Find nonsense words in a text












2















I have a dataset with answers of user if they know a brand or not. Some of the users just answered nonsense, as you can see in my example. 



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad")





spamidenfifier <- function(x) {

  verhaeltnis <- str_count(tolower(x), "[aeoiu]") / str_count(x)

  sequenz <- sum(sequence(rle(as.character(data.frame(strsplit(as.character(x), ""))[,1]))$lengths) >= 3, na.rm = TRUE)

  if(str_count(x) > 4) { weight <- 0.9 }  else {  weight <- 1  } ## Gewicht, weil unwahrscheinlicher bei längerem String

  variation_buchstaben <- (length(unique(data.frame(strsplit(as.character(x), ""))[,1])) / str_count(x) * weight)

  if(verhaeltnis < 0.2 | verhaeltnis > 0.8 | sequenz > 0 | variation_buchstaben < 0.5) {

    return(TRUE)

  } else {

    return(FALSE)

  }

}





sapply(meinstring, spamidenfifier)


Output:



----asdada            no idea                C&A         aaaaaaaaaa                --- adaosdjasodajsdoad 

      TRUE              FALSE              FALSE               TRUE               TRUE              FALSE


My function does not work too bad, however there might be better solutions. Is there a package or better method to identify if a word was just misspelled or a person answered nonsense.
If not, suggestions to improve that function are highly appreciated!



edit: Updated some improvements :-)






r text-mining stringr stringi







share|improve this question




















  • 1





    I think a good first order solution is see if the words can be recognized as real words. You could use a spellchecker such as hunspell and see if that package can recognize the words. If they cannot, the word is probably a bogus word.

    – Paul Hiemstra
    Nov 19 '18 at 16:51
















2















I have a dataset with answers of user if they know a brand or not. Some of the users just answered nonsense, as you can see in my example. 



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad")





spamidenfifier <- function(x) {

  verhaeltnis <- str_count(tolower(x), "[aeoiu]") / str_count(x)

  sequenz <- sum(sequence(rle(as.character(data.frame(strsplit(as.character(x), ""))[,1]))$lengths) >= 3, na.rm = TRUE)

  if(str_count(x) > 4) { weight <- 0.9 }  else {  weight <- 1  } ## Gewicht, weil unwahrscheinlicher bei längerem String

  variation_buchstaben <- (length(unique(data.frame(strsplit(as.character(x), ""))[,1])) / str_count(x) * weight)

  if(verhaeltnis < 0.2 | verhaeltnis > 0.8 | sequenz > 0 | variation_buchstaben < 0.5) {

    return(TRUE)

  } else {

    return(FALSE)

  }

}





sapply(meinstring, spamidenfifier)


Output:



----asdada            no idea                C&A         aaaaaaaaaa                --- adaosdjasodajsdoad 

      TRUE              FALSE              FALSE               TRUE               TRUE              FALSE


My function does not work too bad, however there might be better solutions. Is there a package or better method to identify if a word was just misspelled or a person answered nonsense.
If not, suggestions to improve that function are highly appreciated!



edit: Updated some improvements :-)






r text-mining stringr stringi







share|improve this question




















  • 1





    I think a good first order solution is see if the words can be recognized as real words. You could use a spellchecker such as hunspell and see if that package can recognize the words. If they cannot, the word is probably a bogus word.

    – Paul Hiemstra
    Nov 19 '18 at 16:51














2












2








2


1






I have a dataset with answers of user if they know a brand or not. Some of the users just answered nonsense, as you can see in my example. 



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad")





spamidenfifier <- function(x) {

  verhaeltnis <- str_count(tolower(x), "[aeoiu]") / str_count(x)

  sequenz <- sum(sequence(rle(as.character(data.frame(strsplit(as.character(x), ""))[,1]))$lengths) >= 3, na.rm = TRUE)

  if(str_count(x) > 4) { weight <- 0.9 }  else {  weight <- 1  } ## Gewicht, weil unwahrscheinlicher bei längerem String

  variation_buchstaben <- (length(unique(data.frame(strsplit(as.character(x), ""))[,1])) / str_count(x) * weight)

  if(verhaeltnis < 0.2 | verhaeltnis > 0.8 | sequenz > 0 | variation_buchstaben < 0.5) {

    return(TRUE)

  } else {

    return(FALSE)

  }

}





sapply(meinstring, spamidenfifier)


Output:



----asdada            no idea                C&A         aaaaaaaaaa                --- adaosdjasodajsdoad 

      TRUE              FALSE              FALSE               TRUE               TRUE              FALSE


My function does not work too bad, however there might be better solutions. Is there a package or better method to identify if a word was just misspelled or a person answered nonsense.
If not, suggestions to improve that function are highly appreciated!



edit: Updated some improvements :-)






r text-mining stringr stringi







share|improve this question
















I have a dataset with answers of user if they know a brand or not. Some of the users just answered nonsense, as you can see in my example. 



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad")





spamidenfifier <- function(x) {

  verhaeltnis <- str_count(tolower(x), "[aeoiu]") / str_count(x)

  sequenz <- sum(sequence(rle(as.character(data.frame(strsplit(as.character(x), ""))[,1]))$lengths) >= 3, na.rm = TRUE)

  if(str_count(x) > 4) { weight <- 0.9 }  else {  weight <- 1  } ## Gewicht, weil unwahrscheinlicher bei längerem String

  variation_buchstaben <- (length(unique(data.frame(strsplit(as.character(x), ""))[,1])) / str_count(x) * weight)

  if(verhaeltnis < 0.2 | verhaeltnis > 0.8 | sequenz > 0 | variation_buchstaben < 0.5) {

    return(TRUE)

  } else {

    return(FALSE)

  }

}





sapply(meinstring, spamidenfifier)


Output:



----asdada            no idea                C&A         aaaaaaaaaa                --- adaosdjasodajsdoad 

      TRUE              FALSE              FALSE               TRUE               TRUE              FALSE


My function does not work too bad, however there might be better solutions. Is there a package or better method to identify if a word was just misspelled or a person answered nonsense.
If not, suggestions to improve that function are highly appreciated!



edit: Updated some improvements :-)






r text-mining stringr stringi




r text-mining stringr stringi






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 19 '18 at 18:59







Markus LnGa

















asked Nov 19 '18 at 15:54









Markus LnGaMarkus LnGa

1129




1129








  • 1





    I think a good first order solution is see if the words can be recognized as real words. You could use a spellchecker such as hunspell and see if that package can recognize the words. If they cannot, the word is probably a bogus word.

    – Paul Hiemstra
    Nov 19 '18 at 16:51














  • 1





    I think a good first order solution is see if the words can be recognized as real words. You could use a spellchecker such as hunspell and see if that package can recognize the words. If they cannot, the word is probably a bogus word.

    – Paul Hiemstra
    Nov 19 '18 at 16:51








1




1





I think a good first order solution is see if the words can be recognized as real words. You could use a spellchecker such as hunspell and see if that package can recognize the words. If they cannot, the word is probably a bogus word.

– Paul Hiemstra
Nov 19 '18 at 16:51





I think a good first order solution is see if the words can be recognized as real words. You could use a spellchecker such as hunspell and see if that package can recognize the words. If they cannot, the word is probably a bogus word.

– Paul Hiemstra
Nov 19 '18 at 16:51












1 Answer
1






active

oldest

votes


















1














Just my spontaneous idea:



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad", "+-*-", "*-+-", "adfpdflrraaeea")



grepl('^\W+$|(?:[-!@#$%^&*\[\]()";:_<>.,=+/ ]){2,}|[-!@#$%^&*\[\]()";:_<>.,=+/ ]{3,}|[aeoiu]{3,}',

meinstring , perl = T) & !grepl("iou|zweieiig", meinstring) # add the exceptions in the second grepl.



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



There is no neat perfect solution.






share|improve this answer





















  • 1





    Interesting try, but your solution will find words like delicious (iou) as nonsense.

    – iod
    Nov 19 '18 at 16:28











  • hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

    – Andre Elrico
    Nov 19 '18 at 16:30













  • Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

    – iod
    Nov 19 '18 at 16:35











  • wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

    – Markus LnGa
    Nov 19 '18 at 18:13











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1 Answer
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1














Just my spontaneous idea:



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad", "+-*-", "*-+-", "adfpdflrraaeea")



grepl('^\W+$|(?:[-!@#$%^&*\[\]()";:_<>.,=+/ ]){2,}|[-!@#$%^&*\[\]()";:_<>.,=+/ ]{3,}|[aeoiu]{3,}',

meinstring , perl = T) & !grepl("iou|zweieiig", meinstring) # add the exceptions in the second grepl.



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



There is no neat perfect solution.






share|improve this answer





















  • 1





    Interesting try, but your solution will find words like delicious (iou) as nonsense.

    – iod
    Nov 19 '18 at 16:28











  • hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

    – Andre Elrico
    Nov 19 '18 at 16:30













  • Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

    – iod
    Nov 19 '18 at 16:35











  • wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

    – Markus LnGa
    Nov 19 '18 at 18:13
















1














Just my spontaneous idea:



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad", "+-*-", "*-+-", "adfpdflrraaeea")



grepl('^\W+$|(?:[-!@#$%^&*\[\]()";:_<>.,=+/ ]){2,}|[-!@#$%^&*\[\]()";:_<>.,=+/ ]{3,}|[aeoiu]{3,}',

meinstring , perl = T) & !grepl("iou|zweieiig", meinstring) # add the exceptions in the second grepl.



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



There is no neat perfect solution.






share|improve this answer





















  • 1





    Interesting try, but your solution will find words like delicious (iou) as nonsense.

    – iod
    Nov 19 '18 at 16:28











  • hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

    – Andre Elrico
    Nov 19 '18 at 16:30













  • Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

    – iod
    Nov 19 '18 at 16:35











  • wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

    – Markus LnGa
    Nov 19 '18 at 18:13














1












1








1







Just my spontaneous idea:



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad", "+-*-", "*-+-", "adfpdflrraaeea")



grepl('^\W+$|(?:[-!@#$%^&*\[\]()";:_<>.,=+/ ]){2,}|[-!@#$%^&*\[\]()";:_<>.,=+/ ]{3,}|[aeoiu]{3,}',

meinstring , perl = T) & !grepl("iou|zweieiig", meinstring) # add the exceptions in the second grepl.



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



There is no neat perfect solution.






share|improve this answer















Just my spontaneous idea:



meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad", "+-*-", "*-+-", "adfpdflrraaeea")



grepl('^\W+$|(?:[-!@#$%^&*\[\]()";:_<>.,=+/ ]){2,}|[-!@#$%^&*\[\]()";:_<>.,=+/ ]{3,}|[aeoiu]{3,}',

meinstring , perl = T) & !grepl("iou|zweieiig", meinstring) # add the exceptions in the second grepl.



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



There is no neat perfect solution.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 19 '18 at 16:32

























answered Nov 19 '18 at 16:21









Andre ElricoAndre Elrico

5,72011029




5,72011029








  • 1





    Interesting try, but your solution will find words like delicious (iou) as nonsense.

    – iod
    Nov 19 '18 at 16:28











  • hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

    – Andre Elrico
    Nov 19 '18 at 16:30













  • Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

    – iod
    Nov 19 '18 at 16:35











  • wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

    – Markus LnGa
    Nov 19 '18 at 18:13














  • 1





    Interesting try, but your solution will find words like delicious (iou) as nonsense.

    – iod
    Nov 19 '18 at 16:28











  • hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

    – Andre Elrico
    Nov 19 '18 at 16:30













  • Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

    – iod
    Nov 19 '18 at 16:35











  • wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

    – Markus LnGa
    Nov 19 '18 at 18:13








1




1





Interesting try, but your solution will find words like delicious (iou) as nonsense.

– iod
Nov 19 '18 at 16:28





Interesting try, but your solution will find words like delicious (iou) as nonsense.

– iod
Nov 19 '18 at 16:28













hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

– Andre Elrico
Nov 19 '18 at 16:30







hehe :D, sure also in in german there is zweieiig, Donauauen, Donau-Auen, Treueeid some micro is well needed.

– Andre Elrico
Nov 19 '18 at 16:30















Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

– iod
Nov 19 '18 at 16:35





Yeah, I think a real solution will have to use something like the lexicon package and search through it. Which could be overkill, depending on how big the actual data is.

– iod
Nov 19 '18 at 16:35













wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

– Markus LnGa
Nov 19 '18 at 18:13





wow i understand nothing ;-). But it does not work too bad. Guess it might be better than my solution

– Markus LnGa
Nov 19 '18 at 18:13




















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