Question

我有两个数据集： competitor_data - 包含特定产品的竞争对手以及收集竞争对手价格的价格和日期。

product_price - 每次价格变动的日期。

competitor_data <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
                            crawl_date=c("2014-04-05", "2014-04-22", "2014-05-05", "2014-05-22","2014-06-05", "2014-06-22",
                                   "2014-05-08", "2014-06-17", "2014-06-09", "2014-06-14","2014-07-01", "2014-08-04"),
                            competitor =c("amazon","apple","google","facebook","alibaba","tencent","ebay","bestbuy","gamespot","louis vuitton","gucci","tesla"),
                            competitor_price =c(2.5,2.35,1.99,2.01,2.22,2.52,5.32,5.56,5.01,6.01,5.86,5.96), stringsAsFactors=FALSE)

competitor_data$crawl_date = as.Date(competitor_data$crawl_date)

＃

product_price <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
                                      date=c("2014-05-05", "2014-06-22", "2014-07-05", "2014-08-31","2014-05-03", "2014-02-22",
                                                  "2014-05-21", "2014-06-19", "2014-03-09", "2014-06-22","2014-07-03", "2014-09-08"),
                                    price =c(2.12,2.31,2.29,2.01,2.04,2.09,5.22,5.36,5.21,5.91,5.36,5.56), stringsAsFactors=FALSE)

product_price$date = as.Date(product_price$date)

目标

对于product_price中的给定产品，对于每条记录（日期），查找来自competitor_data的相关crawl_date价格。
将product_price $ price与最低的competitor_data $ competitor_price进行比较。
如果product_price $ price＆lt; = competitor_data $ competitor_price，则创建一个新列以标记1（price_leader）else标志0（price_leader）

我的脚本使用嵌套for循环，但处理5000个唯一的product_id需要24小时：

unique_skus <- unique(product_price$productId)
all_competitive_data <- data.frame()
mid_step_data <- data.frame()

start_time <-Sys.time()
for (i in 1:length(unique_skus)){
  step1 <- subset(product_price, productId == unique_skus[i])
  transact_dates = unique(step1$date)
  for (a in 1:length(transact_dates)){
    step2 <- subset(step1, date ==transact_dates[a])
    step3 <- inner_join(step2,competitor_data, by='productId')
    if (nrow(subset(step3, date > crawl_date)) == 0){
      step3 <- step3[ order(step3$crawl_date , decreasing = FALSE ),]
      competitor_price <- head(step3,1)$competitor_price
      step2$competitor_price = competitor_price
    }
    else {
      step4 <- subset(step3, date > crawl_date)
      step4 <- step4[ order(step4$crawl_date , decreasing = TRUE ),]
      competitor_price <- head(step4,1)$competitor_price
      step2$competitor_price = competitor_price
    }
    step2$price_leader <- ifelse(step2$price <= step2$competitor_price, 1, 0)
    mid_step_data = rbind(mid_step_data,step2)
  }
  all_competitive_data <- rbind(all_competitive_data,mid_step_data)
}
Sys.time()-start_time
all_competitive_data = unique(all_competitive_data)

有没有办法快速使用dplyr来实现这个目标？

Answer 1

competitor_data <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
                              crawl_date=c("2014-04-05", "2014-04-22", "2014-05-05", "2014-05-22","2014-06-05", "2014-06-22",
                                           "2014-05-08", "2014-06-17", "2014-06-09", "2014-06-14","2014-07-01", "2014-08-04"),
                              competitor =c("amazon","apple","google","facebook","alibaba","tencent","ebay","bestbuy","gamespot","louis vuitton","gucci","tesla"),
                              competitor_price =c(2.5,2.35,1.99,2.01,2.22,2.52,5.32,5.56,5.01,6.01,5.86,5.96), stringsAsFactors=FALSE)

competitor_data$crawl_date = as.Date(competitor_data$crawl_date)
#
product_price <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
                            date=c("2014-05-05", "2014-06-22", "2014-07-05", "2014-08-31","2014-05-03", "2014-02-22",
                                   "2014-05-21", "2014-06-19", "2014-03-09", "2014-06-22","2014-07-03", "2014-09-08"),
                            price =c(2.12,2.31,2.29,2.01,2.04,2.09,5.22,5.36,5.21,5.91,5.36,5.56), stringsAsFactors=FALSE)

product_price$date = as.Date(product_price$date)

使用此功能向前向后填充向量

## fill in NAs
f <- function(..., lead = NA) {
  # f(NA, 1, NA, 2, NA, NA, lead = NULL)
  x <- c(lead, c(...))
  head(zoo::na.locf(zoo::na.locf(x, na.rm = FALSE), fromLast = TRUE),
       if (is.null(lead)) length(x) else -length(lead))
}

按产品和日期合并两者。我们用额外的NA来按产品填写第一个价格，这样当我们填写NA时，这将有效地使用之前的价格

然后进行价格和竞争对手价格的比较。最后一步只是进行一些清理以证明它是相同的结果

dd <- merge(product_price, competitor_data,
            by.y = c('productId', 'crawl_date'),
            by.x = c('productId', 'date'), all = TRUE)
dd$competitor_price <-
  unlist(sapply(split(dd$competitor_price, dd$productId), f))
dd$price_leader <- +(dd$price <= dd$competitor_price)
(res1 <- `rownames<-`(dd[!is.na(dd$price_leader), -4], NULL))

#    productId       date price competitor_price price_leader
# 1     banana 2014-02-22  2.09             2.50            1
# 2     banana 2014-05-03  2.04             2.35            1
# 3     banana 2014-05-05  2.12             2.35            1
# 4     banana 2014-06-22  2.31             2.22            0
# 5     banana 2014-07-05  2.29             2.52            1
# 6     banana 2014-08-31  2.01             2.52            1
# 7        fig 2014-03-09  5.21             5.32            1
# 8        fig 2014-05-21  5.22             5.32            1
# 9        fig 2014-06-19  5.36             5.56            1
# 10       fig 2014-06-22  5.91             5.56            0
# 11       fig 2014-07-03  5.36             5.86            1
# 12       fig 2014-09-08  5.56             5.96            1

res0 <- `rownames<-`(all_competitive_data[
  order(all_competitive_data$productId, all_competitive_data$date), ], NULL)

all.equal(res0, res1)
# [1] TRUE

您可以将任何这些步骤更改为dplyr或data.table语法;我没有使用任何一个，但它应该是直截了当的：

library('dplyr')
dd <- full_join(product_price, competitor_data,
                by = c(
                  'productId' = 'productId',
                  'date' = 'crawl_date'
                )
) %>% arrange(productId, date)

dd %>% group_by(productId) %>%
  mutate(
    competitor_price = f(competitor_price),
    price_leader = as.integer(price <= competitor_price)
) %>% filter(!is.na(price_leader)) %>% select(-competitor)

# Source: local data frame [12 x 5]
# Groups: productId [2]
# 
#      productId       date price competitor_price price_leader
#          <chr>     <date> <dbl>            <dbl>        <int>
#   1     banana 2014-02-22  2.09             2.50            1
#   2     banana 2014-05-03  2.04             2.35            1
#   3     banana 2014-05-05  2.12             2.35            1
#   4     banana 2014-06-22  2.31             2.22            0
#   5     banana 2014-07-05  2.29             2.52            1
#   6     banana 2014-08-31  2.01             2.52            1
#   7        fig 2014-03-09  5.21             5.32            1
#   8        fig 2014-05-21  5.22             5.32            1
#   9        fig 2014-06-19  5.36             5.56            1
#   10       fig 2014-06-22  5.91             5.56            0
#   11       fig 2014-07-03  5.36             5.86            1
#   12       fig 2014-09-08  5.56             5.96            1

Answer 2

以下解决方案使用dplyr join进行匹配。（注意：我将“crawl_date”更改为“date”，以便dplyr join自动选择匹配的列。可以使用类似

的内容明确匹配

git push

作为加入的参数。

by=c('productId'='productId', date'='crawl_date')

结果数据框是

competitor_data <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
                              date=c("2014-04-05", "2014-04-22", "2014-05-05", "2014-05-22","2014-06-05", "2014-06-22",
                                           "2014-05-08", "2014-06-17", "2014-06-09", "2014-06-14","2014-07-01", "2014-08-04"),
                              competitor =c("amazon","apple","google","facebook","alibaba","tencent","ebay","bestbuy","ga**strong text**mespot","louis vuitton","gucci","tesla"),
                              competitor_price =c(2.5,2.35,1.99,2.01,2.22,2.52,5.32,5.56,5.01,6.01,5.86,5.96), stringsAsFactors=FALSE)

competitor_data$date = as.Date(competitor_data$date)

product_price <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
                            date=c("2014-05-05", "2014-06-22", "2014-07-05", "2014-08-31","2014-05-03", "2014-02-22",
                                   "2014-05-21", "2014-06-19", "2014-03-09", "2014-06-22","2014-07-03", "2014-09-08"),
                            price =c(2.12,2.31,2.29,2.01,2.04,2.09,5.22,5.36,5.21,5.91,5.36,5.56), stringsAsFactors=FALSE)

product_price$date = as.Date(product_price$date)

require(dplyr)
joined <- product_price %>% left_join(competitor_data)
joined$leader <- as.integer(joined$price <= joined$competitor_price)

joined

R中嵌套For循环的替代方案

2 个答案: