On May 21, 2010, at 12:21 AM, santana sarma wrote:
Hi David,
SORRY - I am trying to be more clearer this time.
Let's say the dataframe has some rows and columns, with unique
rownames and column names. The rest of the data in the dataframe are
just numbers.
So have you tried any of the solutions offered so far?
c(apply(df[1:10,], 1, I))
c(df[ , 1:10]
This will not become any more clear unless you offer an R object. Your
terminology has obviously been contaminated by excessive contact with
spreadsheets. If you want R-answers you need to learn to define R-
objects and employ R-terminology.
--
David.
I wish to concatenate those rows. That means : I wish to concatenate
first 10 rows' values in one row, then next 20 rows's values in the
next row .... and so on.
Similarly, for columns : The first 10 column's values will be one
below the other ... and so on.
Cheers,
= = = = = = = = = = =
On Fri, May 21, 2010 at 3:53 PM, David Winsemius <dwinsem...@comcast.net
> wrote:
On May 20, 2010, at 11:05 PM, santana sarma wrote:
Hi,
I have a dataframe with some 800 rows and 14 columns.
Could you please advise how I can concatenate the rows - one after
another.
Similarly for columns, one below the other.
Not sure exactly what you are after:
unlist might accomplish the second task.
Whether c(apply(df, 1, I)) would be satisfactory for the first task
might depend on whether the columns in the dataframe were all of the
same type. Now that I think of it, both soolutions would force the
types to be that same.
?"c"
?I
?apply
df[1:nrow(df), ] ... would essentially give you the first
request, but it would not be any different than just typing df.
So .... what do intend this process to accomplish?
--
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
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