我有一个data.table由不同操作创建的两个 s，但值是相同的。但是，正如输出所示，它们具有不同的结构：
> 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
有没有一种简单的方法可以rbind对它们进行处理而不是重新调整因素，因为级别“标题”是通过构建对的？
澄清我在问什么：我期待这样的行为。我有两个 data.tables/framesa和b. 两者都有变量x，factor但在a1->“bla”、2->“ble”和b1->“ble”、2->“bla”中。让我们假设结构a$xisbla bla ble ble和b$xis ble ble bla bla。通过做merged <- rbind(a,b)我想得到merged$x的是bla bla ble ble ble ble bla bla，但我正在经历merged$x的是bla bla ble ble bla bla ble ble。
感谢您的建议。