R - List to data.frame using list names

Question

So I have this list:

structure(list(scaf = structure(1L, .Label = "HE638", class = "factor"), 
    pos = 8L, scaf = structure(1L, .Label = "HE638", class = "factor"), 
    pos = 8L, scaf = structure(1L, .Label = "HE638", class = "factor"), 
    pos = 8L), .Names = c("scaf", "pos", "scaf", "pos", "scaf", 
"pos"))

I want to get a data.frame so that the two columns are scaf and pos

This is as far as I've gotten:

do.call(rbind, poor.loci)

Desired result:

scaf     pos
HE638    8
HE638    8
HE638    8

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2013-11-29 18:48:22Z

6

Here are three options to consider:

Option 1

Straightforward selecting of every other element and manually creating your data.frame.

setNames(
  data.frame(unlist(poor.loci[c(TRUE, FALSE)], use.names = FALSE),
             unlist(poor.loci[c(FALSE, TRUE)], use.names = FALSE)),
  unique(names(poor.loci)))
#    scaf pos
# 1 HE638   8
# 2 HE638   8
# 3 HE638   8

Option 2

Convert your list into a "long" data.frame and reshape it to the form you want. Can also be done with the "reshape2" package using melt and dcast instead of stack and reshape.

X <- stack(lapply(poor.loci, as.character))
X$ID <- ave(X$values, X$values, FUN = seq_along)
reshape(X, direction = "wide", idvar="ID", timevar="ind")
#   ID values.scaf values.pos
# 1  1       HE638          8
# 3  2       HE638          8
# 5  3       HE638          8

Option 3

Rearrange your list and convert the rearranged list to a data.frame:

A <- unique(names(poor.loci))
data.frame(
  setNames(
    lapply(A, function(x) unlist(poor.loci[names(poor.loci) %in% x], 
                                 use.names = FALSE)), A))

edited Nov 29, 2013 at 18:48

answered Nov 29, 2013 at 18:38

A5C1D2H2I1M1N2O1R2T1

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2 Comments

flodel Over a year ago

one more: as.data.frame(lapply(split(ll, names(ll)), unlist))

Sven Hohenstein Over a year ago

@flodel Another one without lapply is: as.data.frame(split(unlist(poor.loci), names(poor.loci))).

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