dplyr filter on a vector rather than a dataframe in R

Question

This seems like a simple question, but I have not come across a clean solution for it yet. I have a vector in R and I want to remove certain elements from the vector, however I want to avoid the vector[vector != "thiselement"] notation for a variety of reasons. In particular, here is what I am trying to do:

# this doesnt work
all_states = gsub(" ", "-", tolower(state.name)) %>% filter("alaska")

# this doesnt work either
all_states = gsub(" ", "-", tolower(state.name)) %>% filter(!= "alaska")

# this does work but i want to avoid this approach to filtering
all_states = gsub(" ", "-", tolower(state.name))
all_states = all_states[all_states != "alaska"]

can this be done in a simple manner? Thanks in advance for the help!

EDIT - the reason I'm struggling with this is because I'm only finding things online regarding filtering based on a column of a dataframe, for example:

my_df %>% filter(col != "alaska")

however I'm working with a vector not a dataframe here

Solution 1

Sorry for posting on a 5-month-old question to archive a simpler solution.

Package dplyr can filter character vectors in following ways:

> c("A", "B", "C", "D") %>% .[matches("[^AB]", vars=.)]
[1] "C" "D"
> c("A", "B", "C", "D") %>% .[.!="A"]
[1] "B" "C" "D"

The first approach allows you to filter with regular expression, and the second approach uses fewer words. It works because package dplyr imports package magrittr albeit masks its functions like extract, but not the placeholder ..

Details of placeholder . can be found on within help of forward-pipe operator %>%, and this placeholder has mainly three usage:

• Using the dot for secondary purposes
• Using lambda expressions with %>%
• Using the dot-place holder as lhs

Here we are taking advantage of its 3rd usage.

Solution 2

You may like to try magrittr::extract. e.g.

> library(magrittr)

> c("A", "B", "C", "D") %>% extract(.!="A")
[1] "B" "C" "D"

For more extract-like functions load magrittr package and type ?alises.

Solution 3

Pretty sure dplyr only really operates on data.frames. Here's a two line example coercing the vector to a data.frame and back.

myDf = data.frame(states = gsub(" ", "-", tolower(state.name))) %>% filter(states != "alaska")
all_states = myDf$states

or a gross one liner:

all_states = (data.frame(states = gsub(" ", "-", tolower(state.name))) %>% filter(states != "alaska"))$states

Comments

[Gregor Thomas]

The d in dplyr is for data.frame. "using dplyr to write cleaner code" should mean using dplyr for what it's made for (data frames) and not trying to use it when inappropriate (not data frames).

[Canovice]

got it. yeah maybe im making my life harder than it needs to be. okay thanks

[David Pedack]

yeah, it'd be nice to have 1 tool to use. dplyr ends up looking a lot cleaner than the base R code in my opinion. unfortunately it always ends up a mess with vectors.

[Hielke Walinga]

No documentation for ‘alises’ in specified packages and libraries:

[Łukasz Deryło]

It must've been removed in current version ?extract works now.

[JelenaČuklina]

very unfortunate that tidyr extract means a totally different thing. Love pipes, and this vector function is awesome!

[Jens]

Is there a way to negate the first approach with "matches" ?

[Quar]

@Jens, one could negate via either indexing, such as c("A", "B", "C", "D") %>% .[-matches("[^AB]", vars=.)] , or regular expression itself, such as c("A", "B", "C", "D") %>% .[matches("[AB]", vars=.)] -- perhaps the caveats here is to prepend - instead of ! to the matches selected indices for the negation, because matches returns an integer vector c(3, 4), rather than a boolean mask c(F, F, T, T).

[Jens]

Many thanks! I tried ! and it did not work. Now I know why. -matches worked