Question

假设我有两个数据集。一个包含具有开始/结束日期的促销列表，另一个包含每个程序的月度销售数据。

promotions = data.frame(
    start.date = as.Date(c("2012-01-01", "2012-06-14", "2012-02-01", "2012-03-31", "2012-07-13")), 
    end.date = as.Date(c("2014-04-05", "2014-11-13", "2014-02-25", "2014-08-02", "2014-09-30")), 
    program = c("a", "a", "a", "b", "b"))

sales = data.frame(
    year.month.day = as.Date(c("2013-02-01", "2014-09-01", "2013-08-01", "2013-04-01", "2012-11-01")), 
    program = c("a", "b", "a", "a", "b"), 
    monthly.sales = c(200, 200, 200, 400, 200))

请注意，sales$year.month.day用于表示年/月。包括日，因此R可以更简单地将列视为日期对象的向量，但它与实际销售额无关。

我需要确定每个计划每月发生的促销数量。这是一个产生我想要的输出的循环示例：

sales$count = rep(0, nrow(sales))
sub = list()
for (i in 1:nrow(sales)) {
  sub[[i]] = promotions[which(promotions$program == sales$program[i]),]
  if (nrow(sub[[i]]) > 1) {
    for (j in 1:nrow(sub[[i]])) {
      if (sales$year.month.day[i] %in% seq(from = as.Date(sub[[i]]$start.date[j]), to = as.Date(sub[[i]]$end.date[j]), by = "day")) {
        sales$count[i] = sales$count[i] + 1
      }
    }
  }
}

示例输出：

 sales = data.frame(
    year.month.day = as.Date(c("2013-02-01", "2014-09-01", "2013-08-01", "2013-04-01", "2012-11-01")), 
    program = c("a", "b", "a", "a", "b"), 
    monthly.sales = c(200, 200, 200, 400, 200),
    count = c(3, 1, 3, 3, 2)
)

但是由于我的实际数据集非常大，所以当我在R中运行它时，这个循环会崩溃。

是否有更有效的方法来实现相同的结果？也许是dplyr的东西？

Answer 1

您可以使用sql执行此操作。

library(sqldf)
sqldf("select s.ymd,p.program,s.monthlysales, count(*) from promotions p outer left join sales s on p.program=s.program 
where s.ymd between p.startdate and p.enddate and p.program=s.program group by s.ymd, s.program" )

这将首先加入2数据集，其中销售中的ymd介于促销的开始和结束日期之间，并且两个数据中的程序是相同的。然后它将按ymd分组并计算实例。我已从变量名称中删除了句点。

Answer 2

使用当前开发版data.table中新实现的 non-equi 连接：

require(data.table) # v1.9.7+
setDT(promotions) # convert to data.table by reference
setDT(sales)

ans = promotions[sales, .(monthly.sales, .N), by=.EACHI, allow.cartesian=TRUE, 
        on=.(program, start.date<=year.month.day, end.date>=year.month.day), nomatch=0L]

ans[, end.date := NULL]
setnames(ans, "start.date", "year.month.date")
#    program year.month.date monthly.sales N
# 1:       a      2013-02-01           200 3
# 2:       b      2014-09-01           200 1
# 3:       a      2013-08-01           200 3
# 4:       a      2013-04-01           400 3
# 5:       b      2012-11-01           200 2

请参阅开发版here的安装说明。

Answer 3

可以尝试?data.table::foverlaps

library(data.table)
setDT(sales)[, c("start.date", "end.date") := year.month.day] # Add overlap cols
setkey(sales, program, start.date, end.date) # Key for join
res <- foverlaps(setDT(promotions), sales)[, .N, by = year.month.day] # Count joins
sales[res, count := i.N, on = "year.month.day"] # Update `sales` with results
sales
#    year.month.day program monthly.sales start.date   end.date count
# 1:     2013-02-01       a           200 2013-02-01 2013-02-01     3
# 2:     2013-04-01       a           400 2013-04-01 2013-04-01     3
# 3:     2013-08-01       a           200 2013-08-01 2013-08-01     3
# 4:     2012-11-01       b           200 2012-11-01 2012-11-01     2
# 5:     2014-09-01       b           200 2014-09-01 2014-09-01     1

这基本上是在sales创建间隔列，由program加入+，重叠计数，然后加入sales。如果真的困扰你，可以通过sales[, c("start.date", "end.date") := NULL]删除其他列。 Google foverlaps和data.table了解更多示例

Answer 4

我是哈德利包裹的粉丝：

library(dplyr)
library(lubridate)

发言日期，因此它们的格式与sales数据框的格式相同：

df <- promotions %>% 
    mutate(start.date = floor_date(start.date, unit = "month"),
           end.date = floor_date(end.date, unit = "month"))

展开日期间隔：

df$output <- mapply(function(x,y) seq(x, y, by =  "month"),
       df$start.date,
       df$end.date)

根据日期范围，组和计数展开数据框，并合并到销售日期和计划：

df %>% tidyr::unnest(output) %>% 
    group_by(output, program) %>%
    summarise(prom_num = n()) %>%
    merge(sales, ., 
      by.x = c("year.month.day", "program"),
      by.y = c("output", "program"))

输出：

  year.month.day program monthly.sales prom_num
1     2012-11-01       b           200        2
2     2013-02-01       a           200        3
3     2013-04-01       a           400        3
4     2013-08-01       a           200        3
5     2014-09-01       b           200        1

按日期范围和分类变量组合数据集

4 个答案: