Subset rows in an R dataframe based on partial match of multiple strings

Question

I don't think this exact question has been asked - lots of stuff on subsetting based on one value (i.e., x[grepl("some string", x[["column1"]]),]), but not multiple values/strings.

Here is an example of my data:

#create sample data frame
data = data.frame(id = c(1,2,3,4), phrase = c("dog, frog, cat, moose", "horse, bunny, mouse", "armadillo, cat, bird,", "monkey, chimp, cow"))

#convert the `phrase` column to character string (the dataset I'm working on requires this)
data$phrase = data$phrase

#list of strings to remove rows by
remove_if = c("dog", "cat")

This will give a dataset that looks like:

  id                phrase
1  1 dog, frog, cat, moose
2  2   horse, bunny, mouse
3  3 armadillo, cat, bird,
4  4    monkey, chimp, cow

I want to remove row 1 and row 3 (because row 1 contains "dog" and row 3 contains "cat"), but keep row 2 and row 4.

  id                phrase
1  2   horse, bunny, mouse
2  4    monkey, chimp, cow

In other words, I want to subset data so that it is only (the headers and) row 2 and row 4 (because they contain neither "dog" nor "cat").

Thanks!

deepseefan · Accepted Answer · 2019-10-30 15:41:08Z

2

In case you want to mix it with dplyr and stringr:

library(stringr)
library(dplyr)

data %>%
  filter(str_detect(phrase, paste(remove_if, collapse = "|"), negate = TRUE))
#   id              phrase
# 1  2 horse, bunny, mouse
# 2  4  monkey, chimp, cow

answered Oct 30, 2019 at 15:41

deepseefan

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akrun · Accepted Answer · 2019-10-30 15:31:02Z

1

We can use grepl with subset after pasteing the 'remove_if' to a single string

subset(data, !grepl(paste(remove_if, collapse="|"), phrase))
#    id              phrase
#2  2 horse, bunny, mouse
#4  4  monkey, chimp, cow

edited Oct 30, 2019 at 15:31

answered Oct 30, 2019 at 15:30

akrun

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1 Comment

arranjdavis Over a year ago

Worked as expected on the real dataset - thank you!

Jilber Urbina · Accepted Answer · 2019-10-30 15:30:17Z

1

Use grep

> data[grep(paste0(remove_if, collapse = "|"), data$phrase, invert = TRUE), ]
  id              phrase
2  2 horse, bunny, mouse
4  4  monkey, chimp, cow

answered Oct 30, 2019 at 15:30

Jilber Urbina

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Comments

Ape · Accepted Answer · 2019-10-30 15:50:52Z

1

data[!grepl(paste0("(^|, )(", paste0(remove_if, collapse = "|"), ")(,|$)"), data$phrase),]

# id                    phrase
#  2 caterpillar, bunny, mouse
#  4        monkey, chimp, cow

the regex constructed in this example is "(^|, )(dog|cat)(,|$)", to avoid matching words that contain 'cat' or 'dog', but aren't actually the exact words, e.g. 'caterpillar'

edited Oct 30, 2019 at 15:50

answered Oct 30, 2019 at 15:38

Ape

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Comments

dc37 · Accepted Answer · 2019-10-30 15:34:23Z

0

An another way (maybe not the best one):

data[-unique(unlist(sapply(c(remove_if),function(x){grep(x,data$phrase)}))),]
  id              phrase
2  2 horse, bunny, mouse
4  4  monkey, chimp, cow

answered Oct 30, 2019 at 15:34

dc37

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