R：循环遍历行直到满足条件，然后在下一行重新开始答案

【问题标题】：R: Looping over rows until condition is met, then start over in next rowR：循环遍历行直到满足条件，然后在下一行重新开始
【发布时间】：2019-07-22 08:49:14
【问题描述】：

我有一个表格，每个客户的订单都带有时间戳。我想知道在下一个订单后在时间范围x 内发生了哪些订单，一旦时间范围结束，就从时间范围x 重新开始下一个订单。新列应始终说明哪一个是第一个订单。

最好看下面的例子。

我已经用next尝试了一些for loops，但根本无法使用。

数据如下：

x <- data.frame("Customer" =c(123,123,123,123,123,123,123,567), "Order_nr" = c(1,2,3,4,5,6,7,1), "Order_datetime" = c('2018-11-24 00:00:25','2018-11-24 15:58:23','2018-11-24 19:10:29','2018-11-24 21:29:04','2018-11-24 22:03:59','2018-11-24 22:26:59','2018-11-24 22:36:13','2018-11-24 12:00:55'))
x

| Customer | Order_nr | Order_datetime|
| ------------- |:-------------:| -----:|
| 123      | 1 | 2018-11-24 00:00:25 |
| 123      | 2 | 2018-11-24 15:58:23 |
| 123      | 3 | 2018-11-24 19:10:29 |
| 123      | 4 | 2018-11-24 21:29:04 |
| 123      | 5 | 2018-11-24 22:03:59 |
| 123      | 6 | 2018-11-24 22:26:59 |
| 123      | 7 | 2018-11-24 22:36:1 |
| 567      | 1 | 2018-11-24 12:00:55 |

如果我想知道1小时内的订单，我想在1h bundle first order列中得到结果，如果是3小时，它应该是3h bundle first order列的结果

| Customer | Order_nr | Order_datetime| 3h bundle first order| 3h bundle first order|
| ------------- |:-------------:| -----:|-----:|
| 123      | 1A | 2018-11-24 00:00:25 |1A |1A|
| 123      | 2A | 2018-11-24 15:58:23 |2A |2A|
| 123      | 3A | 2018-11-24 19:10:29 |3A |3A|
| 123      | 4A | 2018-11-24 21:29:04 |4A |3A|
| 123      | 5A | 2018-11-24 22:03:59 |4A |3A|
| 123      | 6A | 2018-11-24 22:26:59 |4A |4A|
| 123      | 7A | 2018-11-24 22:36:1  |5A |4A|
| 567      | 1B | 2018-11-24 12:00:55 |1B |1B|

所以我需要知道订单 4A、5A 和 6A 发生在从订单 4A 开始的 1h 内，例如 1h bundle first order。

【问题讨论】：

标签： r loops

【解决方案1】：

so <- data.frame("Customer" =c(123,123,123,123,123,123,123,567), 
                 "Order_nr" = c(1,2,3,4,5,6,7,1), 
                 "Order_datetime" = c('2018-11-24 00:00:25','2018-11-24 15:58:23',
                                      '2018-11-24 19:10:29','2018-11-24 21:29:04',
                                      '2018-11-24 22:03:59','2018-11-24 22:26:59',
                                      '2018-11-24 22:36:13','2018-11-24 12:00:55'))


learn <- function(date_time, df, hr.within, i){

  subject <- abs(difftime(date_time, df$Order_datetime, units="hours"))

  ifelse(i ==1, 
         thatrow <- which((subject <= hr.within) == TRUE), 
         thatrow <- intersect( which((subject <= hr.within) == TRUE), 
                               which((subject >= hr.within-1) == TRUE)))

  if(identical(thatrow, integer(0))) return()

  else{
    R2 <- df[thatrow, c("Customer", "Order_nr", "Order_datetime")]
    R2$x <- NA
    R2[,"x"] <- paste0(hr.within, "A")
    colnames(R2)[4] <- paste0(hr.within,"A bundle first order")
    return(R2)
  }
}


learn.wrapper <- function(date_time, df, hr.within=seq(1,100,1)){
  learn.out <- list()
  for(i in 1:length(hr.within)){
    learn.out[[i]] <- learn(date_time,so, hr.within[i], i)
  }
  return(rbindlist(learn.out, fill=TRUE))
}

learnery <- learn.wrapper('2018-11-24 19:00:00', df=so) #first argument is the time you want to ref. with
learnery

这假设所有操作都在 100 小时内发生，您可以通过 hr.within=seq(1,100,1) 重置到适当的持续时间窗口，然后重新编译。然后，您可以在查看其输出时自行对结果进行行合并。

【讨论】：

【解决方案2】：

一位同事帮我找到了解决办法：

Order_nr_3h = [] 
order1 = df['Order_nr ']
initial_order = order1[0] ##initials needed so the loop can start with a value
time = df['Order_datetime']
initial = time[0]
customer1 = df['customer']
initial_customer = customer1[0]
for i in range(-1,len(time)-1):
    delta = (time[i+1]-initial).seconds/3600 ## 1h
    if customer1[i+1] != initial_customer: 
        initial_order = order1[i+1]
        initial_customer = customer1[i+1] 
    if delta > 1: 
        initial = time[i+1]
        initial_order = order1[i+1]
        Order_nr_3h.append(initial_order)
    else:
        Order_nr_3h.append(initial_order)
        continue
df['Order_nr_3h'] = Order_nr_3h

然后，我在每个需要的时间间隔（1 小时、3 小时等）周围都有另一个循环

【讨论】：

你要求 R 并且你用 Python 回答了自己？