R:关于表重塑的问题

时间:2011-07-20 15:56:07

标签: r dataframe

我有以下数据框:

id,property1,property2,property3
1,1,0,0  
2,1,1,0  
3,0,0,1  
4,1,1,1

d.f <- structure(list(id = 1:4, property1 = c(1L, 1L, 0L, 1L), property2 = c(0L, 
1L, 0L, 1L), property3 = c(0L, 0L, 1L, 1L)), .Names = c("id", 
"property1", "property2", "property3"), class = "data.frame", row.names = c(NA, 
-4L))

获取以下数据框的最简单方法是什么:

id,properties_list
1,property1
2,property1, property2
3,property3
4,property1, property2, property3

也许像meltreshape这样的花哨选项?

3 个答案:

答案 0 :(得分:2)

这真的不是重塑。使用paste

for(i in seq(1,3) ) {
   tf <- as.logical(d.f[,i+1])
   d.f[,i+1] <- as.character(d.f[,i+1])
   d.f[,i+1][tf] <- colnames(d.f)[i+1]
   d.f[,i+1][!tf] <- " "
}
d.f$property.list <- paste(d.f[,2],d.f[,3],d.f[,4],sep=" ")

与往常一样,如果您首先dput()数据框,您将获得更好的答案:

d.f <- structure(list(id = 1:4, property1 = c(1L, 1L, 0L, 1L), property2 = c(0L, 
1L, 0L, 1L), property3 = c(0L, 0L, 1L, 1L)), .Names = c("id", 
"property1", "property2", "property3"), class = "data.frame", row.names = c(NA, 
-4L))

答案 1 :(得分:2)

实际上这不是一个合适的数据帧,必须具有相同数量的条目的所有行,所以正确的答案是你可能想要一个列表。如果那不是你想要的,那就试试这个:

dfrm[-1] <- t( apply(dfrm[-1], 1, function(x) ifelse(x, names(x), "") )   )
dfrm
  id property1 property2 property3
1  1 property1                    
2  2 property1 property2          
3  3                     property3
4  4 property1 property2 property3

您需要t()因为应用行操作会转换其结果,因为R强加了列主要顺序。

如果您确实需要列表版本,那么这里有一种方法:

 prop_list <- apply(dfrm[-1], 1, function(x)  c(names(x)[ as.logical(x)]  ) )
 names(prop_list) <- dfrm[,1]
 prop_list
$`1`
[1] "property1"

$`2`
[1] "property1" "property2"

$`3`
[1] "property3"

$`4`
[1] "property1" "property2" "property3"

答案 2 :(得分:2)

此解决方案假设您正在寻找类似于gsk3如何解释问题(将属性粘贴在一起)的数据框,但必须避免使用for循环,这只是因为我们如何使用R滚动:

property_list <- apply(d.f[,-1],1,
                    FUN=function(x,nms){paste(nms[as.logical(x)],collapse=",")},
                        nms=colnames(d.f)[-1])

as.data.frame(cbind(d.f$id,property_list))


  V1                 property_list
1  1                     property1
2  2           property1,property2
3  3                     property3
4  4 property1,property2,property3