6

我有数据框

test <- structure(list(
     y2002 = c("freshman","freshman","freshman","sophomore","sophomore","senior"),
     y2003 = c("freshman","junior","junior","sophomore","sophomore","senior"),
     y2004 = c("junior","sophomore","sophomore","senior","senior",NA),
     y2005 = c("senior","senior","senior",NA, NA, NA)), 
              .Names = c("2002","2003","2004","2005"),
              row.names = c(c(1:6)),
              class = "data.frame")
> test
       2002      2003      2004   2005
1  freshman  freshman    junior senior
2  freshman    junior sophomore senior
3  freshman    junior sophomore senior
4 sophomore sophomore    senior   <NA>
5 sophomore sophomore    senior   <NA>
6    senior    senior      <NA>   <NA>

并且我需要创建一个顶点/边列表（用于 igraph），每次学生类别在连续几年发生变化时，而在没有变化时忽略，如

testvertices <- structure(list(
 vertex = 
  c("freshman","junior", "freshman","junior","sophomore","freshman",
    "junior","sophomore","sophomore","sophomore"),
 edge = 
  c("junior","senior","junior","sophomore","senior","junior",
    "sophomore","senior","senior","senior"),
 id =
  c("1","1","2","2","2","3","3","3","4","5")),
                       .Names = c("vertex","edge", "id"),
                       row.names = c(1:10),
                       class = "data.frame")
> testvertices
      vertex      edge id
1   freshman    junior  1
2     junior    senior  1
3   freshman    junior  2
4     junior sophomore  2
5  sophomore    senior  2
6   freshman    junior  3
7     junior sophomore  3
8  sophomore    senior  3
9  sophomore    senior  4
10 sophomore    senior  5

在这一点上，我忽略了 id，我的图表应该按计数加权边缘（即新生 -> 初级 =3）。这个想法是制作一个树形图。我知道它在主要的改造点旁边，但如果你问的话……

4

2 回答 2

3

如果我理解正确，你需要这样的东西：

elist <- lapply(seq_len(nrow(test)), function(i) {
  x <- as.character(test[i,])
  x <- unique(na.omit(x))
  x <- rep(x, each=2)
  x <- x[-1]
  x <- x[-length(x)]
  r <- matrix(x, ncol=2, byrow=TRUE)
  if (nrow(r) > 0) { r <- cbind(r, i) } else { r <- cbind(r, numeric()) }
  r
})

do.call(rbind, elist)

#                              i  
# [1,] "freshman"  "junior"    "1"
# [2,] "junior"    "senior"    "1"
# [3,] "freshman"  "junior"    "2"
# [4,] "junior"    "sophomore" "2"
# [5,] "sophomore" "senior"    "2"
# [6,] "freshman"  "junior"    "3"
# [7,] "junior"    "sophomore" "3"
# [8,] "sophomore" "senior"    "3"
# [9,] "sophomore" "senior"    "4"
#[10,] "sophomore" "senior"    "5"

这不是最有效的解决方案，但我认为它相当具有指导意义。我们为输入矩阵的每一行分别创建边，因此lapply. 要从一行创建边，我们首先删除 NA 和重复项，然后将每个顶点包含两次。最后，我们删除第一个和最后一个顶点。这样我们创建了一个边列表矩阵，我们只需要删除第一个和最后一个顶点并将其格式化为两列（实际上将其保留为向量会更有效，没关系）。

添加额外列时，我们必须小心检查我们的边列表矩阵是否有零行。

该do.call功能只会将所有内容粘合在一起。结果是一个矩阵，您可以根据需要将其转换为数据框，通过as.data.frame()，然后您还可以将第三列转换为数字。如果您愿意，还可以更改列名。

于 2012-09-12T06:23:26.707 回答
1

这是你想要的好吗...

test1<-c(test[[2]],test[[3]],test[[4]])
test2<-c(test[[3]],test[[4]],test[[5]])
df<-data.frame(vertex=test1,edge=test2)
df1<-df[complete.cases(df),]
result<-df1[df1$vertex != df1$edge,]
于 2012-09-11T08:58:35.993 回答