R:避免循环或行应用功能

Question

我关注了两个数据框 df_sales和df_supply.

我希望以这样的方式合并销售到供应,以便我的df_sales表在以下条件下具有来自df_supply的DATE_SUPPLY和QNT_SUPPLY

*条件:DATE_SUPPLY应该是对应"STORE"的相应"ITEM"的最近DATE_SUPPLY,即DATE_SALE <- max(df_supply[df_supply$DATE_SUPPLY <= df_sales$DATE_SALE & df_supply$STORE == df_sales$STORE & df_supply$ITEM == df_sales$ITEM,]$DATE_SUPPLY)*

可以使用行应用功能或仅通过写循环.但我有庞大的数据集,所以不想循环.

df_sales <- data.frame("STORE"=c(1001,1001,1001,1001,1001,1002,1002,1002,1002,1002),"ITEM"=c(13048, 13057, 13082, 13048, 13057, 13145, 13166, 13229, 13057, 13048),"DATE_SALE"=as.Date(c("1/1/2014","1/1/2014","1/2/2014","1/2/2014","1/2/2014","1/2/2014","1/3/2014","1/3/2014","1/3/2014","1/4/2014"),"%m/%d/%Y"),"QNT_SALE"=c(1,1,1,1,1,1,1,1,1,1))

df_sales

   STORE  ITEM  DATE_SALE QNT_SALE
1   1001 13048 2014-01-01        1
2   1001 13057 2014-01-01        1
3   1001 13082 2014-01-02        1
4   1001 13048 2014-01-02        1
5   1001 13057 2014-01-02        1
6   1002 13145 2014-01-02        1
7   1002 13166 2014-01-03        1
8   1002 13229 2014-01-03        1
9   1002 13057 2014-01-03        1
10  1002 13048 2014-01-04        1

df_supply <- data.frame("STORE"=c(1001,1002,1001,1001,1002,1002,1002,1002,1002,1002),"ITEM"=c(13048,13229,13057,13082,13145,13166,13229,13057,13048,13048),"DATE_SUPPLY"=as.Date(c("1/31/2013","1/31/2013","1/31/2013","1/1/2014","1/2/2014","1/2/2014","1/2/2014","1/2/2014","1/3/2014","2/1/2014"),"%m/%d/%Y"),"QNT_SUPPLY"=c(2,1,2,1,1,1,2,3,1,2))
df_supply
   STORE  ITEM DATE_SUPPLY CUM_QNT_SUPPLY
1   1001 13048 2013-01-31          2
2   1002 13229 2013-01-31          1
3   1001 13057 2013-01-31          2
4   1001 13082 2014-01-01          1
5   1002 13145 2014-01-02          1
6   1002 13166 2014-01-02          1
7   1002 13229 2014-01-02          2
8   1002 13057 2014-01-02          3
9   1002 13048 2014-01-03          1
10  1002 13048 2014-02-01          2



Output Required:
Sales Vs Supply
   STORE  ITEM  DATE_SALE QNT_SALE  DATE_SUPPLY QNT_SUPPLY
1   1001 13048 2014-01-01        1  2013-01-31          2
2   1001 13057 2014-01-01        1  2013-01-31          2
3   1001 13082 2014-01-02        1  2014-01-01          1
4   1001 13048 2014-01-02        1  2013-01-31          2
5   1001 13057 2014-01-02        1  2013-01-31          2
6   1002 13145 2014-01-03        1  2014-01-02          1
7   1002 13166 2014-01-03        1  2014-01-02          1
8   1002 13229 2014-01-03        1  2014-01-02          2
9   1002 13057 2014-01-03        1  2014-01-02          3
10  1002 13048 2014-01-04        1  2014-01-03          1

Answer 1

使用滚动连接来自data.table:

require(data.table)
setkey(setDT(df_supply), STORE, ITEM, DATE_SUPPLY)
idx = df_supply[df_sales, roll=Inf, which=TRUE]
cbind(df_sales, df_supply[idx, 3:4, with=FALSE])
#    STORE  ITEM  DATE_SALE QNT_SALE DATE_SUPPLY QNT_SUPPLY
# 1   1001 13048 2014-01-01        1  2013-01-31          2
# 2   1001 13057 2014-01-01        1  2013-01-31          2
# 3   1001 13082 2014-01-02        1  2014-01-01          1
# 4   1001 13048 2014-01-02        1  2013-01-31          2
# 5   1001 13057 2014-01-02        1  2013-01-31          2
# 6   1002 13145 2014-01-02        1  2014-01-02          1
# 7   1002 13166 2014-01-03        1  2014-01-02          1
# 8   1002 13229 2014-01-03        1  2014-01-02          2
# 9   1002 13057 2014-01-03        1  2014-01-02          3
# 10  1002 13048 2014-01-04        1  2014-01-03          1

cbind返回一个全新的对象.相反,如果你想通过引用添加新列来代替df_sales使用:=.在SO上有很多使用它的例子,并在新的HTML插图下进行了解释.