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grabbing from elements of a list without a loop

10 messages · Bert Gunter, David Winsemius, Dimitri Liakhovitski +2 more

arun
# 
Hi,


?mapply(`[`,mylist,list(1,2,0),SIMPLIFY=FALSE)
#[[1]]
#? a
#1 1
#2 2

#[[2]]
?# b
#1 5
#2 6

#[[3]]
#data frame with 0 columns and 2 rows


A.K.

----- Original Message -----
From: Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com>
To: r-help <r-help at r-project.org>
Cc: 
Sent: Tuesday, February 12, 2013 4:33 PM
Subject: [R] grabbing from elements of a list without a loop

Hello!

# I have a list with several data frames:
mylist<-list(data.frame(a=1:2,b=2:3),
? ? ? ? ?  data.frame(a=3:4,b=5:6),data.frame(a=7:8,b=9:10))
(mylist)

# I want to grab only one specific column from each list element
neededcolumns<-c(1,2,0)? # number of the column I need from each element of
the list

# Below, I am doing it using a loop:
newlist<-NULL
for(i in 1:length(mylist) ) {
? newlist[[i]]<-mylist[[i]] [neededcolumns[i]]
}
newlist<-do.call(cbind,newlist)
(newlist)

I was wondering if there is any way to avoid the loop above and make it
faster.
In reality, I have a much longer list, each of my data frames is much
larger and I have to do it MANY-MANY times.
Thanks a lot!

Dimitri Liakhovitski
gfk.com <http://marketfusionanalytics.com/>

??? [[alternative HTML version deleted]]

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Bert Gunter
# 
The answer is essentially no, a loop is required (as best I can see),
although it can be in the form of an apply type call instead.  Also,
your code will fail with a 0 index. Something like this should work:

newlist <- lapply(1:3,function(i)if(!neededcolumns[i])NULL else
mylist[[c(i,neededcolumns[i])]])

## note the use of [[c(i,j)]]  form for selecting columns as an
element from a list of lists

## Note that your cbind call produces a matrix, not a list.

-- Bert


You might wish to check the parallel package, as this looks like the
sort of thing parallellization could be profitably used for; but I
have no experience to offer beyond that suggestion.

-- Bert

On Tue, Feb 12, 2013 at 1:33 PM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        
      

    

    

      
      

        


  

        

      David Winsemius
    

    

      
      #
    

  


  

      
On Feb 12, 2013, at 3:01 PM, Dimitri Liakhovitski wrote:

        

            

      
      
      
    

        
[[1]]
[1] 1 2 3

            

      
      
      
    

        
[[1]]
[1] 1

[[2]]
[1] 2

[[3]]
[1] 3

  
    

      
      
    

  


  


      

    

    

      
      

        
      

    

    

      
      

        
      

    

    

      
      

        


  

        

      arun
    

    

      
      #
    

  


  

      
Hi Dimitri,

?neededcolumns<- c(1,2,0)
mapply(`[`,mylist,lapply(neededcolumns,function(x) x),SIMPLIFY=FALSE)

#[[1]]
?# a
#1 1
#2 2
#
#[[2]]
?# b
#1 5
#2 6

#[[3]]
#data frame with 0 columns and 2 rows

A.K.

        

      
      
    

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      William Dunlap
    

    

      
      #
    

  


  

      
One could try using unlist(mylist, recursive=FALSE) to make a list of all the columns
of the data.frames in mylist and subscripting from that.  E.g., here is some barely tested
code:
  > nCols <- vapply(mylist, ncol, 0L)
  > neededcolumns<-c(1,2,0) # I assume 0 means no column wanted from 3rd df in list
  > i <- neededcolumns + c(0, cumsum(nCols[-length(nCols)]))
  > i <- i[neededcolumns>0]
  > data.frame(unlist(mylist, recursive=FALSE)[i])
    a b
  1 1 5
  2 2 6


Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com