combine filter() and select()
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 <calandra at 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