合并具有重复键值的两个数据集答案

【问题标题】：Merging two datasets with duplicate key values合并具有重复键值的两个数据集
【发布时间】：2020-01-28 06:27:50
【问题描述】：

我有两个数据表x和y，如下图：

x
Sample  code    code_conn   pool_content
Sample1   A1        X1      ASD
Sample1   A1        X3      POS
Sample1   B1        Y1      QAW
Sample1   B1        Y2      QWP
Sample1   B1        Y3      SDE
Sample1   B1        Y4      LKS
Sample1   C1        Z2      JMP
Sample1   C1        Z3      MSK
Sample2   A1        X1      ASD
Sample2   A1        X2      POS
Sample2   B1        Y1      QAW
Sample2   B1        Y2      QWP
Sample2   B1        Y3      LKS
Sample2   B1        Y4      ASE
Sample2   B1        Y5      ASM
Sample2   C1        Z1      MSK
Sample2   D1        P1      EAX
Sample2   D1        P2      EBX
Sample2   D1        P3      ECX
Sample2   D1        P4      EDX
Sample2   D1        P5      MOV
Sample2   D1        P6      ALL

y
code    code_conn
A1      X1
A1      X2
A1      X3
B1      Y1
B1      Y2
B1      Y3
B1      Y4
B1      Y5
C1      Z1
C1      Z2
C1      Z3
C1      Z4
D1      P1
D1      P2
D1      P3
D1      P4
D1      P5
D1      P6

我希望输出如下：

Sample  code    code_conn   pool_content
Sample1 A1      X1          ASD
Sample1 A1      X2          …
Sample1 A1      X3          POS
Sample1 B1      Y1          QAW
Sample1 B1      Y2          QWP
Sample1 B1      Y3          SDE
Sample1 B1      Y4          LKS
Sample1 B1      Y5          …
Sample1 C1      Z2          JMP
Sample1 C1      Z3          MSK
Sample1 C1      Z1          …
Sample1 C1      Z4          …
Sample2 A1      X1          ASD
Sample2 A1      X2          POS
Sample2 A1      X3          …
Sample2 B1      Y1          QAW
Sample2 B1      Y2          QWP
Sample2 B1      Y3          LKS
Sample2 B1      Y4          ASE
Sample2 B1      Y5          ASM
Sample2 C1      Z1          MSK
Sample2 C1      Z2          …
Sample2 C1      Z3          …
Sample2 C1      Z4          …
Sample2 D1      P1          EAX
Sample2 D1      P2          EBX
Sample2 D1      P3          ECX
Sample2 D1      P4          EDX
Sample2 D1      P5          MOV
Sample2 D1      P6          ALL

所以，我基本上想要的是，对于 x 中的每个唯一样本，对于每组代码，我希望将表 x 中表 y 中所有缺失的 code_conn 及其 pool_content 设置为“...”。有没有一种内存高效且快速的方法来做到这一点。 code 和 code_conn 可以是任何字符串（不是特别是字母和数字的序列组合）。我附上了生成下面数据的代码。非常感谢。

x <- data.table("Sample" = c(rep("Sample1", 8), rep("Sample2", 14)), 
           "code" = c("A1", "A1", "B1", "B1", "B1", "B1", "C1", "C1", "A1", "A1", rep("B1", 5), "C1", "C1", rep("D1", 5)), 
           "code_conn" = c("X1","X3","Y1","Y2","Y3","Y4","Z2","Z3","X1","X2","Y1","Y2","Y3","Y4","Y5","Z1","P1","P2","P3","P4","P5","P6"), 
           "pool_content" = c("ASD","POS","QAW","QWP","SDE","LKS","JMP","MSK","ASD","POS","QAW","QWP","LKS","ASE","ASM","MSK","EAX","EBX","ECX","EDX","MOV","ALL"))

y <- data.table("code" = c(rep("A1", 3), rep("B1", 5), rep("C1", 4), rep("D1", 6)), 
                "code_conn" = c("X1","X2","X3","Y1","Y2","Y3","Y4","Y5","Z1","Z2","Z3","Z4","P1","P2","P3","P4","P5","P6"))

【问题讨论】：

merge(x, y, all = T) 应该这样做。它将输入NAs，您可以将其替换为... 或您想要的任何其他字符串。
我尝试过合并，但它给出的行数比预期的要少。
糟糕，我错过了获取所有样本的CJ 步骤。 tidyr::complete 可能是一种更简单的方法。

标签： r merge

【解决方案1】：

也许您可以尝试以下基本 R 解决方案（使用 merge()）

z <- merge(x,cbind(Sample = rep(unique(x$Sample),each = nrow(y)),y),all = T)

这样

> z
    Sample code code_conn pool_content
1  Sample1   A1        X1          ASD
2  Sample1   A1        X2         <NA>
3  Sample1   A1        X3          POS
4  Sample1   B1        Y1          QAW
5  Sample1   B1        Y2          QWP
6  Sample1   B1        Y3          SDE
7  Sample1   B1        Y4          LKS
8  Sample1   B1        Y5         <NA>
9  Sample1   C1        Z1         <NA>
10 Sample1   C1        Z2          JMP
11 Sample1   C1        Z3          MSK
12 Sample1   C1        Z4         <NA>
13 Sample1   D1        P1         <NA>
14 Sample1   D1        P2         <NA>
15 Sample1   D1        P3         <NA>
16 Sample1   D1        P4         <NA>
17 Sample1   D1        P5         <NA>
18 Sample1   D1        P6         <NA>
19 Sample2   A1        X1          ASD
20 Sample2   A1        X2          POS
21 Sample2   A1        X3         <NA>
22 Sample2   B1        Y1          QAW
23 Sample2   B1        Y2          QWP
24 Sample2   B1        Y3          LKS
25 Sample2   B1        Y4          ASE
26 Sample2   B1        Y5          ASM
27 Sample2   C1        Z1          MSK
28 Sample2   C1        Z2         <NA>
29 Sample2   C1        Z3         <NA>
30 Sample2   C1        Z4         <NA>
31 Sample2   D1        P1          EAX
32 Sample2   D1        P2          EBX
33 Sample2   D1        P3          ECX
34 Sample2   D1        P4          EDX
35 Sample2   D1        P5          MOV
36 Sample2   D1        P6          ALL

【讨论】：

【解决方案2】：

我认为 dplyr::left_join 通过示例和代码可以帮助您。对于不存在的值，它将为您提供 NULL 并将其保持与您拥有的行数相同。

【讨论】：