Maybe you are thinking of the levels function rather than the names function?
Which still presumes the column is a factor column, when it might actually be a
character column (in which case you might use the unique function). Again, a
reproducible example would stop the guessing.
--
Sent from my phone. Please excuse my brevity.
On November 6, 2017 11:26:24 AM PST, Matti Viljamaa <mvilja...@kapsi.fi> wrote:
>It’s sometimes faster to ask from someone who has already learnt the
>syntax.
>In this case one has to do e.g.
>
>names(data$somecol)
>
>To get the collection and then iteration through it is almost like in
>Python:
>
>for(i in names(data$somecol)) {
> # do something
>}
>
>> Bert Gunter <bgunter.4...@gmail.com> kirjoitti 6.11.2017 kello 19.55:
>>
>> Time to go through a tutorial or two! -- This forum cannot replace
>such self study.
>>
>> Your query evidences some basic confusion, but ?tapply or the
>equivalent lapply(split(...)) construct are most likely relevant.
>>
>> Cheers,
>> Bert
>>
>>
>>
>> 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 Mon, Nov 6, 2017 at 9:40 AM, mviljamaa <mvilja...@kapsi.fi
><mailto:mvilja...@kapsi.fi>> wrote:
>> How can I do a for loop that does to a data.frame column what:
>>
>> for x in xs:
>>
>> does in Python?
>>
>> Obviously the data.frame column in question holds "levels". What if
>the data.frame is in matrix form?
>>
>> BR, Matti
>>
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