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R: Filter a data frame on multiple partial strings

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This is a blog post about a very specific topic. I wanted to filter a data frame on a set of strings that I wanted to match partially. Let’s dive right in.

Matching partially is fairly easy, and there are many libraries to choose from, with grepl and str_detect the most popular ones. However, partially matching multiple strings is rather difficult, if you don’t want to fall back on traditional for loops. I built a solution using a combination of functions that I wanted to share.

Let’s say we have a data frame that contains some varieties of pears and apples, and I want to select the varieties that contain “esta” and “uji”.

library(stringr)
library(data.table)

df <- data.table(fruit = c('zestar apple',
                           'redlove apple',
                           'kiku apple',
                           'bonne de beugny pear',
                           'chinese white pear',
                           'fuji apple',
                           'envy apple',
                           'asian pear'),
                 count = c(5,8,9,3,2,1,5,7))

to_match_partially <- c('esta','uji') # strings to match          

To explain the code below, I really outdid myself with indentation. Let’s go from the inside to the outside.

  1. I match the column fruit of the data frame df to x, which is specified higher by the lapply function. The result wil be a vector of TRUE’s and FALSE’s.
  2. I convert these to numeric, because in #6 I’ll want to sum over these.
  3. This is where I used lapply to loop over the strings I want to match partially.
  4. The lapply function will return a list of lists that contain a TRUE/FALSE, for each string to match. However, I need them unlisted (huge 1-dimension vector), to create a matrix out of them in step #5.
  5. I convert the vector to a matrix that contains the same number of columns as there are strings to match.
  6. I can now sum each row of that matrix using rowSums. If this number is higher than zero, there was a match.
  7. I convert the row sums to a logical.
  8. Make the match.
selection <- as.logical( # 7
  rowSums( # 6
    matrix( # 5
      unlist( # 4
        lapply(to_match_partially, function(x) { #3
          as.numeric( # 2
            str_detect(df$fruit,x) # 1
          )
        })
      ), 
      ncol = length(to_match_partially), byrow = F)
  )
)

df[selection,] # 8

I can’t make the claim that this is the fastest solution. But if you’ve found another solution, based on a completely different way of thinking, I’d be glad to benchmark it. Let me know in the comments of this post!

I can’t make the claim that this is the fastest solution. But if you’ve found another solution, based on a completely different way of thinking, I’d be glad to benchmark it. Let me know in the comments of this post!

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