如何在特定条件下创建排名变量？答案

【问题标题】：How to create a rank variable under certain conditions?如何在特定条件下创建排名变量？
【发布时间】：2018-05-27 16:46:23
【问题描述】：

我的数据包含时间变量和选择的品牌变量，如下所示。 time表示购物时间，choicebrand表示当时购买的品牌。

有了这些数据，我想创建排名变量，如第三列、第四列等所示。

品牌排名（例如，brand1 - brand3）应基于过去 36 小时。因此，要计算第二行的排名，其商店时间为"2013-09-01 08:54:00 UTC"，排名应基于该时间前 36 小时内的所有chosenbrand 值。（第二行的brand1 不应在 36 小时内）

因此，rank_brand1, rank_brand2, rank_brand3, rank_bran4,,, 是我想要的变量。

如果我想创建 rank_brand5，rank_brand6 也是如此......

有什么简单的方法吗？

另外，如果我要个人做（如果每个客户都有多个购买历史），该怎么做？

数据如下，

          shoptime          chosenbrand  rank_brand1 rank_brand2 rank_brand3, ...
  2013-09-01 08:35:00 UTC      brand1          NA         NA          NA
  2013-09-01 08:54:00 UTC      brand1          1          NA          NA
  2013-09-01 09:07:00 UTC      brand2          1          2          NA
  2013-09-01 09:08:00 UTC      brand3          1          2          3
  2013-09-01 09:11:00 UTC      brand5          1          2          3
  2013-09-01 09:14:00 UTC      brand2          1          2          3
  2013-09-01 09:26:00 UTC      brand6          1          1          3
  2013-09-01 09:26:00 UTC      brand2          1          1          3
  2013-09-01 09:29:00 UTC      brand2          2          1          3
  2013-09-01 09:32:00 UTC      brand4          2          1          3

这是数据代码

dat <- data.frame(shoptime = c("2013-09-01 08:35:00 UTC", "2013-09-01 08:54:00 UTC", "2013-09-01 09:07:00 UTC" ,"2013-09-01 09:08:00 UTC", "2013-09-01 09:11:00 UTC", "2013-09-01 09:14:00 UTC",
                           "2013-09-01 09:26:00 UTC", "2013-09-01 09:26:00 UTC" ,"2013-09-01 09:29:00 UTC", "2013-09-01 09:32:00 UTC"),
                  chosenbrand = c("brand1", "brand1", "brand2", "brand3", "brand5", "brand2", "brand6", "brand2"  ,  "brand2"  ,   "brand4"   ),
                  rank_brand1 = NA,
                  rank_brand2 = NA,
                 rank_brand3 = NA,
                  stringsAsFactors = FALSE)

【问题讨论】：

我需要计算最近 36 小时内的观察结果。有什么想法吗？
请问有人可以帮忙吗？
为什么NA 在第一排是rank_brand2？根据您的band1、band2 和band3 提供的数据，在第一行的 36 小时内可用。
第一行 rank_brand1 应该是 NA。我刚刚更正了。由于第一行之前没有先前的历史记录，因此都应该是 NA。还有第二行的历史
关于您的附加问题，请提出一个单独的问题并提供样本数据，其中包括多个客户的购买历史。谢谢。

标签： r dataframe dplyr data.table plyr

【解决方案1】：

一种可能性是为作业编写一个函数（使用loop）。考虑到 OP 中提供的数据：

library(dplyr)

dat <- data.frame(shoptime = c("2013-09-01 08:35:00 UTC", "2013-09-01 08:54:00 UTC", "2013-09-01 09:07:00 UTC" ,"2013-09-01 09:08:00 UTC", "2013-09-01 09:11:00 UTC", "2013-09-01 09:14:00 UTC",
                               "2013-09-01 09:26:00 UTC", "2013-09-01 09:26:00 UTC" ,"2013-09-01 09:29:00 UTC", "2013-09-01 09:32:00 UTC"),
                  chosenbrand = c("brand1", "brand1", "brand2", "brand3", "brand5", "brand2", "brand6", "brand2"  ,  "brand2"  ,   "brand4"   ),
                  rank_brand1 = NA,
                  rank_brand2 = NA,
                  rank_brand3 = NA,
                  stringsAsFactors = FALSE)

#Write a function that data.frame and calculate rank
Calculate.Rank <- function(x){
  #loop through each row and calculate count for each brand 
  for(i in 1:nrow(x)){
    #DateTime of the current row. 
    currentrow.time <- as.POSIXlt(x$shoptime[i])
    #calculate number of times brand1 appears
    x$rank_brand1[i] <- nrow(filter(x, as.POSIXlt(shoptime) <= currentrow.time & as.POSIXlt(shoptime) >= (currentrow.time-36*60*60) & chosenbrand == "brand1" ))
    #calculate number of times brand2 appears
    x$rank_brand2[i] <- nrow(filter(x, as.POSIXlt(shoptime) <= currentrow.time & as.POSIXlt(shoptime) >= (currentrow.time-36*60*60) & chosenbrand == "brand2" ))    
    #calculate number of times brand3 appears
    x$rank_brand3[i] <- nrow(filter(x, as.POSIXlt(shoptime) <= currentrow.time & as.POSIXlt(shoptime) >= (currentrow.time-36*60*60) & chosenbrand == "brand3" ))

#Replace the 0 values with NA. I dont think this right approach as one can consider those count to be 0 anyway

    if(x$rank_brand1[i] == 0 ){
      x$rank_brand1[i] = NA
    }

    if(x$rank_brand2[i] == 0 ){
      x$rank_brand2[i] = NA
    }
    if(x$rank_brand3[i] == 0 ){
      x$rank_brand3[i] = NA
    }    

  }

  #Now count of brand1, brand2 and brand3 is available now. Lets calculate rank.
  new.x <- data.frame(x[,1:2], t(apply(-x[,3:5], 1, rank, ties.method='min', na.last = "keep")))

  print(new.x)
}

Calculate.Rank(dat)

生成的 data.frame new.x 将如下所示：

                shoptime chosenbrand rank_brand1 rank_brand2 rank_brand3
1  2013-09-01 08:35:00 UTC      brand1           1          NA          NA
2  2013-09-01 08:54:00 UTC      brand1           1          NA          NA
3  2013-09-01 09:07:00 UTC      brand2           1           2          NA
4  2013-09-01 09:08:00 UTC      brand3           1           2           2
5  2013-09-01 09:11:00 UTC      brand5           1           2           2
6  2013-09-01 09:14:00 UTC      brand2           1           1           3
7  2013-09-01 09:26:00 UTC      brand6           2           1           3
8  2013-09-01 09:26:00 UTC      brand2           2           1           3
9  2013-09-01 09:29:00 UTC      brand2           2           1           3
10 2013-09-01 09:32:00 UTC      brand4           2           1           3

【讨论】：

非常感谢，但我的原始数据大约有 100 万行。似乎需要太多时间。你不这么认为吗？
@Johnlegend2 同意。我想首先找到实现目标的方法。现在我们需要一个优化的版本。也许arrange 可以提供帮助。
如果你坚持使用for循环，我建议在调用Calculate.Rank()之前转换一次shoptime。这将避免循环内的重复转换。而且，POSIXlt 类的内存效率远低于POSIXct。

【解决方案2】：

这是一个棘手的问题。下面的解决方案使用 non-equi 连接 以 36 小时为周期聚合，dcast() 将长格式重塑为宽格式，并使用原始 dat 进行第二次连接。可以有任意数量的品牌。

library(data.table)
library(lubridate)

setDT(dat)[, shoptime := as_datetime(shoptime)]
setorder(dat, shoptime) # not required, just for convenience of observers
dat[.(lb = shoptime - hours(36), ub = shoptime), on = .(shoptime >= lb, shoptime < ub), 
    nomatch = 0L, by = .EACHI, 
    .SD[, .N, by = brand][, rank := frank(-N, ties.method="dense")]][
      , dcast(unique(.SD[, -1]), shoptime ~ brand, value.var = "rank")][
        dat, on = "shoptime"]

               shoptime brand1 brand2 brand3 brand5 brand6  brand
 1: 2013-09-01 08:35:00     NA     NA     NA     NA     NA brand1
 2: 2013-09-01 08:54:00      1     NA     NA     NA     NA brand1
 3: 2013-09-01 09:07:00      1     NA     NA     NA     NA brand2
 4: 2013-09-01 09:08:00      1      2     NA     NA     NA brand3
 5: 2013-09-01 09:11:00      1      2      2     NA     NA brand5
 6: 2013-09-01 09:14:00      1      2      2      2     NA brand2
 7: 2013-09-01 09:26:00      1      1      2      2     NA brand6
 8: 2013-09-01 09:26:00      1      1      2      2     NA brand2
 9: 2013-09-01 09:29:00      2      1      3      3      3 brand2
10: 2013-09-01 09:32:00      2      1      3      3      3 brand4

说明

dat[.(lb = shoptime - hours(36), ub = shoptime), on = .(shoptime >= lb, shoptime < ub), 
    nomatch = 0L, by = .EACHI, 
    .SD[, .N, by = brand][, rank := frank(-N, ties.method="dense")]]

返回每 36 小时的汇总结果：

               shoptime            shoptime  brand N rank
 1: 2013-08-30 20:54:00 2013-09-01 08:54:00 brand1 1    1
 2: 2013-08-30 21:07:00 2013-09-01 09:07:00 brand1 2    1
 3: 2013-08-30 21:08:00 2013-09-01 09:08:00 brand1 2    1
 4: 2013-08-30 21:08:00 2013-09-01 09:08:00 brand2 1    2
 5: 2013-08-30 21:11:00 2013-09-01 09:11:00 brand1 2    1
 6: 2013-08-30 21:11:00 2013-09-01 09:11:00 brand2 1    2
 7: 2013-08-30 21:11:00 2013-09-01 09:11:00 brand3 1    2
 8: 2013-08-30 21:14:00 2013-09-01 09:14:00 brand1 2    1
 9: 2013-08-30 21:14:00 2013-09-01 09:14:00 brand2 1    2
10: 2013-08-30 21:14:00 2013-09-01 09:14:00 brand3 1    2
11: 2013-08-30 21:14:00 2013-09-01 09:14:00 brand5 1    2
12: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand1 2    1
13: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand2 2    1
14: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand3 1    2
15: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand5 1    2
16: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand1 2    1
17: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand2 2    1
18: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand3 1    2
19: 2013-08-30 21:26:00 2013-09-01 09:26:00 brand5 1    2
20: 2013-08-30 21:29:00 2013-09-01 09:29:00 brand1 2    2
21: 2013-08-30 21:29:00 2013-09-01 09:29:00 brand2 3    1
22: 2013-08-30 21:29:00 2013-09-01 09:29:00 brand3 1    3
23: 2013-08-30 21:29:00 2013-09-01 09:29:00 brand5 1    3
24: 2013-08-30 21:29:00 2013-09-01 09:29:00 brand6 1    3
25: 2013-08-30 21:32:00 2013-09-01 09:32:00 brand1 2    2
26: 2013-08-30 21:32:00 2013-09-01 09:32:00 brand2 4    1
27: 2013-08-30 21:32:00 2013-09-01 09:32:00 brand3 1    3
28: 2013-08-30 21:32:00 2013-09-01 09:32:00 brand5 1    3
29: 2013-08-30 21:32:00 2013-09-01 09:32:00 brand6 1    3
               shoptime            shoptime  brand N rank

然后，这个中间结果从长格式改成宽格式：

dat[.(lb = shoptime - hours(36), ub = shoptime), on = .(shoptime >= lb, shoptime < ub), 
    nomatch = 0L, by = .EACHI, 
    .SD[, .N, by = brand][, rank := frank(-N, ties.method="dense")]][
      , dcast(unique(.SD[, -1]), shoptime ~ brand, value.var = "rank")]

              shoptime brand1 brand2 brand3 brand5 brand6
1: 2013-09-01 08:54:00      1     NA     NA     NA     NA
2: 2013-09-01 09:07:00      1     NA     NA     NA     NA
3: 2013-09-01 09:08:00      1      2     NA     NA     NA
4: 2013-09-01 09:11:00      1      2      2     NA     NA
5: 2013-09-01 09:14:00      1      2      2      2     NA
6: 2013-09-01 09:26:00      1      1      2      2     NA
7: 2013-09-01 09:29:00      2      1      3      3      3
8: 2013-09-01 09:32:00      2      1      3      3      3

与原始dat 数据框的最终右连接补全缺失的行和列（参见上面的代码和结果）。

数据

dat <- data.frame(
  shoptime = c("2013-09-01 08:35:00 UTC", "2013-09-01 08:54:00 UTC", "2013-09-01 09:07:00 UTC" ,"2013-09-01 09:08:00 UTC", "2013-09-01 09:11:00 UTC", "2013-09-01 09:14:00 UTC",
               "2013-09-01 09:26:00 UTC", "2013-09-01 09:26:00 UTC" ,"2013-09-01 09:29:00 UTC", "2013-09-01 09:32:00 UTC"),
  brand = c("brand1", "brand1", "brand2", "brand3", "brand5", "brand2", "brand6", "brand2"  ,  "brand2"  ,   "brand4"   ),
  stringsAsFactors = FALSE)

【讨论】：

你太棒了。谢谢