On 3/18/20 9:02 PM, Bert Gunter wrote:
Untested in the absence of example data, but I think
combined <- do.call(rbind, lapply(ls2972, function(x)get(x)[[2]]))
Or if you have largish data, use rbindlist() from the data.table package:
combined <- data.table::rbindlist(
lapply(ls2972, function(x) get(x)[[2]])
)
However, it seems you are on the wrong track when you create 2972 lists
in your workspace. (Note: there are no "list files" objects in R. Lists
are objects, not files.) You should have one list of 2972 lists each
having 4 data.frames.
E.g.:
x <- list(
list(
data.frame(),
data.frame(x = 1),
data.frame(),
data.frame()
),
list(
data.frame(),
data.frame(x = 2),
data.frame(),
data.frame()
),
list(
data.frame(),
data.frame(x = 3),
data.frame(),
data.frame()
)
)
keep <- lapply(x, "[[", 2L)
combined <- data.table::rbindlist(keep)
HTH,
Denes
should do it.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Mar 18, 2020 at 12:16 PM Yuan Chun Ding <ycd...@coh.org> wrote:
Hi R users,
I generated 2972 list files in R, each list includes four data frame files
, file names for those list file are VNTR13576, VNTR14689, etc. the second
data frame in each list has the same 11 column names, but different number
of rows.
I can combine two dataframes by
list2972 <-ls(pat="VNTR.*.")
test <-rbind(get(list2972[16])[[2]],get(list2972[166])[[2]] )
I tried to combine all 2972 data frames from those 2972 list files using
do.call or lapply function, but not successful.
Can you help me?
Thank you very much!
Ding
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