[R] Query about extracting subset of datafram

Marc Schwartz marc_schwartz at comcast.net
Wed Jan 24 22:10:51 CET 2007

On Wed, 2007-01-24 at 12:31 -0800, lalitha viswanath wrote:
> Hi
> I have a table read from a mysql database which is of
> the  kind
> clusterid clockrate
> I obtained this table in R as
> clockrates_table <-sqlQuery(channel,"select....");
> I have a function within which I wish to extract the
> clusterid for a given cluster.
> Although I know that there is just one row per
> clusterid in the data frame, I am using subset to
> extract the clockrate.
> clockrate = subset(clockrates_table, clusterid==15,
> select=c(clockrate));

You don't need the ';', though some will argue that it is a personal
preference.  Also, the c(...) around 'clockrate' is not needed when only
one column is being selected.

> Is there any way of extracting the clockrate without
> using subset.

You could use:

  clockrates_table[clockrates_table$clusterid == 15, 

or perhaps:

  with(clockrates_table, clockrates_table[clusterid == 15, clockrate])

See ?with

If you did not need the conditional, you could of course use:





  clockrates_table[, "clockrate"]


  with(clockrates_table, clockrate)

My personal preference is to use subset(), as for me, it makes the code
easier to read.

> In the help section for subset, it mentioned to "see
> also: [,..."
> However I could find no mention for this entry when I
> searched as "?[", etc.

Try:  ?"[" or ?Extract

Note the placement of the quotes in the first case.

> The R manuals also, despite discussing complex
> libraries, techniques etc, dont always seem to provide
> such handy hints/tips and tricks for manipulating
> data, which is a first stumbling block for newbies
> like me.
> I would greatly appreciate if you could point me to
> such resources as well, for future reference.

If you have not yet, reading the Posting Guide, for which there is a
link at the bottom of each e-mail is a good place to start.

Also, see ?RSiteSearch for a function which will enable you to search
the e-mail list archives.


Marc Schwartz

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