CODEWITHSUNDEEP
regexrpattern-matching
Chris
Chris

Reputation: 3441

Count Pattern Matching in R

How would one efficiently count the number of instances of one character string which occur within another character string?

Below is my code to date. It successfully identifies if any instance of the one string occurs in the other string. However, I do not know how to extend it from a TRUE/FALSE relationship to a counting relationship. 

x <- ("Hello my name is Christopher. Some people call me Chris")
y <- ("Chris is an interesting person to be around")
z <- ("Because he plays sports and likes statistics")

lll <- tolower(list(x,y,z))
dict <- tolower(c("Chris", "Hell"))

mmm <- matrix(nrow=length(lll), ncol=length(dict), NA)

for (i in 1:length(lll)) {
for (j in 1:length(dict)) {
    mmm[i,j] <- sum(grepl(dict[j],lll[i]))
}
}
mmm

It yields:

       [,1] [,2]
 [1,]    1    1
 [2,]    1    0
 [3,]    0    0

Since the lower-case string "chris" appears twice in the lll[1] I would like mmm[1,1] to be 2 instead of 1.

Real example is much higher dimension...so would love if code could be vectorized instead of using my brute force for loops.

Upvotes: 6

Views: 8386

Answers (5)

Tyler Rinker
Tyler Rinker

Reputation: 109864

This uses the qdap package. The CRAN version should work fine but you may want the dev version

library(qdap)

termco(c(x, y, z), 1:3, c('chris', 'hell'))

##   3 word.count     chris      hell
## 1 1         10 2(20.00%) 1(10.00%)
## 2 2          8 1(12.50%)         0
## 3 3          7         0         0


termco(c(x, y, z), 1:3, c('chris', 'hell'))$raw

##   3 word.count chris hell
## 1 1         10     2    1
## 2 2          8     1    0
## 3 3          7     0    0

Upvotes: 1

Metrics
Metrics

Reputation: 15458

llll<-rep(lll,length(dict))
dict1<-rep(dict,each=length(lll))


 do.call(rbind,Map(function(x,y)list(y,sum(gregexpr(y,x)[[1]] > 0)), llll,dict1))
                                                        [,1]    [,2]
hello my name is christopher. some people call me chris "chris" 2   
chris is an interesting person to be around             "chris" 1   
because he plays sports and likes statistics            "chris" 0   
hello my name is christopher. some people call me chris "hell"  1   
chris is an interesting person to be around             "hell"  0   
because he plays sports and likes statistics            "hell"  0

You can then use reshape to get what you want.

Upvotes: 1

TheComeOnMan
TheComeOnMan

Reputation: 12875

Instead of sum(grepl(dict[j],lll[i])), try sum(gregexpr(dict[j],lll[i])[[1]] > 0)

Upvotes: 2

Matthew Plourde
Matthew Plourde

Reputation: 44614

You can also do something like this:

count.matches <- function(pat, vec) sapply(regmatches(vec, gregexpr(pat, vec)), length)
mapply(count.matches, c('chris', 'hell'), list(lll))
#      chris hell
# [1,]     2    1
# [2,]     1    0
# [3,]     0    0

Upvotes: 4

Ricardo Saporta
Ricardo Saporta

Reputation: 55350

Two quick tips:

  1. avoid the dual for-loop, you dont need it ;)
  2. use the stringr package
library(stringr)

dict <- setNames(nm=dict)  # simply for neatness
lapply(dict, str_count, string=lll)
# $chris
# [1] 2 1 0
#
# $hell
# [1] 1 0 0

Or as a matrix:

#  sapply(dict, str_count, string=lll)
#      chris hell
# [1,]     2    1
# [2,]     1    0
# [3,]     0    0

Upvotes: 8

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