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]))$lengths) >= 3, na.rm = TRUE)
if(str_count(x) > 4) { weight <- 0.9 } else { weight <- 1 } ## Gewicht, weil unwahrscheinlicher bei längerem String
variation_buchstaben <- (length(unique(data.frame(strsplit(as.character(x), ""))[,1])) / str_count(x) * weight)
if(verhaeltnis < 0.2 | verhaeltnis > 0.8 | sequenz > 0 | variation_buchstaben < 0.5) {
return(TRUE)
} else {
return(FALSE)
}
}
sapply(meinstring, spamidenfifier)
Output:
----asdada no idea C&A aaaaaaaaaa --- adaosdjasodajsdoad
TRUE FALSE FALSE TRUE TRUE FALSE
My function does not work too bad, however there might be better solutions. Is there a package or better method to identify if a word was just misspelled or a person answered nonsense.
If not, suggestions to improve that function are highly appreciated!
edit: Updated some improvements :-)
r text-mining stringr stringi
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
1129
add a comment |
2
I have a dataset with answers of user if they know a brand or not. Some of the users just answered nonsense, as you can see in my example.
meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad")
spamidenfifier <- function(x) {
verhaeltnis <- str_count(tolower(x), "[aeoiu]") / str_count(x)
sequenz <- sum(sequence(rle(as.character(data.frame(strsplit(as.character(x), ""))[,1]))$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
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
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
add a comment |
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]))$lengths) >= 3, na.rm = TRUE)
if(str_count(x) > 4) { weight <- 0.9 } else { weight <- 1 } ## Gewicht, weil unwahrscheinlicher bei längerem String
variation_buchstaben <- (length(unique(data.frame(strsplit(as.character(x), ""))[,1])) / str_count(x) * weight)
if(verhaeltnis < 0.2 | verhaeltnis > 0.8 | sequenz > 0 | variation_buchstaben < 0.5) {
return(TRUE)
} else {
return(FALSE)
}
}
sapply(meinstring, spamidenfifier)
Output:
----asdada no idea C&A aaaaaaaaaa --- adaosdjasodajsdoad
TRUE FALSE FALSE TRUE TRUE FALSE
My function does not work too bad, however there might be better solutions. Is there a package or better method to identify if a word was just misspelled or a person answered nonsense.
If not, suggestions to improve that function are highly appreciated!
edit: Updated some improvements :-)
r text-mining stringr stringi
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
1129
I have a dataset with answers of user if they know a brand or not. Some of the users just answered nonsense, as you can see in my example.
meinstring <- c("----asdada", "no idea", "C&A", "aaaaaaaaaa", "---", "adaosdjasodajsdoad")
spamidenfifier <- function(x) {
verhaeltnis <- str_count(tolower(x), "[aeoiu]") / str_count(x)
sequenz <- sum(sequence(rle(as.character(data.frame(strsplit(as.character(x), ""))[,1]))$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
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
1129
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
1129
edited Nov 19 '18 at 18:59
Markus LnGa
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
1129
asked Nov 19 '18 at 15:54
Markus LnGaMarkus LnGa
1129
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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.
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
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 iszweieiig, Donauauen, Donau-Auen, Treueeidsome micro is well needed.
– Andre Elrico
Nov 19 '18 at 16:30
Yeah, I think a real solution will have to use something like thelexiconpackage 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
add a comment |
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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.
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
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 iszweieiig, Donauauen, Donau-Auen, Treueeidsome micro is well needed.
– Andre Elrico
Nov 19 '18 at 16:30
Yeah, I think a real solution will have to use something like thelexiconpackage 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
add a comment |
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.
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
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 iszweieiig, Donauauen, Donau-Auen, Treueeidsome micro is well needed.
– Andre Elrico
Nov 19 '18 at 16:30
Yeah, I think a real solution will have to use something like thelexiconpackage 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
add a comment |
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.
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
5,72011029
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.
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
5,72011029
edited Nov 19 '18 at 16:32
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
5,72011029
answered Nov 19 '18 at 16:21
Andre ElricoAndre Elrico
5,72011029
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 iszweieiig, Donauauen, Donau-Auen, Treueeidsome micro is well needed.
– Andre Elrico
Nov 19 '18 at 16:30
Yeah, I think a real solution will have to use something like thelexiconpackage 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
add a comment |
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 iszweieiig, Donauauen, Donau-Auen, Treueeidsome micro is well needed.
– Andre Elrico
Nov 19 '18 at 16:30
Yeah, I think a real solution will have to use something like thelexiconpackage 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
add a comment |
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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