R 帮助 - 如何计算一个新列，该列对按两个日期过滤的不同表中的列求和？

Question

我有两个 tbls。

工作订单：

# A tibble: 6 x 6
  cleaningindex wonumber createddate nextdate  
  <chr>            <int> <date>      <date>    
1 1                 2093 2017-01-11  2017-02-09
2 2                 2514 2017-02-09  2017-03-03
3 3                 2904 2017-03-03  2017-03-24
4 4                 3070 2017-03-24  2017-06-06
5 5                 3669 2017-06-06  2017-07-17
6 6                 3997 2017-07-17  2017-08-24

和批收据：

# A tibble: 6 x 9
  datetimeindex `Batch-Num` `Receipt-Num` `Receipt-Date`      `Receipt-Time` Quantity   datetime  
  <chr>               <int>         <int> <dttm>              <chr>             <dbl>   <chr>    <chr>      
1 1                   99241         88678 2017-01-11 00:00:00 00:57:55          46500   2017-01~ 
2 2                   99322         88689 2017-01-11 00:00:00 05:09:29          45800   2017-01~ 
3 3                   99323         88703 2017-01-11 00:00:00 05:29:51          45000   2017-01~ 
4 4                   99242         88704 2017-01-11 00:00:00 13:04:20          44600   2017-01~ 
5 5                   99243         88711 2017-01-11 00:00:00 13:08:36          45000   2017-01~ 
6 6                   99353         88733 2017-01-12 00:00:00 03:47:23          45225   2017-01~ 

我需要做的是我需要在第一个表中创建一个名为“quantity”的新列，它是 batchreceipts 表中“Quantity”列的总和，经过过滤，以便 workorders 表中的数量列仅对数量求和基于batchreceipts$`Receipt-Date` >= workorders$`createddate` AND batchreceipts$`Receipt-Date <= workorders$`nextdate`

我没有找到任何资源来建议一种方法来构建这样的新列添加。 任何人都可以提供指导吗？

** 编辑以显示我需要/预期的 output 将是什么（我添加了数量列并输入了一些任意值）：

工作订单：

# A tibble: 6 x 6
  cleaningindex wonumber createddate nextdate   quantity
  <chr>            <int> <date>      <date>        <int>
1 1                 2093 2017-01-11  2017-02-09   800000
2 2                 2514 2017-02-09  2017-03-03   925000
3 3                 2904 2017-03-03  2017-03-24  1200000 
4 4                 3070 2017-03-24  2017-06-06   715000
5 5                 3669 2017-06-06  2017-07-17   945000
6 6                 3997 2017-07-17  2017-08-24   400000

Answer 1

一种解决方案是将mapply()与 function 一起使用，该 function 根据它们是否在“收据日期”在“createdate”和“nextdate”之间的范围内连续排列您的“数量”：

quantity_sum <- function(date1, date2) {
  rdate <- batchreceipts$`Receipt-Date`
  matching_rows <- (rdate >= date1) & (rdate <= date2)
  sum(batchreceipts$Quantity[matching_rows])
}

workorders$`Quantity Sum` <- mapply(quantity_sum, workorders$createddate, workorders$nextdate)

这给了我们：

# A tibble: 6 x 5
  cleaningindex wonumber createddate nextdate   `Quantity Sum`
          <dbl>    <dbl> <date>      <date>              <dbl>
1             1     2093 2017-01-11  2017-02-09         272125
2             2     2514 2017-02-09  2017-03-03              0
3             3     2904 2017-03-03  2017-03-24              0
4             4     3070 2017-03-24  2017-06-06              0
5             5     3669 2017-06-06  2017-07-17              0
6             6     3997 2017-07-17  2017-08-24              0

注意：将来，请提供代码以重现您的数据。 我在这里使用tribble()但更好的方法是从dput(workorders)复制 output 。

library(dplyr)
workorders <- tribble(
~cleaningindex, ~wonumber, ~createddate, ~nextdate, 
1,                 2093, "2017-01-11",  "2017-02-09",
2,                 2514, "2017-02-09",  "2017-03-03",
3,                 2904, "2017-03-03",  "2017-03-24",
4,                 3070, "2017-03-24",  "2017-06-06",
5,                 3669, "2017-06-06",  "2017-07-17",
6,                 3997, "2017-07-17",  "2017-08-24"
)

batchreceipts <- tribble(
~datetimeindex, ~`Batch-Num`, ~`Receipt-Num`, ~`Receipt-Date`, ~`Receipt-Time`, ~Quantity, 
1, 99241, 88678, "2017-01-11 00:00:00", "00:57:55", 46500,
2, 99322, 88689, "2017-01-11 00:00:00", "05:09:29", 45800,
3, 99323, 88703, "2017-01-11 00:00:00", "05:29:51", 45000,
4, 99242, 88704, "2017-01-11 00:00:00", "13:04:20", 44600,
5, 99243, 88711, "2017-01-11 00:00:00", "13:08:36", 45000,
6, 99353, 88733, "2017-01-12 00:00:00", "03:47:23", 45225
)

workorders[[3]] <- as.Date(workorders[[3]])
workorders[[4]] <- as.Date(workorders[[4]])
batchreceipts[[4]] <- as.Date(batchreceipts[[4]])