Hi Jeff,

The code you show is exactly what I usually do, in base R; but I wanted
to play with tidyverse to learn it (and also understand when it makes
sense and when it doesn't).

And yes, of course, in the example I gave, I end up with a 1-cell
tibble, which could be better extracted as a length-1 vector. But my
real goal is not to end up with a single value or even a single column.
I just thought that simplifying my example was the best approach to ask
for advice.

But thank you for letting me know that what I'm doing is pointless!

Ivan

--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeological Research Centre and
Museum for Human Behavioural Evolution
Schloss Monrepos
56567 Neuwied, Germany
+49 (0) 2631 9772-243
https://www.researchgate.net/profile/Ivan_Calandra

On 19/08/2020 19:27, Jeff Newmiller wrote:
> The whole point of dplyr primitives is to support data frames... that is, 
> lists of columns. When you pare your data frame down to one column you are 
> almost certainly using the wrong tool for the job.
>
> So, sure, your code works... and it even does what you wanted in the dplyr 
> style, but what a pointless exercise.
>
> grep( "a", mytbl$file, value=TRUE )
>
> On August 19, 2020 7:56:32 AM PDT, Ivan Calandra <calan...@rgzm.de> wrote:
>> Dear useRs,
>>
>> I'm new to the tidyverse world and I need some help on basic things.
>>
>> I have the following tibble:
>> mytbl <- structure(list(files = c("a", "b", "c", "d", "e", "f"), prop =
>> 1:6), row.names = c(NA, -6L), class = c("tbl_df", "tbl", "data.frame"))
>>
>> I want to subset the rows with "a" in the column "files", and keep only
>> that column.
>>
>> So I did:
>> myfile <- mytbl %>%
>>   filter(grepl("a", files)) %>%
>>   select(files)
>>
>> It works, but I believe there must be an easier way to combine filter()
>> and select(), right?
>>
>> Thank you!
>> Ivan

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