R中嵌套For循環的替代方法
[英]Alternative to Nested For Loop in R
問題描述
我有兩個數據集：competitor_data-包含給定產品的競爭對手以及收集競爭對手價格的價格和日期。
product_price-每次價格更改的日期。
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)
目的
- 對於product_price中的給定產品，對於每個記錄（日期），請從competitor_data中找到相關的crawl_date價格。
- 將product_price $ price與最低的競爭對手數據$ competitor_price進行比較。
- 如果product_price $ price <= competitor_data $ competitor_price，則創建一個新列以標記1（price_leader）否則標記0（price_leader）
我的以下腳本使用嵌套的for循環，但處理5000個唯一的product_id需要24小時以上：
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$competitor_price, 1, 0) mid_step_data = rbind(mid_step_data,step2) } all_competitive_data <- rbind(all_competitive_data,mid_step_data) } Sys.time()-start_time all_competitive_data = unique(all_competitive_data)
有沒有辦法使用dplyr快速完成此任務？
2 個解決方案
解決方案1
3 已采納 2017-01-03 05:01:52
competitor_data <- data.frame(productId=c('banana', 'banana','banana', 'banana','banana', 'banana','fig', 'fig','fig', 'fig','fig', 'fig'),
crawl_date=c("2014-04-05", "2014-04-22", "2014-05-05", "2014-05-22","2014-06-05", "2014-06-22",
"2014-05-08", "2014-06-17", "2014-06-09", "2014-06-14","2014-07-01", "2014-08-04"),
competitor =c("amazon","apple","google","facebook","alibaba","tencent","ebay","bestbuy","gamespot","louis vuitton","gucci","tesla"),
competitor_price =c(2.5,2.35,1.99,2.01,2.22,2.52,5.32,5.56,5.01,6.01,5.86,5.96), stringsAsFactors=FALSE)
competitor_data$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$date)
使用此功能可以用向前和向后的NA填充向量
## fill in NAs
f <- function(..., lead = NA) {
# f(NA, 1, NA, 2, NA, NA, lead = NULL)
x <- c(lead, c(...))
head(zoo::na.locf(zoo::na.locf(x, na.rm = FALSE), fromLast = TRUE),
if (is.null(lead)) length(x) else -length(lead))
}
合並兩個產品和日期。 我們用額外的NA來填充產品的第一個價格，因此當我們填寫NA時，這將有效地使用先前的價格
然后進行價格與競爭對手價格的比較。 最后只是進行一些清理以證明它是相同的結果
dd <- merge(product_price, competitor_data,
by.y = c('productId', 'crawl_date'),
by.x = c('productId', 'date'), all = TRUE)
dd$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
您可以將任何這些步驟更改為dplyr或data.table語法； 我都不使用任何一個，但是應該很簡單：
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
解決方案2
0 2017-01-03 03:58:43
以下解決方案使用dplyr join進行匹配。 （注意：我將“ crawl_date”更改為“ date”，以便dplyr join會自動選擇匹配的列。可以使用類似以下的方法來明確匹配
by=c('productId'='productId', date'='crawl_date')
作為要加入的參數。
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
結果數據幀是
productId date price competitor competitor_price leader
1 banana 2014-05-05 2.12 google 1.99 0
2 banana 2014-06-22 2.31 tencent 2.52 1
3 banana 2014-07-05 2.29 <NA> NA NA
4 banana 2014-08-31 2.01 <NA> NA NA
5 banana 2014-05-03 2.04 <NA> NA NA
6 banana 2014-02-22 2.09 <NA> NA NA
7 fig 2014-05-21 5.22 <NA> NA NA
8 fig 2014-06-19 5.36 <NA> NA NA
9 fig 2014-03-09 5.21 <NA> NA NA
10 fig 2014-06-22 5.91 <NA> NA NA
11 fig 2014-07-03 5.36 <NA> NA NA
12 fig 2014-09-08 5.56 <NA> NA NA
>
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