Okay, all day on this and I send the msg. and almost immediately discover
that:
dat <- ldply(result)
solves the problem.
On Mon, Apr 21, 2014 at 5:40 PM, Timothy W. Cook <tim@mlhim.org> wrote:
> I am processing an arbitrary number of XML files and extracting specific
> nodes. I then create a dataframe for each nodeset.
> I return a list containing these dataframes.
>
> Example:
>
> > str(result)List of 2
> $ :'data.frame': 1 obs. of 5 variables:
> ..$ data-name : chr "Etiologic diagnosis of Acute Febrile
Hemorrhagic Syndrome"
> ..$ valid-time-begin: chr "2014-04-28T01:31:49Z"
> ..$ valid-time-end : chr "3014-04-28T01:31:48Z"
> ..$ DvString-dv : chr "Typhoid fever"
> ..$ language : chr "en-US"
> $ :'data.frame': 1 obs. of 5 variables:
> ..$ data-name : chr "Etiologic diagnosis of Acute Febrile
Hemorrhagic Syndrome"
> ..$ valid-time-begin: chr "2014-04-29T01:02:00Z"
> ..$ valid-time-end : chr "3014-04-29T01:01:59Z"
> ..$ DvString-dv : chr "Dengue"
> ..$ language : chr "en-US"
>
>
> I would like to have each of the list items as a row in one dataframe. My
> attempts so far have been unsuccessful. I have tried t(), merge(),
> lapply(), etc. Short of some elaborate *for loop* (I know is frowned
> upon) I am out of answers.
>
> I would think this is a fairly common need, maybe even a FAQ that I
> haven't found yet.
>
> Thanks in advance.
>
>
>
>
>
>
>
> --
>
> ===========================================> Timothy Cook
> LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook
> MLHIM http://www.mlhim.org
>
>
--
===========================================Timothy Cook
LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook
MLHIM http://www.mlhim.org
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