我正在尝试找到一种将新产品与我拥有历史数据的产品相匹配的方法。然后，我将使用预览年份产品的历史数据对新产品进行一些预测。
请考虑以下数据子集：
# A tibble: 13 x 11
prdct_id prdct_grp_1 prdct_grp_2 prdct_grp_3 prdct_grp_4 Start_season January February March April sales_total
<dbl> <chr> <chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1.00 WUW SW BH B21 2017 2.00 10.0 5.00 4.00 21.0
2 2.00 WUW SW BK R21 2017 7.00 9.00 4.00 5.00 25.0
3 3.00 MUW NW UW P1 2018 6.00 8.00 10.0 6.00 32.0
4 4.00 LNG KW LW L1 2016 8.00 9.00 12.0 7.00 36.0
5 5.00 QKQ MZ KA AQ 2013 10.0 8.67 16.7 8.00 43.3
6 6.00 MUW NW UW P1 2019 0 0 0 0 0
7 7.00 WUW SW BK R21 2019 0 0 0 0 0
8 8.00 LNG NW UW P2 2014 15.1 8.67 28.7 11.0 63.4
9 9.00 QKQ KW LW L2 2016 16.8 8.67 32.7 12.0 70.1
10 10.0 WUW MZ KA AQ 2017 18.5 8.67 36.7 13.0 76.8
11 11.0 QKQ MZ KA AQ 2019 0 0 0 0 0
12 12.0 WUW MZ KA AQ 2019 0 0 0 0 0
13 13.0 MUW NW UW P1 2019 0 0 0 0 0
prdct_grp代表产品组（例如prdct_grp_1=WUW表示该产品在“女士内衣”中，而prdct_grp_2=SW将指定该产品在“泳装”组中，依此类推） ）。如果某个产品与prdct_grp中的(1-4)相同，那么我将假定它们的销售额非常相似。
我希望获得以下结果
# A tibble: 3 x 11
new_prdct_id prdct_grp_1 prdct_grp_2 prdct_grp_3 prdct_grp_4 Start_s January February March April sales_total
<chr> <chr> <chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 6~3 MUW NW UW P1 2019 6.00 8.00 10.0 6.00 32.0
2 7~2 WUW SW BK R21 2019 7.00 9.00 4.00 5.00 25.0
3 11~5 QKQ MZ KA AQ 2019 10.0 9.00 17.0 8.00 43.0
我使用tidyverse获得了想要的结果，但结果不是很好。
如果一个产品与一个以上的产品匹配或与另一个具有2019年开始季节的产品匹配是另一个问题。我该如何处理？
谢谢您的帮助。
最佳 A
答案 0 :(得分：1)
下面是可能的dplyr解决方案以及详细的注释。请始终通过提供dput()输出或至少一个用于创建数据集的代码段来确保您的问题是可重现的。
# import required package
library(dplyr)
# reproduce your data frame (or at least something similar to it)
# please give more details next time
prdct_df <- data_frame(
prdct_id = 1:13,
prdct_grp_1 = c("WUW", "WUW", "MUW", "LNG", "QKQ", "MUW", "WUW", "LNG", "QKQ", "WUW", "QKQ", "WUW", "MUW"),
prdct_grp_2 = c("SW", "SW", "NW", "KW", "MZ", "NW", "SW", "NW", "KW", "MZ", "MZ", "MZ", "NW"),
prdct_grp_3 = c("BH", "BK", "UW", "LW", "KA", "UW", "BK", "UW", "LW", "KA", "KA", "KA", "UW"),
prdct_grp_4 = c("B21", "R21", "P1", "L1", "AQ", "P1", "R21", "P2", "L2", "AQ", "AQ", "AQ", "P1"),
Start_season = c(2017, 2017, 2018, 2016, 2013, 2019, 2019, 2014, 2016, 2017, 2019, 2019, 2019),
January = c(2, 7, 6 , 8, 10, 0, 0, 15.1, 16.8, 18.5, 0, 0, 0),
February = c(10, 9, 8, 9, 8.67, 0, 0, 8.86, 8.67, 8.67, 0, 0, 0),
March = c(4, 5, 10, 12, 16.7, 0, 0, 28.7, 32.7, 36.7, 0, 0, 0),
April = c(4, 5, 6, 7, 8, 0, 0, 11, 12, 13, 0, 0, 0),
sales_total = c(21, 25, 32, 36, 43.3, 0, 0, 63.4, 70.1, 76.8, 0, 0, 0)
)
# define new season in case you have additional seasons in the furture
new_prdct_seasons <- 2019 # with new seasons: c(2019, 2020, 2012) and so on
# keep the historical and new data separate (optional but clean)
# filter your data to separate new products
new_prdct_df <- prdct_df %>%
filter(Start_season %in% new_prdct_seasons)
# filter your data to separate old products
old_prdct_df <- prdct_df %>%
filter(!(Start_season %in% new_prdct_seasons))
# match the new and old products to get the data frame you want
final_df <- old_prdct_df %>%
inner_join(
# only the first 6 columns are needed from new product data frame
new_prdct_df[1:6],
# inner join by product group features
by = c("prdct_grp_1", "prdct_grp_2", "prdct_grp_3", "prdct_grp_4")
) %>%
# reorder the columns and change their names when necessary
select(
new_prdct_id = 12,
old_prdct_id = 1,
2:5,
Start_season = 13,
7:11
)
# we obtained the data frame you asked for
# note that we avoided matches among new products by keeping new and old products in distinct data frames
final_df
# # A tibble: 5 x 12
# new_prdct_id old_prdct_id prdct_grp_1 prdct_grp_2 prdct_grp_3 prdct_grp_4 Start_season January
# <int> <int> <chr> <chr> <chr> <chr> <dbl> <dbl>
# 1 7 2 WUW SW BK R21 2019 7
# 2 6 3 MUW NW UW P1 2019 6
# 3 13 3 MUW NW UW P1 2019 6
# 4 11 5 QKQ MZ KA AQ 2019 10
# 5 12 10 WUW MZ KA AQ 2019 18.5
# # ... with 4 more variables: February <dbl>, March <dbl>, April <dbl>, sales_total <dbl>
# you can also exclude matches with more than one old product if needed
final_df[-3, ] # this removes the match 13-3 as there is already 6-3