Re: [R] Combine list element by column name to make a dataframe
forgot to cc to list have a look at https://stat.ethz.ch/pipermail/r-help/2012-January/300275.html and other messages in the sequence if you use Marc Schwartz's list2df you with have to transpose it with t() Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Mohammad Tanvir Ahamed via R-help Sent: Monday, 6 April 2015 16:06 To: r-help@r-project.org Subject: [R] Combine list element by column name to make a dataframe Hi ,� I have a example list like follow� lst-list(setNames(c(1,10,50,60,70,80),c(id,id1,math,phy,che,bio)),setNames(c(2,20,45),c(id,id1,phy)),setNames(c(3,30,75),c(id,id1,bio))) My expected outcome :� - df-rbind(c(1,10,50,60,70,80),c(2,20,NA,45,NA,NA),c(3,30,NA,NA,NA,75)) colnames(df)-c(id,id1,math,phy,che,bio) row.names(df) - NULL df Any suggestion will be appreciated .� Thanks in advance. � Best regards ...� Tanvir Ahamed G�teborg, Sweden [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combine list element by column name to make a dataframe
Maybe something like the appended?
-- Mike
lst - list(setNames(c(1,10,50,60,70,80),
c(id,id1,math,phy,che,bio)),
setNames(c(2,20,45),
c(id,id1,phy)),
setNames(c(3,30,75),
c(id,id1,bio)))
lst
df - rbind(c(1,10,50,60,70,80),
c(2,20,NA,45,NA,NA),
c(3,30,NA,NA,NA,75))
colnames(df)-c(id,id1,math,phy,che,bio)
row.names(df) - NULL
df
allNames - unique(unlist(lapply(lst, names)))
allNames
newLst - lapply(lst, function(element) {
element[allNames]
})
newLst
df2 - do.call(rbind, newLst)
df2
all.equal(df, df2)
On Sun, Apr 5, 2015 at 11:05 PM, Mohammad Tanvir Ahamed via R-help
r-help@r-project.org wrote:
Hi ,
I have a example list like follow
lst-list(setNames(c(1,10,50,60,70,80),c(id,id1,math,phy,che,bio)),setNames(c(2,20,45),c(id,id1,phy)),setNames(c(3,30,75),c(id,id1,bio)))
My expected outcome :
-
df-rbind(c(1,10,50,60,70,80),c(2,20,NA,45,NA,NA),c(3,30,NA,NA,NA,75))
colnames(df)-c(id,id1,math,phy,che,bio)
row.names(df) - NULL
df
Any suggestion will be appreciated .
Thanks in advance.
Best regards
...
Tanvir Ahamed
Göteborg, Sweden
[[alternative HTML version deleted]]
__
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Re: [R] Combine list element by column name to make a dataframe
On 06 Apr 2015, at 08:05 , Mohammad Tanvir Ahamed via R-help
r-help@r-project.org wrote:
Hi ,�
I have a example list like follow�
lst-list(setNames(c(1,10,50,60,70,80),c(id,id1,math,phy,che,bio)),setNames(c(2,20,45),c(id,id1,phy)),setNames(c(3,30,75),c(id,id1,bio)))
My expected outcome :�
-
df-rbind(c(1,10,50,60,70,80),c(2,20,NA,45,NA,NA),c(3,30,NA,NA,NA,75))
colnames(df)-c(id,id1,math,phy,che,bio)
row.names(df) - NULL
df
Any suggestion will be appreciated .�
Hmm, in principle this looks like a merge() problem, if you first convert each
list element to a data frame. That could be painful to get right though.
You could try something like this:
nm - Reduce(union, lapply(lst, names)) # or just type it in
nm
[1] id id1 math phy che bio
blank - setNames(rep(NA_real_, length(nm)), nm)
fill1 - function(x, blank) {blank[names(x)] - x; blank}
lapply(lst, fill1, blank)
[[1]]
id id1 math phy che bio
1 10 50 60 70 80
[[2]]
id id1 math phy che bio
2 20 NA 45 NA NA
[[3]]
id id1 math phy che bio
3 30 NA NA NA 75
do.call(rbind,lapply(lst, fill1, blank))
id id1 math phy che bio
[1,] 1 10 50 60 70 80
[2,] 2 20 NA 45 NA NA
[3,] 3 30 NA NA NA 75
(NB, this is a matrix, not a data frame, but so is your df!)
Notice that this does not do a proper merge on the id fields, i.e. if you have
two different records with different grades (say one with che and another
with phy) on the same id, you get two records, not one. However, that might
well be what you wanted.
(It is tempting to use
fill1 - function(x) `[-`(blank, names(x), x)
lapply(lst, fill1)
which does seem to work, but should probably be avoided because of the risk of
destructive modification.)
Thanks in advance.
�
Best regards
...�
Tanvir Ahamed
G�teborg, Sweden
[[alternative HTML version deleted]]
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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