C                5
    2017-04-06     C                6
    2017-04-09     C                9

我尝试将 rollapply 与 partial=T 一起使用，但结果似乎不正确。

 input %>% 
     group_by(Group) %>% 
     arrange(Date) %>% mutate("3DaysSumPerGroup"=rollapply(data=Score,width=3,align="right",FUN=sum,partial=T,fill=NA,rm.na=T))
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1 回答 1

4

这是使用新的非 equi 连接和by = .EACHIdata.table (v1.9.8+) 中的功能的（据说是有效的）解决方案

library(data.table) #v1.10.4

## Convert to a proper date class, and add another column in order to define the range
setDT(input)[, c("Date", "Date2") := {
  Date = as.IDate(Date)
  Date2 = Date - 2L
  .(Date, Date2)
}]

## Run a non-equi join against the unique Date/Group combination in input
## Sum the Scores on the fly
## You can ignore the second Date column 

input[unique(input, by = c("Date", "Group")), ## This removes the dupes
      on = .(Group, Date <= Date, Date >= Date2), ## The join condition
      .(Score = sum(Score)), ## sum the scores
      keyby = .EACHI] ## Run the sum by each row in unique(input, by = c("Date", "Group"))

#     Group       Date       Date Score
#  1:     A 2017-04-01 2017-03-30     1
#  2:     A 2017-04-02 2017-03-31     3
#  3:     A 2017-04-04 2017-04-02     6
#  4:     A 2017-04-07 2017-04-05     7
#  5:     B 2017-04-02 2017-03-31     4
#  6:     B 2017-04-05 2017-04-03     5
#  7:     B 2017-04-08 2017-04-06     8
#  8:     C 2017-04-02 2017-03-31     2
#  9:     C 2017-04-03 2017-04-01     5
# 10:     C 2017-04-06 2017-04-04     6
# 11:     C 2017-04-09 2017-04-07     9
于 2017-06-07T08:22:03.523 回答