Question

我有这样的数据：

DT = data.table(Brand = c('Apple', 'Apple'),
                Time1 = c('2015-11', '2016-01'),
                value1 = c(119.7268, 336.8033),
                vaule2 = c(3380, 7710))

我想生成以下新数据：

Brand    Time1         Time2        LapseMonth     value1      value2
Apple    2015-11-01    2015-11-01       0          119.7268    3380
Apple    2015-11-01    2015-12-01       1          286.2842    0
Apple    2015-11-01    2016-01-01       2          286.2842    0
Apple    2015-11-01    2016-02-01       3          267.8142    0
Apple    2015-11-01    2016-03-01       4          286.2842    0
Apple    2015-11-01    2016-04-01       5          277.0492    0
Apple    2015-11-01    2016-05-01       6          286.2842    0
Apple    2015-11-01    2016-06-01       7          277.0492    0
Apple    2015-11-01    2016-07-01       8          286.2842    0
Apple    2015-11-01    2016-08-01       9          286.2842    0
Apple    2015-11-01    2016-09-01       10         277.0492    0
Apple    2015-11-01    2016-10-01       11         286.2842    0
Apple    2015-11-01    2016-11-01       12         157.3224    0
Apple    2016-01-01    2016-01-01       0          336.8033    7710
Apple    2016-01-01    2016-02-01       1          610.9016    0
Apple    2016-01-01    2016-03-01       2          653.0328    0
Apple    2016-01-01    2016-04-01       3          631.9672    0
Apple    2016-01-01    2016-05-01       4          653.0328    0
Apple    2016-01-01    2016-06-01       5          631.9672    0
Apple    2016-01-01    2016-07-01       6          653.0328    0
Apple    2016-01-01    2016-08-01       7          653.0328    0
Apple    2016-01-01    2016-09-01       8          631.9672    0
Apple    2016-01-01    2016-10-01       9          653.0328    0
Apple    2016-01-01    2016-11-01       10         631.9672    0
Apple    2016-01-01    2016-12-01       11         653.0328    0

我在这里解释新数据：
1.我将生成2个新列（Time2＆amp; LapseMonth）
我计算value1
3.最重要的是：
如果Time1 2015 且LapseMonth 12 ，value1 = value2 * days_in_month(Time2) / 366 - 原始值1。
见上文，157.3224 = 3380 * 30/366 - 119.7268。

这是我的代码：

DT[ , Time1 := as.Date(paste(Time1, 01, sep = "-"), "%Y/%m/%d")]
DT[ , rep := ifelse(year(Time1)==2016, 12-month(Time1)+1, 13)][rep(1:.N,rep)]
DT[ , LapseMonth := seq_len(.N)-1, by = Brand,Time1,value2) ]
DT[ , Time2:= Time1 - days(mday(Time1)-1) + months(LapseMonth)]
DT[ , value1 := ifelse(Time1==Time2,value1,value2*days_in_month(Time2)/366)]
DT[ , value2 := ifelse(Time1==Time2,value2,0)]

ifelse 2015 时，我不知道value1如何使用Time1＆amp; LapseMonth 12 任何的想法？

DT[ , value1:=if(Time1==Time2 & LapseMonth==12) value2*days_in_month(time2)/366-value1]

然而，我收到了一些警告：

Warning message:
In if (PurshasedDate == EXPMTH & LapseMonth == 12) WP * days_in_month(EXPMTH)/366 -  :
the condition has length > 1 and only the first element will be used

Answer 1

我仍然不理解输入到输出的映射，但无论如何。您的主要问题是如何使用ifelse来确定当月的天数。答案是：不要。

相反，只需使用查找表直接获取天数：

monthdays = data.table(month = sprintf('%02d', 1:12),
                       ndays = c(31, 29, 31, 30, 31, 30, 
                                 31, 31, 30, 31, 30, 31),
                       key = 'month')

DT[ , {
  Time2 = seq.Date(Time1, as.Date('2016-12-01'), by = 'month')
  Time2 = Time2[seq_len(min(13L, length(Time2)))]
  LapseMonth = seq_along(Time2) - 1L
  value1 = value2*monthdays[format(Time2, '%m'), ndays]/366 - value1
  .(Brand = Brand, Time2 = Time2,
    LapseMonth = LapseMonth,
    value1 = value1,
    value2 = c(value2, rep(0, length(LapseMonth) - 1)))
  }, by = Time1]
#          Time1 Brand      Time2 LapseMonth   value1 value2
#  1: 2015-11-01 Apple 2015-11-01          0 157.3224   3380
#  2: 2015-11-01 Apple 2015-12-01          1 166.5574      0
#  3: 2015-11-01 Apple 2016-01-01          2 166.5574      0
#  4: 2015-11-01 Apple 2016-02-01          3 148.0874      0
#  5: 2015-11-01 Apple 2016-03-01          4 166.5574      0
#  6: 2015-11-01 Apple 2016-04-01          5 157.3224      0
#  7: 2015-11-01 Apple 2016-05-01          6 166.5574      0
#  8: 2015-11-01 Apple 2016-06-01          7 157.3224      0
#  9: 2015-11-01 Apple 2016-07-01          8 166.5574      0
# 10: 2015-11-01 Apple 2016-08-01          9 166.5574      0
# 11: 2015-11-01 Apple 2016-09-01         10 157.3224      0
# 12: 2015-11-01 Apple 2016-10-01         11 166.5574      0
# 13: 2015-11-01 Apple 2016-11-01         12 157.3224      0
# 14: 2016-01-01 Apple 2016-01-01          0 316.2295   7710
# 15: 2016-01-01 Apple 2016-02-01          1 274.0983      0
# 16: 2016-01-01 Apple 2016-03-01          2 316.2295      0
# 17: 2016-01-01 Apple 2016-04-01          3 295.1639      0
# 18: 2016-01-01 Apple 2016-05-01          4 316.2295      0
# 19: 2016-01-01 Apple 2016-06-01          5 295.1639      0
# 20: 2016-01-01 Apple 2016-07-01          6 316.2295      0
# 21: 2016-01-01 Apple 2016-08-01          7 316.2295      0
# 22: 2016-01-01 Apple 2016-09-01          8 295.1639      0
# 23: 2016-01-01 Apple 2016-10-01          9 316.2295      0
# 24: 2016-01-01 Apple 2016-11-01         10 295.1639      0
# 25: 2016-01-01 Apple 2016-12-01         11 316.2295      0
#          Time1 Brand      Time2 LapseMonth   value1 value2

Answer 2

我自己弄清楚，我在这里发布答案。也许其他用户将来会有类似的问题。

DT = data.table(Brand = c('Apple', 'Apple'),
            Time1 = c('2015-11', '2016-01'),
            value1 = c(119.7268, 336.8033),
            value2 = c(3380, 7710))
DT$Time1 <- ymd( paste( DT$Time1, 01, sep = "-"))
DT<-as.data.table(DT)
DT <-DT[,Time1:=as.Date(Time1,"%Y/%m/%d")]
DT <- DT[,rep := ifelse(year(Time1)==2016, 12-month(Time1)+1, 13)][rep(1:.N,rep)]
DT <- DT[, LapseMonth := seq_len(.N)-1, by =. (Brand,Time1,value2) ]
DT <- DT[, Time2:= Time1 - days(mday(Time1)-1) + months(LapseMonth)]
DT <- DT[, value1 := ifelse(Time1==Time2,value1,ifelse(LapseMonth==12, value2*days_in_month(Time2)/366-value1, value2*days_in_month(Time2)/366))] 
DT <- DT[, value2 := ifelse(Time1==Time2,value2,0)]

因此，我得到了结果。

ifelse有三个条件

2 个答案: