我有一个关于data.table
“非连接”成语的问题,灵感来自迭代器的问题。这是一个例子:
library(data.table)
dt1 <- data.table(A1=letters[1:10], B1=sample(1:5,10, replace=TRUE))
dt2 <- data.table(A2=letters[c(1:5, 11:15)], B2=sample(1:5,10, replace=TRUE))
setkey(dt1, A1)
setkey(dt2, A2)
data.table
s 看起来像这样
> dt1 > dt2
A1 B1 A2 B2
[1,] a 1 [1,] a 2
[2,] b 4 [2,] b 5
[3,] c 2 [3,] c 2
[4,] d 5 [4,] d 1
[5,] e 1 [5,] e 1
[6,] f 2 [6,] k 5
[7,] g 3 [7,] l 2
[8,] h 3 [8,] m 4
[9,] i 2 [9,] n 1
[10,] j 4 [10,] o 1
要查找 中的哪些行dt2
具有相同的键dt1
,请将which
选项设置为TRUE
:
> dt1[dt2, which=TRUE]
[1] 1 2 3 4 5 NA NA NA NA NA
马修在这个答案中建议,“不加入”成语
dt1[-dt1[dt2, which=TRUE]]
dt1
将那些具有未出现在中的索引的行作为子集dt2
。在我的data.table
v1.7.1 机器上,我收到一个错误:
Error in `[.default`(x[[s]], irows): only 0's may be mixed with negative subscripts
相反,使用选项nomatch=0
,“非加入”有效
> dt1[-dt1[dt2, which=TRUE, nomatch=0]]
A1 B1
[1,] f 2
[2,] g 3
[3,] h 3
[4,] i 2
[5,] j 4
这是预期的行为吗?