我想做这样的事情：
> colname <- "a"
> dt <- data.table(colname = c(1,2,3,4))
a
1: 1
2: 2
3: 3
4: 4
但相反，我明白了：
> colname <- "a"
> dt <- data.table(colname = c(1,2,3,4))
colname
1: 1
2: 2
3: 3
4: 4
除了创建data.table并在之后分配名称之外，还有其他方法吗？
答案 0 :(得分：2)
我们可以使用setnames。 setnames的优点是，它可以通过提供old和new参数中的名称来命名所有列或列的子集
dt <- data.table(c(1,2,3,4))
setnames(dt, colname)
我们可以在创建对象本身时包装setnames
dt <- setnames(data.table(c(1,2,3,4)), colname)[]
dt
# a
#1: 1
#2: 2
#3: 3
#4: 4
答案 1 :(得分：2)
不使用@Entity(tableName = "TB_CLASSES")
class TbClass(
@ColumnInfo(name = "ID")
@PrimaryKey(autoGenerate = true) var id: Int =-1,
@ColumnInfo(name = "NAME") var name: String="",
@ColumnInfo(name = "CAPTION") var caption: String="",
@ColumnInfo(name = "TYPE") var type: String="")
@Entity(tableName = "TB_FEATURE_DISA")
class TbFeatureDisa(
@ColumnInfo(name = "ID")
@PrimaryKey(autoGenerate = true) var id: Int=-1,
@ColumnInfo(name = "FEATURE_ID") var featureId: Int=-1,
@ColumnInfo(name = "DISA_ID") var disaId: Int=-1,
@ColumnInfo(name = "IS_SOLVED") var isSolved: Int=-1,
@ColumnInfo(name = "DISA_LEVEL") var disaLevel: Double=-0.1,
@ColumnInfo(name = "RESOLVER_USER_NAME") var resolverUserName: String="",
@ColumnInfo(name = "RESOLVE_TIME") var resolveTime: Date? =null,
@ColumnInfo(name = "REPORT_VIST_ID") var reportVisitId: Int=-1,
@ColumnInfo(name = "REPORT_FAULT_ID") var reportFaultId: Int=-1,
@ColumnInfo(name = "SOLVE_VIST_ID") var solveVisitId: Int=-1,
@ColumnInfo(name = "SOLVE_SERVICE_ID") var solveServiceId: Int=-1,
@ColumnInfo(name = "SOLVE_FUALT_ID") var solveFaultId: Int=-1)
的另一种方法是使用基函数data.table
names如果您有一个包含多列的数据框，只需将列名称向量传递给函数> colnames <- "a"
> dt <- data.table(c(1,2,3,4))
> dt
V1
1: 1
2: 2
3: 3
4: 4
> names(dt) <- colnames
> dt
a
1: 1
2: 2
3: 3
4: 4
。