如何将相同的唯一ID分配给r中两个数据框之间的匹配观察值？

Question

我有一个实际的问题时，我有两个（或更多）数据帧和要指派的唯一标识号内的每个每个匹配的观察和跨两个数据集如：

#1. Create dataframe df1: 

a1 <- c(1, 1, 1, 1, 2, 2, 2, 2, 1, 1) 
b1 <- c(1, 5, 3, 2, 3, 4, 5, 1, 5, 2) 
c1 <- c("white", "red", "black", "white", "red", 
     "white", "black", "silver", "red", "green") 
df1 <- data.frame(a1, b1, c1) 
df1 

    a1 b1  c1 
1 1 1 white 
2 1 5 red 
3 1 3 black 
4 1 2 white 
5 2 3 red 
6 2 4 white 
7 2 5 black 
8 2 1 silver 
9 1 5 red 
10 1 2 green 

#2. Create dataframe df2: 

a2 <- c(2, 2, 1, 1, 2, 2, 2, 2, 2, 2) 
b2 <- c(3, 1, 3, 2, 1, 3, 4, 5, 3, 5) 
c2 <- c("black", "blue", "black", "white", "silver", 
     "green", "green", "red", "blue", "white") 
df2 <- data.frame(a2, b2, c2) 
df2 

    a2 b2  c2 
1 2 3 black 
2 2 1 blue 
3 1 3 black 
4 1 2 white 
5 2 1 silver 
6 2 3 green 
7 2 4 green 
8 2 5 red 
9 2 3 blue 
10 2 5 white 

#3. Assign unique IDs to each observation in df1: 

library(data.table) 
df1.2 <- data.table(df1, key="a1,b1,c1") 
df1.2[, id:=.GRP, by=key(df1.2)] 
df1.2 <- as.data.frame(df1.2) 
df1.2 

    a1 b1  c1 id 
1 1 1 white 1 
2 1 2 green 2 
3 1 2 white 3 
4 1 3 black 4 
5 1 5 red 5 
6 1 5 red 5 
7 2 1 silver 6 
8 2 3 red 7 
9 2 4 white 8 
10 2 5 black 9 

#4. The problematic part!! Assign identical unique IDs to matching observations of df2 as compared to df1.2 
#and assign other unique IDs to all other non-matching obs of df2. 
#Name the resulting dataframe as df2.2 
#My expected result will ideally look as follows: 

df2.2 

    a2 b2  c2 id 
1 2 3 black 10 
2 2 1 blue 11 
3 1 3 black 4 
4 1 2 white 3 
5 2 1 silver 6 
6 2 3 green 12 
7 2 4 green 13 
8 2 5 red 14 
9 2 3 blue 15 
10 2 5 white 16 

任何帮助的如何获得df2.2将非常感激。谢谢。

Answer 1

一种简单的方式来处理，这是做一个哈希：

library(dplyr) 
library(digest) 

df1 %>% 
    rowwise() %>% 
    do(data.frame(., id=digest(paste(.$a1,.$b1,.$c1), algo="md5"), 
        stringsAsFactors=FALSE)) %>% ungroup() 

df2 %>% 
    rowwise() %>% 
    do(data.frame(., id=digest(paste(.$a2,.$b2,.$c2), algo="md5"), 
       stringsAsFactors=FALSE)) %>% ungroup() 

这将产生对df1如下：

a1 b1  c1        id 
1 1 1 white b86fbb78b27f7db2ee50af2d68cce452 
2 1 5 red 68d47f544832989834517630e4a2764c 
3 1 3 black 724e37192140cb2009cf3d982f2be1e4 
4 1 2 white f731b8b38255b8c312543283f8e1c634 
5 2 3 red 2d50b86902056a51faad04d2c566faf2 
6 2 4 white 9396667cd51d1e1b61b0b22a7767d3d9 
7 2 5 black 9ba1f3e04c61c006d3c5382fcad098e6 
8 2 1 silver 38dcd29d200c8b33cd38ac78ef9dd751 
9 1 5 red 68d47f544832989834517630e4a2764c 
10 1 2 green 7d9b1aadfd79de142b234b83d7867b9b 

和df2如下：

a2 b2  c2        id 
1 2 3 black d285febc8ab08e99b11609b98f077e66 
2 2 1 blue bfa0405276406ac4bc596daf957dfa11 
3 1 3 black 724e37192140cb2009cf3d982f2be1e4 
4 1 2 white f731b8b38255b8c312543283f8e1c634 
5 2 1 silver 38dcd29d200c8b33cd38ac78ef9dd751 
6 2 3 green 67eefe9ee2d82486ded30a268289296b 
7 2 4 green d773f58cf144eab15ef459e326494a2f 
8 2 5 red 0724318a9f59d3960edfe4e90f9c4eff 
9 2 3 blue 6883420cc137ba45b773f642176e9ce6 
10 2 5 white 5dea9e63b5fbfb31fb81260cb5a5f41c 

Answer 2

通过编写生成唯一ID的函数，然后将其应用于df1和df2的组合，您可以完成所需的任务。

# Inspiration: http://stackoverflow.com/questions/24119599/how-to-assign-a-unique-id-number-to-each-group-of-identical-values-in-a-column 
unique.id <- function(x) as.numeric(factor(x)) 

(df1.info <- do.call(paste, df1)) 
# [1] "1 1 white 1" "1 5 red 5" "1 3 black 4" "1 2 white 3" "2 3 red 11" 
# [6] "2 4 white 13" "2 5 black 14" "2 1 silver 7" "1 5 red 5" "1 2 green 2" 
df2.info <- do.call(paste, df2) 
ids <- unique.id(c(df1.info, df2.info)) 
df1$id <- head(ids, nrow(df1)) 
df1 
# a1 b1  c1 id 
# 1 1 1 white 1 
# 2 1 5 red 5 
# 3 1 3 black 4 
# 4 1 2 white 3 
# 5 2 3 red 11 
# 6 2 4 white 13 
# 7 2 5 black 14 
# 8 2 1 silver 7 
# 9 1 5 red 5 
# 10 1 2 green 2 
df2$id <- tail(ids, nrow(df2)) 
df2 
# a2 b2  c2 id 
# 1 2 3 black 8 
# 2 2 1 blue 6 
# 3 1 3 black 4 
# 4 1 2 white 3 
# 5 2 1 silver 7 
# 6 2 3 green 10 
# 7 2 4 green 12 
# 8 2 5 red 15 
# 9 2 3 blue 9 
# 10 2 5 white 16 

Answer 3

假设您的列都是一样的，你的最简单的方法可能是：

df.all <- rbind(df1, df2) 

（您可能需要重命名的列是相同的。）

现在你已经想通了同样的伎俩与数据表，整个数据集。然后拆分数据集中备份：

df1 <- df.all[1:nrow(df1),] 
df2 <- df.all[- (1:nrow(df1)),] 

注：我不是说数据表关键是产生独特的组合数的理想方式！但是你已经写出来了。