I am writing a piece of code in which i have to find only complete words for example if i have
String str = "today is tuesday";
and I'm searching for "t" then I should not find any word.
Can anybody tell how can I write such a program in java?
I use a regexps for such tasks. In your case it should look something like this:
String str = "today is tuesday";
return str.matches(".*?\\bt\\b.*?"); // returns "false"
String str = "today is t uesday";
return str.matches(".*?\\bt\\b.*?"); // returns "true"
A short explanation:
. matches any character, *? is for zero or more times, \b is a word boundary.
More information on regexps can be found here or specifically for java here
String sentence = "Today is Tuesday";
Set<String> words = new HashSet<String>(
Arrays.asList(sentence.split(" "))
);
System.out.println(words.contains("Tue")); // prints "false"
System.out.println(words.contains("Tuesday")); // prints "true"
Each contains(word) query is O(1), so short of implementing your own sophisticated dictionary data structure, this is the fastest most practical solution if you have many words to look for in a text.
This uses String.split to separate out the words from the sentence on the " " delimiter. Other possible variations, depending on how the problem is defined, is to use \b, the word boundary anchor. The problem is considerably more difficult if you must take every grammatical features of natural languages into consideration (e.g. "can't" is split by \b into "can" and "t").
Case insensitivity can be easily introduced by using the traditional case normalization trick: split and hash sentence.toLowerCase() instead, and see if it contains(word.toLowerCase()).
See also
regular-expressions.info -- Anchors
Wikipedia -- String searching algorithm
Wikipedia -- Patricia Trie
String[] tokens = str.split(" ");
for(String s: tokens) {
if ("t".equals(s)) {
// t exists
break;
}
}
String[] words = str.split(" ");
Arrays.sort(words);
Arrays.binarySearch(words, searchedFor);
String str = "today is tuesday";
StringTokenizer stringTokenizer = new StringTokenizer(str);
bool exists = false;
while (stringTokenizer.hasMoreTokens()) {
if (stringTokenizer.nextToken().equals("t")) {
exists = true;
break;
}
}
use a regex like "\bt\b".
you can do that by putting a regex which should end with a space.
I would recommend you use the "split" functionality for String with spaces as separators, then go through these elements one by one and make a direct comparison.
I would suggest using this regex pattern1 = ".\bt\b." instead of pattern2 = ".?\bt\b.?" . Pattern1 will help you to match the complete String if 't' occurs in that string rather than the pattern2 which just reaches the string "t" you are searching for and ignores rest of the string. There is not much difference in two approaches and for your particular use case of returning true/false will run fine both the ways. The one I suggested will help you to improvise the regex in case you make further changes in your use case
Related
How can I delete everything after first empty space in a string which user selects? I was reading this how to remove some words from a string in java. Can this help me in my case?
You can use replaceAll with a regex \s.* which match every thing after space:
String str = "Hello java word!";
str = str.replaceAll("\\s.*", "");
output
Hello
regex demo
Like #Coffeehouse Coder mention in comment, This solution will replace every thing if the input start with space, so if you want to avoid this case, you can trim your input using string.trim() so it can remove the spaces in start and in end.
Assuming that there is no space in the beginning of the string.
Follow these steps-
Split the string at space. It will create an array.
Get the first element of that array.
Hope this helps.
str = "Example string"
String[] _arr = str.split("\\s");
String word = _arr[0];
You need to consider multiple white spaces and space in the beginning before considering the above code.
I am not native to JAVA Programming but have an idea that it has split function for string.
And the reference you cited in the question is bit complex, while you can achieve the desired thing very easily.
P.S. In future if you make a mind to get two words or three, splitting method is better (assuming you have already dealt with multiple white-spaces) else substring is better.
A simple way to do it can be:
System.out.println("Hello world!".split(" ")[0]);
// Taking 'str' as your string
// To remove the first space(s) of the string,
str = str.trim();
int index = str.indexOf(" ");
String word = str.substring(0, index);
This is just one method of many.
str = str.replaceAll("\\s+", " "); // This replaces one or more spaces with one space
String[] words = str.split("\\s");
String first = words[0];
The simplest solution in my opinion would be to just locate the index which the user wants it to be cut off at and then call the substring() method from 0 to the index they wanted. Set that = to a new string and you have the string they want.
If you want to replace the string then just set the original string = to the result of the substring() method.
Link to substring() method: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int)
There are already 5 perfectly good answers, so let me add a sixth one. Variety is the spice of life!
private static final Pattern FIRST_WORD = Pattern.compile("\\S+");
public static String firstWord(CharSequence text) {
Matcher m = FIRST_WORD.matcher(text);
return m.find() ? m.group() : "";
}
Advantages over the .split(...)[0]-type answers:
It directly does exactly what is being asked, i.e. "Find the first sequence of non-space characters." So the self-documentation is more explicit.
It is more efficient when called on multiple strings (e.g. for batch processing a large list of strings) because the regular expression is compiled only once.
It is more space-efficient because it avoids unnecessarily creating a whole array with references to each word when we only need the first.
It works without having to trim the string.
(I know this is probably too late to be of any use to the OP but I'm leaving it here as an alternative solution for future readers.)
This would be more efficient
String str = "Hello world!";
int spaceInd = str.indexOf(' ');
if(spaceInd != -1) {
str = str.substring(0, spaceInd);
}
System.out.println(String.format("[%s]", str));
I want to replace words in a string, but I am having little difficulties. Here is what I want to do. I have string:
String a = "I want to replace some words in this string";
It should work like some kind of a translator. I am doing this with String.replaceAll(), but it doesn't work completely because of this. Let's say I am translating from English to German, than this should be the output (Ich means I in German).
String toTranslate = "I";
String translated = "Ich";
a = a.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
Now the output of the String a will be this:
"ich want to replace some words ich**n** **th**ich**s** **str**ich**ng**"
How to replace just the words, not the subwords in the words?
replaceAll uses regex, so you may add word boundaries or look-around mechanisms to check if there are no non-space characters surrounding word you want to replace.
String toTranslate = "I";
String translated = "Ich";
a = a.replaceAll("(?<!\\S)"+toTranslate.toLowerCase()+"(?!\\S)", translated.toLowerCase());
You can also add quotation mechanism to escape any regex metacharacters like + * ( inside word you want to replace. BTW you don't need to change your string to lower case, simply add case-insensitive flag to regex (?i).
a = a.replaceAll("(?i)(?<!\\S)"+Pattern.quote(toTranslate)+"(?!\\S)", translated.toLowerCase());
Use split(" ") for getting each word in the sentence. And then use replaceAll on each word.
String a = "I want to replace some words in this string";
String toTranslate = "I";
String translated = "Ich";
String newString[]=a.split(" ");
for (String string : newString) {
string=string.replaceAll(toTranslate, toTranslate.toLowerCase());//Adding this line ensures you dont miss any uppercase toTranslate
string=string.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
System.out.println("after translation ="+string);
}
String toTranslate = "I ";
String translated = "Ich ";
a = a.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
If you add a space after the "I" it should replace it when it comes to the word "Ich" but if your word ends in a "I" then thats another problem
If you assume that I will always be capitalized in English as it should be then
a = a.replaceAll(toTranslate, translated);
will work, otherwise you need to replace both cases
a = a.replaceAll(toTranslate, translated);
a = a.replaceAll("([^a-zA-Z])("+toTranslate.toLowerCase()+")([^a-zA-Z])", "$1"+translated.toLowerCase()+"$3");
Here is a working example
Yes, the word boundaries are the solution. I just did this in the regex:
text.replaceAll("\\b" + parts1[i] + "\\b", map.element.value);
Don't be confused with the second argument it's string (from Hash table).
You can use RegEx's word bound, which is \b
String toTranslate = "\\bI\\b";
String translated = "Ich";
a = a.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
This should ensure I is separated entirely into its own word
Edit: I misread the question and realized you want whole words. See above, as I have accounted for that
I have a String that I have to parse for different keywords.
For example, I have the String:
"I will come and meet you at the 123woods"
And my keywords are
'123woods'
'woods'
I should report whenever I have a match and where. Multiple occurrences should also be accounted for.
However, for this one, I should get a match only on '123woods', not on 'woods'. This eliminates using String.contains() method. Also, I should be able to have a list/set of keywords and check at the same time for their occurrence. In this example, if I have '123woods' and 'come', I should get two occurrences. Method execution should be somewhat fast on large texts.
My idea is to use StringTokenizer but I am unsure if it will perform well. Any suggestions?
The example below is based on your comments. It uses a List of keywords, which will be searched in a given String using word boundaries. It uses StringUtils from Apache Commons Lang to build the regular expression and print the matched groups.
String text = "I will come and meet you at the woods 123woods and all the woods";
List<String> tokens = new ArrayList<String>();
tokens.add("123woods");
tokens.add("woods");
String patternString = "\\b(" + StringUtils.join(tokens, "|") + ")\\b";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
If you are looking for more performance, you could have a look at StringSearch: high-performance pattern matching algorithms in Java.
Use regex + word boundaries as others answered.
"I will come and meet you at the 123woods".matches(".*\\b123woods\\b.*");
will be true.
"I will come and meet you at the 123woods".matches(".*\\bwoods\\b.*");
will be false.
Hope this works for you:
String string = "I will come and meet you at the 123woods";
String keyword = "123woods";
Boolean found = Arrays.asList(string.split(" ")).contains(keyword);
if(found){
System.out.println("Keyword matched the string");
}
http://codigounico.blogspot.com/
How about something like Arrays.asList(String.split(" ")).contains("xx")?
See String.split() and How can I test if an array contains a certain value.
Got a way to match Exact word from String in Android:
String full = "Hello World. How are you ?";
String one = "Hell";
String two = "Hello";
String three = "are";
String four = "ar";
boolean is1 = isContainExactWord(full, one);
boolean is2 = isContainExactWord(full, two);
boolean is3 = isContainExactWord(full, three);
boolean is4 = isContainExactWord(full, four);
Log.i("Contains Result", is1+"-"+is2+"-"+is3+"-"+is4);
Result: false-true-true-false
Function for match word:
private boolean isContainExactWord(String fullString, String partWord){
String pattern = "\\b"+partWord+"\\b";
Pattern p=Pattern.compile(pattern);
Matcher m=p.matcher(fullString);
return m.find();
}
Done
public class FindTextInLine {
String match = "123woods";
String text = "I will come and meet you at the 123woods";
public void findText () {
if (text.contains(match)) {
System.out.println("Keyword matched the string" );
}
}
}
Try to match using regular expressions. Match for "\b123wood\b", \b is a word break.
The solution seems to be long accepted, but the solution could be improved, so if someone has a similar problem:
This is a classical application for multi-pattern-search-algorithms.
Java Pattern Search (with Matcher.find) is not qualified for doing that. Searching for exactly one keyword is optimized in java, searching for an or-expression uses the regex non deterministic automaton which is backtracking on mismatches. In worse case each character of the text will be processed l times (where l is the sum of the pattern lengths).
Single pattern search is better, but not qualified, too. One will have to start the whole search for every keyword pattern. In worse case each character of the text will be processed p times where p is the number of patterns.
Multi pattern search will process each character of the text exactly once. Algorithms suitable for such a search would be Aho-Corasick, Wu-Manber, or Set Backwards Oracle Matching. These could be found in libraries like Stringsearchalgorithms or byteseek.
// example with StringSearchAlgorithms
AhoCorasick stringSearch = new AhoCorasick(asList("123woods", "woods"));
CharProvider text = new StringCharProvider("I will come and meet you at the woods 123woods and all the woods", 0);
StringFinder finder = stringSearch.createFinder(text);
List<StringMatch> all = finder.findAll();
A much simpler way to do this is to use split():
String match = "123woods";
String text = "I will come and meet you at the 123woods";
String[] sentence = text.split();
for(String word: sentence)
{
if(word.equals(match))
return true;
}
return false;
This is a simpler, less elegant way to do the same thing without using tokens, etc.
You can use regular expressions.
Use Matcher and Pattern methods to get the desired output
You can also use regex matching with the \b flag (whole word boundary).
To Match "123woods" instead of "woods" , use atomic grouping in regular expresssion.
One thing to be noted is that, in a string to match "123woods" alone , it will match the first "123woods" and exits instead of searching the same string further.
\b(?>123woods|woods)\b
it searches 123woods as primary search, once it got matched it exits the search.
Looking back at the original question, we need to find some given keywords in a given sentence, count the number of occurrences and know something about where. I don't quite understand what "where" means (is it an index in the sentence?), so I'll pass that one... I'm still learning java, one step at a time, so I'll see to that one in due time :-)
It must be noticed that common sentences (as the one in the original question) can have repeated keywords, therefore the search cannot just ask if a given keyword "exists or not" and count it as 1 if it does exist. There can be more then one of the same. For example:
// Base sentence (added punctuation, to make it more interesting):
String sentence = "Say that 123 of us will come by and meet you, "
+ "say, at the woods of 123woods.";
// Split it (punctuation taken in consideration, as well):
java.util.List<String> strings =
java.util.Arrays.asList(sentence.split(" |,|\\."));
// My keywords:
java.util.ArrayList<String> keywords = new java.util.ArrayList<>();
keywords.add("123woods");
keywords.add("come");
keywords.add("you");
keywords.add("say");
By looking at it, the expected result would be 5 for "Say" + "come" + "you" + "say" + "123woods", counting "say" twice if we go lowercase. If we don't, then the count should be 4, "Say" being excluded and "say" included. Fine. My suggestion is:
// Set... ready...?
int counter = 0;
// Go!
for(String s : strings)
{
// Asking if the sentence exists in the keywords, not the other
// around, to find repeated keywords in the sentence.
Boolean found = keywords.contains(s.toLowerCase());
if(found)
{
counter ++;
System.out.println("Found: " + s);
}
}
// Statistics:
if (counter > 0)
{
System.out.println("In sentence: " + sentence + "\n"
+ "Count: " + counter);
}
And the results are:
Found: Say
Found: come
Found: you
Found: say
Found: 123woods
In sentence: Say that 123 of us will come by and meet you, say, at the woods of 123woods.
Count: 5
If you want to identify a whole word in a string and change the content of that word you can do this way. Your final string stays equals, except the word you treated. In this case "not" stays "'not'" in final string.
StringBuilder sb = new StringBuilder();
String[] splited = value.split("\\s+");
if(ArrayUtils.isNotEmpty(splited)) {
for(String valor : splited) {
sb.append(" ");
if("not".equals(valor.toLowerCase())) {
sb.append("'").append(valor).append("'");
} else {
sb.append(valor);
}
}
}
return sb.toString();
I'm trying to perform some super simple parsing o log files, so I'm using String.split method like this:
String [] parts = input.split(",");
And works great for input like:
a,b,c
Or
type=simple, output=Hello, repeat=true
Just to say something.
How can I escape the comma, so it doesn't match intermediate commas?
For instance, if I want to include a comma in one of the parts:
type=simple, output=Hello, world, repeate=true
I was thinking in something like:
type=simple, output=Hello\, world, repeate=true
But I don't know how to create the split to avoid matching the comma.
I've tried:
String [] parts = input.split("[^\,],");
But, well, is not working.
You can solve it using a negative look behind.
String[] parts = str.split("(?<!\\\\), ");
Basically it says, split on each ", " that is not preceeded by a backslash.
String str = "type=simple, output=Hello\\, world, repeate=true";
String[] parts = str.split("(?<!\\\\), ");
for (String s : parts)
System.out.println(s);
Output:
type=simple
output=Hello\, world
repeate=true
(ideone.com link)
If you happen to be stuck with the non-escaped comma-separated values, you could do the following (similar) hack:
String[] parts = str.split(", (?=\\w+=)");
Which says split on each ", " which is followed by some word-characters and an =
(ideone.com link)
I'm afraid, there's no perfect solution for String.split. Using a matcher for the three parts would work. In case the number of parts is not constant, I'd recommend a loop with matcher.find. Something like this maybe
final String s = "type=simple, output=Hello, world, repeat=true";
final Pattern p = Pattern.compile("((?:[^\\\\,]|\\\\.)*)(?:,|$)");
final Matcher m = p.matcher(s);
while (m.find()) System.out.println(m.group(1));
You'll probably want to skip the spaces after the comma as well:
final Pattern p = Pattern.compile("((?:[^\\\\,]|\\\\.)*)(?:,\\s*|$)");
It's not really complicated, just note that you need four backslashes in order to match one.
Escaping works with the opposite of aioobe's answer (updated: aioobe now uses the same construct but I didn't know that when I wrote this), negative lookbehind
final String s = "type=simple, output=Hello\\, world, repeate=true";
final String[] tokens = s.split("(?<!\\\\),\\s*");
for(final String item : tokens){
System.out.println("'" + item.replace("\\,", ",") + "'");
}
Output:
'type=simple'
'output=Hello, world'
'repeate=true'
Reference:
Pattern: Special Constructs
I think
input.split("[^\\\\],");
should work. It will split at all commas that are not preceeded with a backslash.
BTW if you are working with Eclipse, I can recommend the QuickRex Plugin to test and debug Regexes.
I am not a beginner to regular expressions, but their use in perl seems a bit different than in Java.
Anyways, I basically have a dictionary of shorthand words and their definitions. I want to iterate over words in the dictionary and replace them with their meanings. what is the best way to do this in JAVA?
I have seen String.replaceAll(), String.replace(), as well as the Pattern/Matcher classes. I wish to do a case insensitive replacement along the lines of:
word =~ s/\s?\Q$short_word\E\s?/ \Q$short_def\E /sig
While I am at it, do you think that it is best to extract all the words from the string and then apply my dictionary or just apply the dictionary to the string? I know that I need to be careful, because the shorthand words could match parts of other shorthand meanings.
Hopefully this all makes sense.
Thanks.
Clarification:
Dictionary is something like:
lol:laugh out loud, rofl:rolling on the floor laughing, ll:like lemons
string is:
lol, i am rofl
replaced text:
laugh out loud, i am rolling on the floor laughing
notice how the ll wasnt added anywhere
The danger is false positives inside of normal words. "fell" != "felikes lemons"
One way is to split the words on whitespace (do multiple spaces need to be conserved?) then loop over the List performing the 'if contains() { replace } else { output original } idea above.
My output class would be a StringBuffer
StringBuffer outputBuffer = new StringBuffer();
for(String s: split(inputText)) {
outputBuffer.append( dictionary.contains(s) ? dictionary.get(s) : s);
}
Make your split method smart enough to return word delimiters also:
split("now is the time") -> now,<space>,is,<space>,the,<space><space>,time
Then you don't have to worry about conserving white space - the loop above will just append anything that isn't a dictionary word to the StringBuffer.
Here's a recent SO thread on retaining delimiters when regexing.
If you insist on using regex, this would work (taking Zoltan Balazs' dictionary map approach):
Map<String, String> substitutions = loadDictionaryFromSomewhere();
int lengthOfShortestKeyInMap = 3; //Calculate
int lengthOfLongestKeyInMap = 3; //Calculate
StringBuffer output = new StringBuffer(input.length());
Pattern pattern = Pattern.compile("\\b(\\w{" + lengthOfShortestKeyInMap + "," + lengthOfLongestKeyInMap + "})\\b");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
String candidate = matcher.group(1);
String substitute = substitutions.get(candidate);
if (substitute == null)
substitute = candidate; // no match, use original
matcher.appendReplacement(output, Matcher.quoteReplacement(substitute));
}
matcher.appendTail(output);
// output now contains the text with substituted words
If you plan to process many inputs, pre-compiling the pattern is more efficient than using String.split(), which compiles a new Pattern each call.
(edit) Compiling all of the keys into a single pattern yields a more efficient approach, like so:
Pattern pattern = Pattern.compile("\\b(lol|rtfm|rofl|wtf)\\b");
// rest of the method unchanged, don't need the shortest/longest key stuff
This allows the regex engine to skip over any words that happen to be short enough but aren't in the list, saving you a lot of map accesses.
The first thing, that comes into my mind is this:
...
// eg: lol -> laugh out loud
Map<String, String> dictionatry;
ArrayList<String> originalText;
ArrayList<String> replacedText;
for(String string : originalText) {
if(dictionary.contains(string)) {
replacedText.add(dictionary.get(string));
} else {
replacedText.add(string);
}
...
Or you could use a StringBuffer instead of the replacedText.