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Split a list

5 messages · Juliet Ndukum, Weidong Gu, R. Michael Weylandt +2 more

Weidong Gu
# 
It would be nice if you could provide a sample. However, if the data
in the list have the same colnames, you can combine them by

df<-do.call('rbind',your_list_data_frame)

Then you can do what you want on the dataframe instead of a list

HTH

Weidong Gu
On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum <jpntsang at yahoo.com> wrote:
R. Michael Weylandt
# 
Comments inline:
On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum <jpntsang at yahoo.com> wrote:
Assuming the third column always takes a constant value for each
data.frame, something this will work:

lst <- list(a = data.frame(1:3, 4:6, 7), b = data.frame(1:3, 4:6, 8),
c = data.frame(1:3, 4:6, 7))
idx <- sapply(lst, function(x) x[1,3] == 7)

lst1 = lst[idx]
lst2 = lst[!idx]
Similar line of argument:

idx <- sapply(lst, function(x) range(x[,2]) > 20)
lst1 = lst[idx]
David Winsemius
#
On Oct 14, 2011, at 9:26 AM, Weidong Gu wrote:

            

      
      
      
    

        
That is certainly true.

            

      
      
      
    

        
Perhaps something like:
list_of_firsts <- lapply(dflist, function(x) X[ , X[,3]=="first"] )
list_of_seconds <- lapply( dflist, function(x) X[ , X[,3]=="second"])

Ow with subset (but without that missing example it is more difficult  
to show the true value of subset:

subset(X, select= X[,3]=="first")

            

      
      
      
    

        
Same as above with an inequality sign.

            

      
      
      
    

        
David Winsemius, MD
West Hartford, CT

  
  


  


      

    

    

      
      

        


  

        

      Dennis Murphy
    

    

      
      #
    

  


  

      
Hi:

Following the lead of others, here's a reproducible example that I
believe achieves what you want.

# Q1:
L <- lapply(1:3, function(n)
    data.frame(x = rnorm(6), y = rnorm(6), g = rep(1:2, each = 3)))

# Using David's suggestion:
L1 <- lapply(L, function(d) subset(d, g == 1L))
L2 <- lapply(L, function(d) subset(d, g == 2L))

# Q2:
# Let range > 2 to retain in this small example:
# Find the range of the second column of each list component:
sapply(L, function(x) diff(range(x[, 2], na.rm = TRUE)))

# The code retains the data frame if the range of the second
# column is > 2, otherwise it is set to NULL:
lapply(L, function(d) if(diff(range(d[, 2], na.rm = TRUE)) > 2) d else NULL)

# If you want to collapse the result into a data frame, the base R
approach would be
do.call('rbind', lapply(L, function(d) if(diff(range(d[, 2], na.rm =
TRUE)) > 2) d else NULL))

# An equivalent way to do all of this in the plyr package is:
library('plyr')
L1 <- llply(L, function(d) subset(d, g == 1L))
L2 <- llply(L, function(d) subset(d, g == 2L))

ldply(L, function(d) if(diff(range(d[, 2], na.rm = TRUE)) > 2) d else NULL)

There are advantages to naming the list components if this is what you
have in mind, since both ldply() and the rbind from do.call() will
output indicators of which component data frame each observation
belongs; ldply() uses an .id variable to designate the list component
name whereas do.call(rbind, ...) uses rownames to distinguish
observations. For this example, try

names(L) <- paste('d', 1:3, sep = '')

and run the code above again to see the difference.

HTH,
Dennis

        
On Fri, Oct 14, 2011 at 6:06 AM, Juliet Ndukum <jpntsang at yahoo.com> wrote: