If you learn to use dput() to provide useful examples in your posts, you
are more likely to receive useful help. It is rather difficult to make much
sense of your messy text, though some brave soul(s) may try to help.
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 Fri, Nov 2, 2018 at 8:00 AM Ek Esawi <esaw...@gmail.com> wrote:
> Hi All,
>
> I have a list that is made up of nested lists, as shown below. I want
> to remove all rows in each sub-list that start with an empty space,
> that’s the first entry of a row is blank; for example, on
> [[1]][[1]][[1]] Remove row 4,on [[1]][[1]][[3]] remove row 5, on
> [[1]][[2]][[1]] remove row 6, etc.. All rows start with 2 digits/ 2
> digits. My formula works on individual sublist but not the whole
> list.. I know my indexing is wrong, but don’t know how to fix it.
>
>
> > FF
>
> [[1]]
> [[1]][[1]]
> [[1]][[1]][[1]]
> [,1] [,2] [,3] [,4] [,5]
> [1,] "30/20" "AAAAAAAA" “ “ "-89"
> [2,] "02/20" "AAAAAAAA” “ “ "-98"
> [3,] "02/20" “AAAAAAA” “ “ "-84"
> [4,] “ “ “ “ “
> [[1]][[1]][[2]]
> [,1] [,2]
> [1,] "02/23" “AAAAAAAA” : 29" “
> [2,] "02/23" “AAAAAAAA” ." “
> [3,] "02/23" “AAAAAAAA” " “
> [4,] "02/23" “AAAAAAAA” "
> [[1]][[1]][[3]]
> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
> [1,] "01/09" “AAAAAAAA" “ “ “ "53"
> [2,] "01/09" “AAAAAAAA” " “ “ “ "403"
> [3,] "01/09" “AAAAAAAA” " “ “ “ "83"
> [4,] "01/09" “AAAAAAAA” " “ “ “ "783"
> [5,] “ “ “AAAAAAAA” 3042742181" “ “ “ “
> [[1]][[2]]
> [[1]][[2]][[1]]
> [,1] [,2] [,3] [,4] [,5]
> [1,] “ “ “ “ “AAAAAAAA” "
> [2,] "Standard Purchases" “ “ “ "
> [3,] "24/90 "AAAAAAAA” “ "243" "
> [4,] "24/90 "AAAAAAAA” " "143" "
> [5,] "24/91 "AAAAAAAA” " “ "143" “
> [6,] “ “ “ “ "792"
> [[1]][[2]][[2]]
> [,1] [,2]
> [1,] "02/23" “AAAAAAAA”: 31" “
> [2,] "02/23" “AAAAAAAA”." “
> [3,] "02/23" “AAAAAAAA” " “
> [4,] "02/23" “AAAAAAAA”
> [5,] "02/23" “AAAAAAAA”
> [6,] "02/23" “AAAAAAAA” 20"
> [7,] "02/23" “AAAAAAAA” “
> [8,] "02/23" “AAAAAAAA” "33"
> [[1]][[3]]
> [[1]][[3]][[1]]
> [,1] [,2]
> [1,] "02/23" “AAAAAAAA”: 28" “
> [2,] "02/23" “AAAAAAAA”." “
> [3,] "02/23" “AAAAAAAA” " “
> [4,] "02/23" “AAAAAAAA” "
> [[1]][[3]][[2]]
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
> [1,] "02/23" “AAAAAAAA” " “ “ "53" "
> [2,] "02/24" “AAAAAAAA” " “ “ "
> [3,] “ “ “ “ “ “ “ “ "1,241"
> [4,] "02/24" "AAAAAAAA” “ "33”
>
> My Formula,:
>
> G <- lapply(FF, function(x) lapply(x, function (y) lapply(y,
> function(z) z[grepl("^[0-9][0-9]/",z[,1]),])))
>
> The error: Error in z[, 1] : incorrect number of dimensions
>
>
>
> Thanks in advance--EK
>
> ______________________________________________
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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