在R中，它们通常比替代向量化解决方案慢得多。

In R it is always preferable to avoid loops wherever possible, as they are usually much slower than alternative vectorized solutions.

此操作可以使用data.table连接完成。基本上，当您运行

This operation can be done with a data.table join. Basically, when you run

dt1[dt2];

您正在执行两个data.tables之间的右连接。 dt1 的预设键列决定要加入哪些列。如果 dt1 没有预置键，则操作失败。但您可以指定参数以手动选择键列：

you are performing a right-join between the two data.tables. The preset key columns of dt1 determine which columns to join on. If dt1 has no preset key, the operation fails. But you can specify the on argument to manually select the key columns on-the-fly:

key <- paste0('V',1:5);
dt1[dt2,on=key];

（另一种方法当然是使用 setkey ）或 setkeyv（）。）

(The alternative of course is to preset a key, using either setkey() or setkeyv().)

上述操作实际上只是返回合并表包含来自 dt1 和 dt2 的数据，这不是您想要的。但我们可以使用data.table索引函数的 j 参数和：= 就地赋值语法将 dt2 的 id 列分配给 id dt1 的列。因为我们有名称冲突，我们必须使用 i.id 引用 id 列> dt2 ，而未修改的名称 id 仍指向 id c $ c> dt1 。这只是data.table提供的用于消除冲突名称的歧义的机制。因此，您正在寻找：

The above operation will actually just return a merged table containing data from both dt1 and dt2, which is not what you want. But we can make use of the j argument of the data.table indexing function and the := in-place assignment syntax to assign the id column of dt2 to the id column of dt1. Because we have a name conflict, we must use i.id to reference the id column of dt2, while the unmodified name id still refers to the id column of dt1. This is simply the mechanism provided by data.table for disambiguating conflicting names. Hence, you're looking for:

dt1[dt2,on=key,id:=i.id];

下面是一个仅使用两个键列和仅几行数据我还生成了键以包括一些不匹配的行，只是为了演示不匹配的行将其操作不会改变其id。

Here's an example that uses only two key columns and just a few rows of data (for simplicity). I also generated the keys to include some non-matching rows, just to demonstrate that the non-matching rows will have their ids left untouched by the operation.

set.seed(1L);
dt1 <- data.table(id=1:12,expand.grid(V1=1:3,V2=1:4),blah1=rnorm(12L));
dt2 <- data.table(id=13:18,expand.grid(V1=1:2,V2=1:3),blah2=rnorm(6L));
dt1;
##     id V1 V2      blah1
##  1:  1  1  1 -0.6264538
##  2:  2  2  1  0.1836433
##  3:  3  3  1 -0.8356286
##  4:  4  1  2  1.5952808
##  5:  5  2  2  0.3295078
##  6:  6  3  2 -0.8204684
##  7:  7  1  3  0.4874291
##  8:  8  2  3  0.7383247
##  9:  9  3  3  0.5757814
## 10: 10  1  4 -0.3053884
## 11: 11  2  4  1.5117812
## 12: 12  3  4  0.3898432
dt2;
##    id V1 V2       blah2
## 1: 13  1  1 -0.62124058
## 2: 14  2  1 -2.21469989
## 3: 15  1  2  1.12493092
## 4: 16  2  2 -0.04493361
## 5: 17  1  3 -0.01619026
## 6: 18  2  3  0.94383621
key <- paste0('V',1:2);
dt1[dt2,on=key,id:=i.id];
dt1;
##     id V1 V2      blah1
##  1: 13  1  1 -0.6264538
##  2: 14  2  1  0.1836433
##  3:  3  3  1 -0.8356286
##  4: 15  1  2  1.5952808
##  5: 16  2  2  0.3295078
##  6:  6  3  2 -0.8204684
##  7: 17  1  3  0.4874291
##  8: 18  2  3  0.7383247
##  9:  9  3  3  0.5757814
## 10: 10  1  4 -0.3053884
## 11: 11  2  4  1.5117812
## 12: 12  3  4  0.3898432