CODEWITHSUNDEEP
rdataframesapply
Saul Garcia
Saul Garcia

Reputation: 940

Convert a function into an apply, sapply (data.frame)

So I just built this function which basically takes two strings (a text, and a set of keywords). Then it has to find how many keywords are contained by the text string, if any. I been trying to apply the code on a data frame with no success.

Function is working:

something=function(text,keywords){
  kw = unlist(strsplit(keywords, ","))
  c=0
  for(i in length(kw)){
    if(grepl(kw[i],text)==0){
      c=c+1
    } else {c}
  }
  return(c)
}

Where if I imput:

> something("this planetarium is the shit","planetarium,amazing")
[1] 1

But what if my data frame was df

     keyword         text_clean
1    planetarium     Man this planetarium is the shit
2 musee,africain     rt lyonmangels reste encore places franceangels tour lyon organisons  investissons pme

My output expected is:

  df.1
1  1
2  0

Any insight? I was trying this code:

substng<-function(text, keywords){

  vector = laply(text,function(text,keywords){
    kw = unlist(strsplit(keywords, ","))
    c=0
    for(i in length(kw)){
      if(grepl(kw[i],text)==0){
        c=c+1
      } else {c}
    }
    return(c)
  })
  vector.df= as.data.frame(vector)
}

df <- read.table(header = TRUE, stringsAsFactors = FALSE, text = "keyword         text_clean
planetarium     'Man this planetarium is the shit'
musee,africain     'rt lyonmangels reste encore places franceangels tour lyon organisons  investissons pme'")

df$count = substng(df$text_clean,df$keyword)

Upvotes: 0

Views: 130

Answers (1)

mjm
mjm

Reputation: 66

I think stri_count in the stringi package can accomplish this.

Use "pattern|amazing" as the pattern/regex. Pipe = "or".

https://cran.r-project.org/web/packages/stringi/stringi.pdf

Upvotes: 0

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