我有两个data.table由不同的操作创建，但值是相同的。但是，它们具有不同的结构，如输出所示：
> str(resultstabledata)
Classes ‘data.table’ and 'data.frame': 234 obs. of 7 variables:
$ pT : Factor w/ 4 levels "Insignificant",..: 4 4 4 4 4 4 1 1 3 3 ...
$ shape : Factor w/ 2 levels "Base","Comparison": 1 1 1 1 1 1 1 1 1 1 ...
$ resid : chr "minzer" "minzer" "minzer" "minzer" ...
$ asset : Factor w/ 3 levels "crude","gold",..: 1 1 2 2 3 3 3 3 3 3 ...
$ base : Factor w/ 3 levels "mlp","elman",..: 1 1 1 1 1 1 1 1 1 1 ...
$ compared: Factor w/ 2 levels "arfima.roll",..: 1 2 1 2 1 2 1 2 1 2 ...
$ N : int 409 317 194 353 206 178 317 333 47 46 ...
- attr(*, ".internal.selfref")=<externalptr>
> str(a)
'data.frame': 7 obs. of 7 variables:
$ pT : Factor w/ 4 levels "Insignificant",..: 1 2 2 3 3 4 4
$ shape : Factor w/ 1 level "Comparison": 1 1 1 1 1 1 1
$ resid : Factor w/ 2 levels "minzer","diff": 1 1 2 1 2 1 2
$ asset : Factor w/ 1 level "crude": 1 1 1 1 1 1 1
$ base : Factor w/ 1 level "mlp": 1 1 1 1 1 1 1
$ compared: Factor w/ 1 level "lame": 1 1 1 1 1 1 1
$ N : num 0 0 0 0 0 0 0
是否有一种简单的方法如何对它们进行rbind而不是重新解释这些因素，因为“字幕”这个级别是按照构造正确的？
澄清我的要求：
我期待这样的行为。我有两个data.tables / frames a和b。两者都有变量x factor，但在a 1-＆gt;“bla”，2-＆gt;“ble”和b 1-＆gt;“ble ”，2→‘BLA’。我们假设结构a$x是bla bla ble ble而b$x是ble ble bla bla。通过执行merged <- rbind(a,b)，我希望merged$x为bla bla ble ble ble ble bla bla，但我遇到merged$x为bla bla ble ble bla bla ble ble。
感谢您的建议。