Question

我正在寻找一种方法来重塑以下示例data

data <- structure(list(id = c(2L, 5L, 7L), name = structure(1:3, .Label = c("Test1","Test10", "Test8"), class = "factor"), source = structure(c(1L,3L, 2L), .Label = c("A", "T", "Z"), class = "factor")), .Names = c("id", "name", "source"), class = "data.frame", row.names = c(NA, -3L))

 id   name source
1  2  Test1      A
2  5 Test10      Z
3  7  Test8      T

进入以下结构

row.names   1.1    2.1     3.1
id            2      5       7
name      Test1 Test10   Test8
source        A      Z       T

如何根据data2的名称添加第二个data.frame，data（仅包含匹配名称的数据）？

data2 <- structure(list(name = structure(1L, .Label = "adddata", class = "factor"), Test1 = 10L, Test10 = 12L, Test8 = 17L, Test12 = 7L), .Names = c("name", "Test1", "Test10", "Test8", "Test12"), class = "data.frame", row.names = c(NA, -1L))

data2
   name   Test1 Test10 Test8 Test12
1 adddata    10     12    17      7

所以最终会出现类似下面的data.frame，其中只包含匹配的名称（data2中的Test12被遗漏）是这样的

datanew
  row.names   1.1    2.1   3.1
1        id     2      5     7
2      name Test1 Test10 Test8
3    source     A      Z     T
4   adddata    10     12    17

修改

我刚刚意识到我的输入数据包含这样的嵌套列表。有没有办法实现这个？

data <- structure(list(`1.1` = structure(list(id = structure(2, .Dim = c(1L, 1L)), name = structure("Test1", .Dim = c(1L, 1L)), source = structure("A", .Dim = c(1L, 1L))), .Names = c("id", "name", "source")), `2.1` = structure(list(id = structure(5, .Dim = c(1L, 1L)), name = structure("Test10", .Dim = c(1L, 1L)), source = structure("Z", .Dim = c(1L, 1L))), .Names = c("id", "name", "source")), `3.1` = structure(list(id = structure(7, .Dim = c(1L, 1L)), name = structure("Test8", .Dim = c(1L, 1L)), source = structure("T", .Dim = c(1L, 1L))), .Names = c("id", "name", "source"))), .Names = c("1.1", "2.1", "3.1"), class = "data.frame", row.names = c("id", "name", "source"))

'data.frame':   3 obs. of  3 variables:
 $ 1.1:List of 3
  ..$ id    : num [1, 1] 2
  ..$ name  : chr [1, 1] "Test1"
  ..$ source: chr [1, 1] "A"
 $ 2.1:List of 3
  ..$ id    : num [1, 1] 5
  ..$ name  : chr [1, 1] "Test10"
  ..$ source: chr [1, 1] "Z"
 $ 3.1:List of 3
  ..$ id    : num [1, 1] 7
  ..$ name  : chr [1, 1] "Test8"
  ..$ source: chr [1, 1] "T"

Answer 1

您可以将第一个数据集（'数据'）和rbind输出（'d1'）转换为'data2'中的列，我们使用列'名称之间的match进行子集化'in'data'和'data2'的列名

  d1 <- as.data.frame(t(data), stringsAsFactors=FALSE)
  res <- rbind(d1, setNames(data2[match(data$name, names(data2))], names(d1)))
  rownames(res)[4] <- as.character(data2$name)
  res
  #           V1     V2    V3
  #id          2      5     7
  #name    Test1 Test10 Test8
  #source      A      Z     T
  #adddata    10     12    17

或者其他选项是从data.table

加入

  library(data.table)#v1.9.5+
  DT <- setDT(data)[melt(data2, id.var='name', value.name='adddata', 
           variable.name='name')[-1], on='name', nomatch=0]
  DT
  #   id   name source adddata
  #1:  2  Test1      A      10
  #2:  5 Test10      Z      12
  #3:  7  Test8      T      17

我会保留这种格式而不是转置它，因为列是不同的类。如果我们进行转置，则数字和非数字元素会在列中混合在一起，而类将为factor或character

 t(DT)

更新

根据已修改的“数据”，我们可unlist list并转换为'data.frame'。然后，我们可以像以前一样使用这些步骤。

 data <- setNames(as.data.frame(matrix(unlist(data), ncol=3, 
               byrow=TRUE)), row.names(data))
 DT <- setDT(data)[melt(data2, id.var='name', value.name='adddata', 
            variable.name='name')[-1], on='name', nomatch=0]
 DT
 #   id   name source adddata
 #1:  2  Test1      A      10
 #2:  5 Test10      Z      12
 #3:  7  Test8      T      17

R：从列到行重新整形数据，并根据名称添加其他数据

1 个答案:

更新