detecting any element in a vector of strings, appearing anywhere in any of several character variables in a dataframe

Just add a word break marker before and after:

zz$v5 <- grepl( paste0( "\\b(", paste0( alarm.words, collapse="|" ), ")\\b" ), do.call( paste, zz[ , 2:3 ] ) ) )
---------------------------------------------------------------------------
Jeff Newmiller                        The     .....       .....  Go Live...
DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live Go...
                                      Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
/Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
---------------------------------------------------------------------------
Sent from my phone. Please excuse my brevity.

On July 9, 2015 10:12:23 AM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote:

Jeff:

Well, it would be much better (no loops!) except, I think, for one
issue: "red" would match "barred" and I don't think that this is what
is wanted: the matches should be on whole "words" not just string
patterns.

So you would need to fix up the matching pattern to make this work,
but it may be a little tricky, as arbitrary whitespace characters,
e.g. " " or "\n" etc. could be in the strings to be matched separating
the words or ending the "sentence."  I'm sure it can be done, but I'll
leave it to you or others to figure it out.

Of course, if my diagnosis is wrong or silly, please point this out.

Cheers,
Bert


Bert Gunter

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
  -- Clifford Stoll


On Thu, Jul 9, 2015 at 9:34 AM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us> wrote:

I think grep is better suited to this:

zz$v5 <- grepl( paste0( alarm.words, collapse="|" ), do.call( paste,

zz[ , 2:3 ] ) ) )

---------------------------------------------------------------------------

Jeff Newmiller                        The     .....       .....  Go

Live...

DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live

Go...

                                      Live:   OO#.. Dead: OO#..

Playing

Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
/Software/Embedded Controllers)               .OO#.       .OO#.

rocks...1k

---------------------------------------------------------------------------

Sent from my phone. Please excuse my brevity.

On July 9, 2015 8:51:10 AM PDT, Bert Gunter <bgunter.4567 at gmail.com>

wrote:

Here's a way to do it that uses %in% (i.e. match() ) and uses only a
single, not a double, loop. It should be more efficient.

sapply(strsplit(do.call(paste,zz[,2:3]),"[[:space:]]+"),

+       function(x)any(x %in% alarm.words))

[1] FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE FALSE  TRUE  TRUE

The idea is to paste the strings in each row (do.call allows an
arbitrary number of columns) into a single string and then use
strsplit to break the string into individual "words" on whitespace.
Then the matching is vectorized with the any( %in% ... ) call.

Cheers,
Bert
Bert Gunter

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
  -- Clifford Stoll


On Thu, Jul 9, 2015 at 6:05 AM, John Fox <jfox at mcmaster.ca> wrote:

Dear Chris,

If I understand correctly what you want, how about the following?

rows <- apply(zz[, 2:3], 1, function(x) any(sapply(alarm.words,

grepl, x=x)))

zz[rows, ]

          v1                              v2                v3 v4
3  -1.022329                    green turtle    ronald weasley  2
6   0.336599              waffle the hamster        red sparks  1
9  -1.631874 yellow giraffe with a long neck gandalf the white  1
10  1.130622                      black bear  gandalf the grey  2

I hope this helps,
 John

------------------------------------------------
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/


On Wed, 08 Jul 2015 22:23:37 -0400
 "Christopher W. Ryan" <cryan at binghamton.edu> wrote:

Running R 3.1.1 on windows 7

I want to identify as a case any record in a dataframe that

contains

any

of several keywords in any of several variables.

Example:

# create a dataframe with 4 variables and 10 records
v2 <- c("white bird", "blue bird", "green turtle", "quick brown

fox",

"big black dog", "waffle the hamster", "benny likes food a lot",

"hello

world", "yellow giraffe with a long neck", "black bear")
v3 <- c("harry potter", "hermione grainger", "ronald weasley",

"ginny

weasley", "dudley dursley", "red sparks", "blue sparks", "white

dress

robes", "gandalf the white", "gandalf the grey")
zz <- data.frame(v1=rnorm(10), v2=v2, v3=v3, v4=rpois(10,

lambda=2),

stringsAsFactors=FALSE)
str(zz)
zz

# here are the keywords
alarm.words <- c("red", "green", "turtle", "gandalf")

# For each row/record, I want to test whether the string in v2 or

the

string in v3 contains any of the strings in alarm.words. And then

if

so,

set zz$v5=TRUE for that record.

# I'm thinking the str_detect function in the stringr package

ought

to

be able to help, perhaps with some use of apply over the rows, but

I

obviously misunderstand something about how str_detect works

library(stringr)

str_detect(zz[,2:3], alarm.words)    # error: the target of the

search

                                     # must be a vector, not

multiple

                                     # columns

str_detect(zz[1:4,2:3], alarm.words) # same error

str_detect(zz[,2], alarm.words)      # error, length of

alarm.words

                                     # is less than the number of
                                     # rows I am using for the
                                     # comparison

str_detect(zz[1:4,2], alarm.words)   # works as hoped when
length(alarm.words)                  # confining nrows
                                     # to the length of

alarm.words

str_detect(zz, alarm.words)          # obviously not right

# maybe I need apply() ?
my.f <- function(x){str_detect(x, alarm.words)}

apply(zz[,2], 1, my.f)     # again, a mismatch in lengths
                           # between alarm.words and that
                           # in which I am searching for
                           # matching strings

apply(zz, 2, my.f)         # now I'm getting somewhere
apply(zz[1:4,], 2, my.f)   # but still only works with 4
                           # rows of the dataframe


# perhaps %in% could do the job?

Appreciate any advice.

--Chris Ryan