汇总列表中具有相同列 ID 的行

Question

假设我们有以下列表，其中包含一些矩阵（可能很多，但为简单起见，我现在放两个）

> rel$E
      i   j   value             
 [1,] "3" "5" "0.070711136732969"  
 [2,] "3" "6" "0.0555555555555557" 
 [3,] "1" "3" "0.0178395371187627" 
 [4,] "1" "2" "0.00488002262797937"
 [5,] "2" "5" "0.0272189957856598" 
 [6,] "3" "4" "0.0438244348035453" 
 [7,] "2" "4" "0.0128664170608579" 
 [8,] "3" "3" "0"                  
 [9,] "2" "3" "0.0167431138932832" 
[10,] "2" "2" "0"                  
[11,] "4" "5" "0.0180208592355387" 
[12,] "2" "6" "0.028063474878704"  
[13,] "1" "5" "0.00937210021651717"
[14,] "1" "4" "0.0033698603568658" 
> rel$D
      i   j   value             
 [1,] "1" "3" "0.0398765637816322" 
 [2,] "1" "1" "0"                  
 [3,] "1" "4" "0.00452411512576561"
 [4,] "3" "4" "0.0193536780493677" 
 [5,] "1" "2" "0.00289466496926153"
 [6,] "2" "5" "0.0283053038069326" 
 [7,] "3" "3" "0"                  
 [8,] "3" "6" "0.0862179235688977" 
 [9,] "1" "5" "0.0242662144621697" 
[10,] "1" "6" "0.00584795321637427"
[11,] "4" "5" "0.0174208488656519" 
[12,] "2" "3" "0.0443079152300233" 
[13,] "2" "2" "0"                  
[14,] "2" "4" "0.0131264776661371" 
[15,] "3" "5" "0.0952553775375157" 

总而言之，我想要实现的是矩阵的所有values 列具有相同的i 和j。

例如对于i=3、j=5，总和应该是0.070711136732969 + 0.0952553775375157。

但是正如您所见，有一些对存在于一个矩阵中，但不存在于另一个矩阵中。

对于这种情况i=1、j=6，总和应该是0 + 0.00584795321637427，因为这对不存在于第一个矩阵中。

是否有任何有效的（lapply，apply）方法来生成具有列i, j, sum 的最终矩阵？但是不使用很多 for 循环？我尝试使用 for 循环来处理它，但最终代码变得难以阅读和更改。

Answer 1

你需要rbind和aggregate，即

aggregate(value ~ i+j, do.call(rbind, l2), sum)

给出，

   i j       value
1  1 1 0.000000000
2  1 2 0.007774688
3  2 2 0.000000000
4  1 3 0.057716101
5  2 3 0.061051029
6  3 3 0.000000000
7  1 4 0.007893975
8  2 4 0.025992895
9  3 4 0.063178113
10 1 5 0.033638315
11 2 5 0.055524300
12 3 5 0.165966514
13 4 5 0.035441708
14 1 6 0.005847953
15 2 6 0.028063475
16 3 6 0.141773479

数据

dput(l2)
list(structure(c(3, 3, 1, 1, 2, 3, 2, 3, 2, 2, 4, 2, 1, 1, 5, 
6, 3, 2, 5, 4, 4, 3, 3, 2, 5, 6, 5, 4, 0.070711136732969, 0.0555555555555557, 
0.0178395371187627, 0.00488002262797937, 0.0272189957856598, 
0.0438244348035453, 0.0128664170608579, 0, 0.0167431138932832, 
0, 0.0180208592355387, 0.028063474878704, 0.00937210021651717, 
0.0033698603568658), .Dim = c(14L, 3L), .Dimnames = list(NULL, 
    c("i", "j", "value"))), structure(c(1, 1, 1, 3, 1, 2, 3, 
3, 1, 1, 4, 2, 2, 2, 3, 3, 1, 4, 4, 2, 5, 3, 6, 5, 6, 5, 3, 2, 
4, 5, 0.0398765637816322, 0, 0.00452411512576561, 0.0193536780493677, 
0.00289466496926153, 0.0283053038069326, 0, 0.0862179235688977, 
0.0242662144621697, 0.00584795321637427, 0.0174208488656519, 
0.0443079152300233, 0, 0.0131264776661371, 0.0952553775375157
), .Dim = c(15L, 3L), .Dimnames = list(NULL, c("i", "j", "value"
))))