I want to filter a string.
Basically when someone types a message, I want certain words to be filtered out, like this:
User types: hey guys lol omg -omg mkdj*Omg*ndid
I want the filter to run and:
Output: hey guys lol - mkdjndid
And I need the filtered words to be loaded from an ArrayList that contains several words to filter out. Now at the moment I am doing if(message.contains(omg)) but that doesn't work if someone types zomg or -omg or similar.
Use replaceAll with a regex built from the bad word:
message = message.replaceAll("(?i)\\b[^\\w -]*" + badWord + "[^\\w -]*\\b", "");
This passes your test case:
public static void main( String[] args ) {
List<String> badWords = Arrays.asList( "omg", "black", "white" );
String message = "hey guys lol omg -omg mkdj*Omg*ndid";
for ( String badWord : badWords ) {
message = message.replaceAll("(?i)\\b[^\\w -]*" + badWord + "[^\\w -]*\\b", "");
}
System.out.println( message );
}
try:
input.replaceAll("(\\*?)[oO][mM][gG](\\*?)", "").split(" ")
Dave gave you the answer already, but I will emphasize the statement here. You will face a problem if you implement your algorithm with a simple for-loop that just replaces the occurrence of the filtered word. As an example, if you filter the word ass in the word 'classic' and replace it with 'butt', the resultant word will be 'clbuttic' which doesn't make any sense. Thus, I would suggest using a word list,like the ones stored in Linux under /usr/share/dict/ directory, to check if the word is valid or it needs filtering.
I don't quite get what you are trying to do.
I ran into this same problem and solved it in the following way:
1) Have a google spreadsheet with all words that I want to filter out
2) Directly download the google spreadsheet into my code with the loadConfigs method (see below)
3) Replace all l33tsp33k characters with their respective alphabet letter
4) Replace all special characters but letters from the sentence
5) Run an algorithm that checks all the possible combinations of words within a string against the list efficiently, note that this part is key - you don't want to loop over your ENTIRE list every time to see if your word is in the list. In my case, I found every combination within the string input and checked it against a hashmap (O(1) runtime). This way the runtime grows relatively to the string input, not the list input.
6) Check if the word is not used in combination with a good word (e.g. bass contains *ss). This is also loaded through the spreadsheet
6) In our case we are also posting the filtered words to Slack, but you can remove that line obviously.
We are using this in our own games and it's working like a charm. Hope you guys enjoy.
https://pimdewitte.me/2016/05/28/filtering-combinations-of-bad-words-out-of-string-inputs/
public static HashMap<String, String[]> words = new HashMap<String, String[]>();
public static void loadConfigs() {
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(new URL("https://docs.google.com/spreadsheets/d/1hIEi2YG3ydav1E06Bzf2mQbGZ12kh2fe4ISgLg_UBuM/export?format=csv").openConnection().getInputStream()));
String line = "";
int counter = 0;
while((line = reader.readLine()) != null) {
counter++;
String[] content = null;
try {
content = line.split(",");
if(content.length == 0) {
continue;
}
String word = content[0];
String[] ignore_in_combination_with_words = new String[]{};
if(content.length > 1) {
ignore_in_combination_with_words = content[1].split("_");
}
words.put(word.replaceAll(" ", ""), ignore_in_combination_with_words);
} catch(Exception e) {
e.printStackTrace();
}
}
System.out.println("Loaded " + counter + " words to filter out");
} catch (IOException e) {
e.printStackTrace();
}
}
/**
* Iterates over a String input and checks whether a cuss word was found in a list, then checks if the word should be ignored (e.g. bass contains the word *ss).
* #param input
* #return
*/
public static ArrayList<String> badWordsFound(String input) {
if(input == null) {
return new ArrayList<>();
}
// remove leetspeak
input = input.replaceAll("1","i");
input = input.replaceAll("!","i");
input = input.replaceAll("3","e");
input = input.replaceAll("4","a");
input = input.replaceAll("#","a");
input = input.replaceAll("5","s");
input = input.replaceAll("7","t");
input = input.replaceAll("0","o");
ArrayList<String> badWords = new ArrayList<>();
input = input.toLowerCase().replaceAll("[^a-zA-Z]", "");
for(int i = 0; i < input.length(); i++) {
for(int fromIOffset = 1; fromIOffset < (input.length()+1 - i); fromIOffset++) {
String wordToCheck = input.substring(i, i + fromIOffset);
if(words.containsKey(wordToCheck)) {
// for example, if you want to say the word bass, that should be possible.
String[] ignoreCheck = words.get(wordToCheck);
boolean ignore = false;
for(int s = 0; s < ignoreCheck.length; s++ ) {
if(input.contains(ignoreCheck[s])) {
ignore = true;
break;
}
}
if(!ignore) {
badWords.add(wordToCheck);
}
}
}
}
for(String s: badWords) {
Server.getSlackManager().queue(s + " qualified as a bad word in a username");
}
return badWords;
}
Related
I have a List of Strings containing names and surnames and i have a free text.
List<String> names; // contains: "jon", "snow", "arya", "stark", ...
String text = "jon snow and stark arya";
I have to find all the names and surnames, possibly with a Java Regex (so using Pattern and Matcher objects). So i want something like:
List<String> foundNames; // contains: "jon snow", "stark arya"
I have done this 2 possible ways but without using Regex, they are not static beacause part of a NameFinder class that have a list "names" that contains all the names.
public List<String> findNamePairs(String text) {
List<String> foundNamePairs = new ArrayList<String>();
List<String> names = this.names;
text = text.toLowerCase();
for (String name : names) {
String nameToSearch = name + " ";
int index = text.indexOf(nameToSearch);
if (index != -1) {
String textSubstring = text.substring(index + nameToSearch.length());
for (String nameInner : names) {
if (name != nameInner && textSubstring.startsWith(nameInner)) {
foundNamePairs.add(name + " " + nameInner);
}
}
}
}
removeDuplicateFromList(foundNamePairs);
return foundNamePairs;
}
or in a worse (very bad) way (creating all the possible pairs):
public List<String> findNamePairsInTextNotOpt(String text) {
List<String> foundNamePairs = new ArrayList<String>();
text = text.toLowerCase();
List<String> pairs = getNamePairs(this.names);
for (String name : pairs) {
if (text.contains(name)) {
foundNamePairs.add(name);
}
}
removeDuplicateFromList(foundNamePairs);
return foundNamePairs;
}
You can create a regex using the list of names and then use find to find the names. To ensure you don't have duplicates, you can check if the name is already in the list of found names. The code would look like this.
List<String> names = Arrays.asList("jon", "snow", "stark", "arya");
String text = "jon snow and Stark arya and again Jon Snow";
StringBuilder regexBuilder = new StringBuilder();
for (int i = 0; i < names.size(); i += 2) {
regexBuilder.append("(")
.append(names.get(i))
.append(" ")
.append(names.get(i + 1))
.append(")");
if (i != names.size() - 2) regexBuilder.append("|");
}
System.out.println(regexBuilder.toString());
Pattern compile = Pattern.compile(regexBuilder.toString(), Pattern.CASE_INSENSITIVE);
Matcher matcher = compile.matcher(text);
List<String> found = new ArrayList<>();
int start = 0;
while (matcher.find(start)) {
String match = matcher.group().toLowerCase();
if (!found.contains(match)) found.add(match);
start = matcher.end();
}
for (String s : found) System.out.println("found: " + s);
If you want to be case sensitive just remove the flag in Pattern.compile(). If all matches have the same capitalization you can omit the toLowerCase() in the while loop as well.
But make sure that the list contains a multiple of 2 as list elements (name and surname) as the for-loop will throw an IndexOutOfBoundsException otherwise. Also the order matters in my code. It will only find the name pairs in the order they occur in the list. If you want to have both orders, you can change the regex generation accordingly.
Edit: As it is unknown whether a name is a surname or name and which build a name/surname pair, the regex generation must be done differently.
StringBuilder regexBuilder = new StringBuilder("(");
for (int i = 0; i < names.size(); i++) {
regexBuilder.append("(")
.append(names.get(i))
.append(")");
if (i != names.size() - 1) regexBuilder.append("|");
}
regexBuilder.append(") ");
regexBuilder.append(regexBuilder);
regexBuilder.setLength(regexBuilder.length() - 1);
System.out.println(regexBuilder.toString());
This regex will match any of the given names followed by a space and then again any of the names.
I am writing a program which is the opposite of Auto Correct. The logic is that the user enters a sentence, when a button is pressed, the grammatical opposite of the sentence entered by the user should be displayed. I am roughly able to get the code. I used the matcher logic.But i am not able to get the desired output. I am linking the code with this question. Can anyone help me please?
imageButton.setOnClickListener(new View.OnClickListener() {
#Override
public void onClick(View v) {
String input = editText.getText().toString();
String store = input;
String store1 [] = store.split(" ");
String correct[] = {"is","shall","this","can","will"};
String wrong [] = {"was","should","that","could","would"};
String output = "";
for (int i=0;i<store1.length;i++) {
for(int j=0;j<correct.length;j++){
if(store1[i].matches(correct[j])) {
output = input.replace(store1[i], wrong[j]);
//store1[i] = wrong[j];
}
else
{
input.replace(store1[i], store1[i]);
//store1[i] = store1[i];
}
}
mTextView.setText(output);
}
}});
By looking at your code, I've found some redundancy and unused variable. Shown as below.
String store = input;
String store1 [] = store.split(" ");
As shown below, I did some cleanup for you and implement your logic using Map interface. The wrong value must be the Key of the map so that we can easily determine is the word a wrong value using Map.containKeys(word). If key is found then we concatenate the output variable with the correct word.
String input = editText.getText().toString().split(" ");
Map<String, String> pairs = new HashMap<>();
pairs.put("is", "was");
pairs.put("shall", "should");
pairs.put("this", "that");
pairs.put("can", "could");
pairs.put("will", "would");
String output = "";
for (String word : input) {
if (pairs.containsKey(word)) {
output = output + pairs.get(word) + " ";
} else {
output = output + word + " ";
}
}
mTextView.setText(output.trim());
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
Regular expression not matching subwords in phrase
My program displays the matching results, but I want to sort the results as complete match (100%), half a match and so on.
My text file contains the following line:
Red car
Red
Car
So If I search for: “red car”. I get the following results
Red car
Red
Car
So what I want to do is to sort the found results as follows:
"red car" 100% match
"red" 40% match
"car" 40% match
Any help is appreciated.
Any help is appreciated. My code is as follows:
public static void main(String[] args) {
// TODO code application logic here
String strLine;
try{
// Open the file that is the first
// command line parameter
FileInputStream fstream = new FileInputStream("C:\\textfile.txt"");
// Get the object of DataInputStream
DataInputStream in = new DataInputStream(fstream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
Scanner input = new Scanner (System.in);
System.out.print("Enter Your Search: "); // String key="red or yellow";
String key = input.nextLine();
while ((strLine = br.readLine()) != null) {
Pattern p = Pattern.compile(key); // regex pattern to search for
Matcher m = p.matcher(strLine); // src of text to search
boolean b = false;
while(b = m.find()) {
System.out.println( " " + m.group()); // returns index and match
// Print the content on the console
}
}
//Close the input stream
in.close();
}catch (Exception e){//Catch exception if any
System.err.println("Error: " + e.getMessage());
}
}
Assuming you are searching for "Red" or "Yellow", and or is the only logical operator you need (no 'and' or 'xor') and you don't want to use any wildcards or regular-expressions in what you search for, then I would simply loop through, trying to match each String in turn against the line. In pseudo-code, something like:
foreach (thisLine: allLinesInTheFile) {
numOfCharsMatching = 0
foreach (thisString: allSearchStrings) {
if (thisLine.contains(thisString) {
numOfCharsMatching = numOfCharsMatching + thisString.length
}
}
score = ( numOfCharsMatching / thisLine.length ) * 100
}
If you don't want spaces to count in your score, then you'd need to remove them from the thisString.length (and not allow them in your search terms)
One other problem is the numOfCharsMatching will be incorrect if matches can overlap (i.e. if searching for 'row' or 'brown' in 'brown row' it will say that there are 11 characters matching, longer than the length of the string. You could use a BitSet to track which characters have been involved in a match, something like:
foreach (thisLine: allLinesInTheFile) {
whichCharsMatch = new BitSet()
foreach (thisString: allSearchStrings) {
if (thisLine.contains(thisString) {
whichCharsMatch.set(startPositionOfMatch, endPositionOfMatch, true)
}
}
score = ( numOfCharsMatching / thisLine.length ) * 100
}
Have a look at the BitSet javadoc, particularly the set and cardinality methods
How can I parse a String str = "abc, \"def,ghi\"";
such that I get the output as
String[] strs = {"abc", "\"def,ghi\""}
i.e. an array of length 2.
Should I use regular expression or Is there any method in java api or anyother opensource
project which let me do this?
Edited
To give context about the problem, I am reading a text file which has a list of records one on each line. Each record has list of fields separated by delimiter(comma or semi-colon). Now I have a requirement where I have to support text qualifier some thing excel or open office supports. Suppose I have record
abc, "def,ghi"
In this , is my delimiter and " is my text qualifier such that when I parse this string I should get two fields abc and def,ghi not {abc,def,ghi}
Hope this clears my requirement.
Thanks
Shekhar
The basic algorithm is not too complicated:
public static List<String> customSplit(String input) {
List<String> elements = new ArrayList<String>();
StringBuilder elementBuilder = new StringBuilder();
boolean isQuoted = false;
for (char c : input.toCharArray()) {
if (c == '\"') {
isQuoted = !isQuoted;
// continue; // changed according to the OP comment - \" shall not be skipped
}
if (c == ',' && !isQuoted) {
elements.add(elementBuilder.toString().trim());
elementBuilder = new StringBuilder();
continue;
}
elementBuilder.append(c);
}
elements.add(elementBuilder.toString().trim());
return elements;
}
This question seems appropriate: Split a string ignoring quoted sections
Along that line, http://opencsv.sourceforge.net/ seems appropriate.
Try this -
String str = "abc, \"def,ghi\"";
String regex = "([,]) | (^[\"\\w*,\\w*\"])";
for(String s : str.split(regex)){
System.out.println(s);
}
Try:
List<String> res = new LinkedList<String>();
String[] chunks = str.split("\\\"");
if (chunks.length % 2 == 0) {
// Mismatched escaped quotes!
}
for (int i = 0; i < chunks.length; i++) {
if (i % 2 == 1) {
res.addAll(Array.asList(chunks[i].split(",")));
} else {
res.add(chunks[i]);
}
}
This will only split up the portions that are not between escaped quotes.
Call trim() if you want to get rid of the whitespace.
I want to tokenize a string like this
String line = "a=b c='123 456' d=777 e='uij yyy'";
I cannot split based like this
String [] words = line.split(" ");
Any idea how can I split so that I get tokens like
a=b
c='123 456'
d=777
e='uij yyy';
The simplest way to do this is by hand implementing a simple finite state machine. In other words, process the string a character at a time:
When you hit a space, break off a token;
When you hit a quote keep getting characters until you hit another quote.
Depending on the formatting of your original string, you should be able to use a regular expression as a parameter to the java "split" method: Click here for an example.
The example doesn't use the regular expression that you would need for this task though.
You can also use this SO thread as a guideline (although it's in PHP) which does something very close to what you need. Manipulating that slightly might do the trick (although having quotes be part of the output or not may cause some issues). Keep in mind that regex is very similar in most languages.
Edit: going too much further into this type of task may be ahead of the capabilities of regex, so you may need to create a simple parser.
line.split(" (?=[a-z+]=)")
correctly gives:
a=b
c='123 456'
d=777
e='uij yyy'
Make sure you adapt the [a-z+] part in case your keys structure changes.
Edit: this solution can fail miserably if there is a "=" character in the value part of the pair.
StreamTokenizer can help, although it is easiest to set up to break on '=', as it will always break at the start of a quoted string:
String s = "Ta=b c='123 456' d=777 e='uij yyy'";
StreamTokenizer st = new StreamTokenizer(new StringReader(s));
st.ordinaryChars('0', '9');
st.wordChars('0', '9');
while (st.nextToken() != StreamTokenizer.TT_EOF) {
switch (st.ttype) {
case StreamTokenizer.TT_NUMBER:
System.out.println(st.nval);
break;
case StreamTokenizer.TT_WORD:
System.out.println(st.sval);
break;
case '=':
System.out.println("=");
break;
default:
System.out.println(st.sval);
}
}
outputs
Ta
=
b
c
=
123 456
d
=
777
e
=
uij yyy
If you leave out the two lines that convert numeric characters to alpha, then you get d=777.0, which might be useful to you.
Assumptions:
Your variable name ('a' in the assignment 'a=b') can be of length 1 or more
Your variable name ('a' in the assignment 'a=b') can not contain the space character, anything else is fine.
Validation of your input is not required (input assumed to be in valid a=b format)
This works fine for me.
Input:
a=b abc='123 456' &=777 #='uij yyy' ABC='slk slk' 123sdkljhSDFjflsakd#*#&=456sldSLKD)#(
Output:
a=b
abc='123 456'
&=777
#='uij yyy'
ABC='slk slk'
123sdkljhSDFjflsakd#*#&=456sldSLKD)#(
Code:
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexTest {
// SPACE CHARACTER followed by
// sequence of non-space characters of 1 or more followed by
// first occuring EQUALS CHARACTER
final static String regex = " [^ ]+?=";
// static pattern defined outside so that you don't have to compile it
// for each method call
static final Pattern p = Pattern.compile(regex);
public static List<String> tokenize(String input, Pattern p){
input = input.trim(); // this is important for "last token case"
// see end of method
Matcher m = p.matcher(input);
ArrayList<String> tokens = new ArrayList<String>();
int beginIndex=0;
while(m.find()){
int endIndex = m.start();
tokens.add(input.substring(beginIndex, endIndex));
beginIndex = endIndex+1;
}
// LAST TOKEN CASE
//add last token
tokens.add(input.substring(beginIndex));
return tokens;
}
private static void println(List<String> tokens) {
for(String token:tokens){
System.out.println(token);
}
}
public static void main(String args[]){
String test = "a=b " +
"abc='123 456' " +
"&=777 " +
"#='uij yyy' " +
"ABC='slk slk' " +
"123sdkljhSDFjflsakd#*#&=456sldSLKD)#(";
List<String> tokens = RegexTest.tokenize(test, p);
println(tokens);
}
}
Or, with a regex for tokenizing, and a little state machine that just adds the key/val to a map:
String line = "a = b c='123 456' d=777 e = 'uij yyy'";
Map<String,String> keyval = new HashMap<String,String>();
String state = "key";
Matcher m = Pattern.compile("(=|'[^']*?'|[^\\s=]+)").matcher(line);
String key = null;
while (m.find()) {
String found = m.group();
if (state.equals("key")) {
if (found.equals("=") || found.startsWith("'"))
{ System.err.println ("ERROR"); }
else { key = found; state = "equals"; }
} else if (state.equals("equals")) {
if (! found.equals("=")) { System.err.println ("ERROR"); }
else { state = "value"; }
} else if (state.equals("value")) {
if (key == null) { System.err.println ("ERROR"); }
else {
if (found.startsWith("'"))
found = found.substring(1,found.length()-1);
keyval.put (key, found);
key = null;
state = "key";
}
}
}
if (! state.equals("key")) { System.err.println ("ERROR"); }
System.out.println ("map: " + keyval);
prints out
map: {d=777, e=uij yyy, c=123 456, a=b}
It does some basic error checking, and takes the quotes off the values.
This solution is both general and compact (it is effectively the regex version of cletus' answer):
String line = "a=b c='123 456' d=777 e='uij yyy'";
Matcher m = Pattern.compile("('[^']*?'|\\S)+").matcher(line);
while (m.find()) {
System.out.println(m.group()); // or whatever you want to do
}
In other words, find all runs of characters that are combinations of quoted strings or non-space characters; nested quotes are not supported (there is no escape character).
public static void main(String[] args) {
String token;
String value="";
HashMap<String, String> attributes = new HashMap<String, String>();
String line = "a=b c='123 456' d=777 e='uij yyy'";
StringTokenizer tokenizer = new StringTokenizer(line," ");
while(tokenizer.hasMoreTokens()){
token = tokenizer.nextToken();
value = token.contains("'") ? value + " " + token : token ;
if(!value.contains("'") || value.endsWith("'")) {
//Split the strings and get variables into hashmap
attributes.put(value.split("=")[0].trim(),value.split("=")[1]);
value ="";
}
}
System.out.println(attributes);
}
output:
{d=777, a=b, e='uij yyy', c='123 456'}
In this case continuous space will be truncated to single space in the value.
here attributed hashmap contains the values
import java.io.*;
import java.util.Scanner;
public class ScanXan {
public static void main(String[] args) throws IOException {
Scanner s = null;
try {
s = new Scanner(new BufferedReader(new FileReader("<file name>")));
while (s.hasNext()) {
System.out.println(s.next());
<write for output file>
}
} finally {
if (s != null) {
s.close();
}
}
}
}
java.util.StringTokenizer tokenizer = new java.util.StringTokenizer(line, " ");
while (tokenizer.hasMoreTokens()) {
String token = tokenizer.nextToken();
int index = token.indexOf('=');
String key = token.substring(0, index);
String value = token.substring(index + 1);
}
Have you tried splitting by '=' and creating a token out of each pair of the resulting array?