匹配 df2 中的 df1$col1 并将 df2$col1 中的输出值匹配到 df1$col2答案

【问题标题】：Match in df1$col1 in df2 and output value in df2$col1 into df1$col2匹配 df2 中的 df1$col1 并将 df2$col1 中的输出值匹配到 df1$col2
【发布时间】：2021-03-26 15:50:11
【问题描述】：

我有两个数据框。

symbols <- c("Santa", "Elves", "Candy Cane", "Reindeers", "Cats",
             "Turkey", "Mashed Potatoes", "Cranberry Sauce", "Dogs",
             "Eggs", "Chocolates with cream", "Bunnies", "Flowers", "Donut")


df1 <- data.frame(symbols)

df1
                 symbols
1                  Santa
2                  Elves
3             Candy Cane
4              Reindeers
5                   Cats
6                 Turkey
7        Mashed Potatoes
8        Cranberry Sauce
9                   Dogs
10                  Eggs
11 Chocolates with cream
12               Bunnies
13               Flowers
14                 Donut

holiday <- c("Christmas", "Thanksgiving", "Easter")
v1 <- c("Santa", "Turkey", "Eggs")
v2 <- c("Elves", "Mashed Potatoes", "Chocolates with cream")
v3 <- c("Candy Canes", "Cranberry Sauce", "Bunnies")
v4 <- c("Reindeers", NA, "Flowers")

df2 <- data.frame(holiday, v1, v2, v3, v4)

df2
       holiday     v1
1    Christmas  Santa
2 Thanksgiving Turkey
3       Easter   Eggs
                     v2              v3
1                 Elves     Candy Canes
2       Mashed Potatoes Cranberry Sauce
3 Chocolates with cream         Bunnies
         v4
1 Reindeers
2      <NA>
3   Flowers

如果 df1$symbols 中的任何内容与 df2 中的任何值（df2$holiday、df2$v1、df2$v2、df2$v3、df2$v4）匹配，我希望它将 df2$holiday 值输出到df1 中的新列。

理想情况下，我会有一个如下所示的 df1：

    df1
                 symbols      holiday
1                  Santa    Christmas
2                  Elves    Christmas
3             Candy Cane    Christmas
4              Reindeers    Christmas
5                   Cats         <NA>
6                 Turkey Thanksgiving
7        Mashed Potatoes Thanksgiving
8        Cranberry Sauce Thanksgiving
9                   Dogs         <NA>
10                  Eggs       Easter
11 Chocolates with cream       Easter
12               Bunnies       Easter
13               Flowers       Easter
14                 Donut         <NA>

我认为我可以做到的一种方法是将 df2 拆分，然后为每一列执行 left_join：

df2_v1 <- data.frame(df2$holiday, df2$v1)
df2_v2 <- data.frame(df2$holiday, df2$v2)
df2_v3 <- data.frame(df2$holiday, df2$v3)
df2_v4 <- data.frame(df2$holiday, df2$v4)

Then I can use left_join for each df1 with df2_v#. For example:

df1_x <- left_join(df1, df2_v1, by = c("symbols" = "df2.v1"))

然后我可以合并或使用一些 ifelse 逻辑来获得一个干净的 df1$holiday 列，但如果 df2 中有更多列，这将非常耗时。

有更快的方法吗？

【问题讨论】：

标签： r dataframe match

【解决方案1】：

library( data.table )
setDT(df1);setDT(df2)
#join on molten df2
df1[ melt(df2, id.vars = "holiday"), 
     holiday := i.holiday, on = .(symbols = value)]

#                  symbols      holiday
# 1:                 Santa    Christmas
# 2:                 Elves    Christmas
# 3:            Candy Cane         <NA>
# 4:             Reindeers    Christmas
# 5:                  Cats         <NA>
# 6:                Turkey Thanksgiving
# 7:       Mashed Potatoes Thanksgiving
# 8:       Cranberry Sauce Thanksgiving
# 9:                  Dogs         <NA>
#10:                  Eggs       Easter
#11: Chocolates with cream       Easter
#12:               Bunnies       Easter
#13:               Flowers       Easter
#14:                 Donut         <NA>

【讨论】：