Question

示例数据：

    df1 <- structure(list(Name = structure(c(3L, 2L, 1L), .Label = c("Bob", 
"Joe", "Mike"), class = "factor"), Location = structure(c(1L, 
1L, 2L), .Label = c("CA", "WA"), class = "factor"), Title = structure(c(2L, 
3L, 1L), .Label = c("CEO", "Manager", "VP"), class = "factor"), 
    Class = structure(c(1L, 2L, 2L), .Label = c("Class1", "Class2"
    ), class = "factor"), Month = c(1, 2, 3), Class.1 = structure(c(3L, 
    2L, 1L), .Label = c("Class1", "Class2", "Class4"), class = "factor"), 
    Month.1 = c(3, 3, 2), Objective = structure(1:3, .Label = c("Obj1", 
    "Obj2", "Obj3"), class = "factor"), Month.2 = c(2, 7, 7), 
    Category = c("x", "y", "z"), Objective.1 = structure(c(3L, 
    2L, 1L), .Label = c("Obj1", "Obj7", "Obj9"), class = "factor"), 
    Month.3 = c(4, 5, 5), Category2 = c("z", "r", "q")), .Names = c("Name", 
"Location", "Title", "Class", "Month", "Class.1", "Month.1", 
"Objective", "Month.2", "Category", "Objective.1", "Month.3", 
"Category2"), class = "data.frame", row.names = c(NA, -3L))

  Name Location   Title  Class Month Class.1 Month.1 Objective Month.2 Category Objective.1 Month.3 Category2
1 Mike       CA Manager Class1     1  Class4       3      Obj1       2        x        Obj9       4         z
2  Joe       CA      VP Class2     2  Class2       3      Obj2       7        y        Obj7       5         r
3  Bob       WA     CEO Class2     3  Class1       2      Obj3       7        z        Obj1       5         q

我想收集每个观察一行的表格：

Name Location Title Variable(Class/Objective) Value

我已经使用gather，spread等在堆栈上尝试了一些相似的示例，但我无法弄清楚如何保持Class-Month和Objective -Month团结在一起。

在我的真实数据集中，有100列，其中包含8个ID列。而不仅仅是Class-Month或Objective-Month对，列的前半部分是四个一组，后半部分是8个组。一组四个的例子是Class-Month-Cost -date。

Mike的示例输出：

  Name Location   Title Variable Value Value.2
1 Mike       CA Manager   Class1     1    <NA>
2 Mike       CA Manager   Class4     3    <NA>
3 Mike       CA Manager     Obj1     2       x
4 Mike       CA Manager     Obj9     4       z

Answer 1

重复的值很好，但是您需要指定哪些组合在一起（在示例中，＆＃34; Class＆＃34;＆＃34; Objective＆＃34;）以获得OP＆＃39输出：

library(data.table)
melt(setDT(df1), 
  meas = patterns("Class|Objective", "Month", "Category")
)[order(Name)]

    Name Location   Title variable value1 value2 value3
 1:  Bob       WA     CEO        1 Class2      3      z
 2:  Bob       WA     CEO        2 Class1      2      q
 3:  Bob       WA     CEO        3   Obj3      7     NA
 4:  Bob       WA     CEO        4   Obj1      5     NA
 5:  Joe       CA      VP        1 Class2      2      y
 6:  Joe       CA      VP        2 Class2      3      r
 7:  Joe       CA      VP        3   Obj2      7     NA
 8:  Joe       CA      VP        4   Obj7      5     NA
 9: Mike       CA Manager        1 Class1      1      x
10: Mike       CA Manager        2 Class4      3      z
11: Mike       CA Manager        3   Obj1      2     NA
12: Mike       CA Manager        4   Obj9      4     NA

如果您具有相同的重复列名称或使用check.names=TRUE消除歧义，那将无关紧要，因为patterns只匹配名称中的模式。有关如何根据需要指定模式的详细信息，请参阅?regex。

melt的其他参数（参见?melt.data.table）可用于为结果中的列提供自定义名称（而不是＆＃34; value1＆＃34;，＆＃34; value2＆＃ 34;，...）。

R：收集重复的列

1 个答案: