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R: Question about table reshaping

开发者 https://www.devze.com 2023-03-22 04:50 出处:网络
I have the following data frame: 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,

I have the following data frame:

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, 1开发者_StackOverflow社区L, 1L)), .Names = c("id", 
"property1", "property2", "property3"), class = "data.frame", row.names = c(NA, 
-4L))

What is the least cumbersome way to get the following data frame:

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

Maybe something like melt or reshape with fancy options?


This solution assumes you're looking for a data frame similar to how gsk3 interpreted the question (pasting the properties together) but with the obligatory avoidance of a for loop, just cause that's how we roll with R:

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


This isn't a reshape at all, really. Use 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=" ")

As always, you'll get better answers if you dput() your dataframe first:

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))


That is not actually a proper dataframe which of necessity has all rows with the same number of entries, so the correct answer is you may want a list. If that's not really what you want, then try this:

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

You need the t() because apply row operations transpose their results because of column-major order that R imposes.

If you do want the list version then here's one approach:

 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"
0

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