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counting number of occurrences of words in a text java

So I'm building a TreeMap from scratch and I'm trying to count the number of occurrences of every word in a text using Java. The text is read from a text file, but I can easily read it from there. I really don't know how to count every word, can someone help?
Imagine the text is something like:
Over time, computer engineers take advantage of each other's work and invent algorithms for new things. Algorithms combine with other algorithms to utilize the results of other algorithms, in turn producing results for even more algorithms.
Output:
Over 1
time 1
computer 1
algotitms 5
...
If possible I want to ignore if it's upper or lower case, I want to count them both together.
EDIT: I don't want to use any sort of Map (hashMap i.e.) or something similiar to do this.

Break down the problem as follows (this is one potential solution - not THE solution):
Split the text into words (create list or array or words).
Remove punctuation marks.
Create your map to collect results.
Iterate over your list of words and add "1" to the value of each encountered key
Display results (Iterate over the map's EntrySet)
Split the text into words
My preference is to split words by using space as a delimiter. The reason being is that, if you split using non-word characters, you may missed on some hyphenated words. I know that the use of hyphenation is being reduced, there are still plenty of words that fall under this rule; for example, middle-aged. If a word such as this is encountered, it MIGHT have to be treated as one word and not two.
Remove punctuation marks
Because of the decision above, you will need to first remove punctuation characters that might attached to your words. Keep in mind that if you use a regular expression to split the words, you might be able to accomplish this step at the same time you are doing the step above. In fact, that would be preferred so that you don't have to iterate over twice. Do both of these in a single pass. While you at it, call toLowerCase() on the input string to eliminate the ambiguity between capitalized words and lowercase words.
Create your map to collect results
This is where you are going to collect your count. Using the TreeMap implementation of the Java Map. One thing to be aware about this particular implementation is that the map is sorted according to the natural ordering of its keys. In this case, since the keys are the words from the inputted text, the keys will be arranged in alphabetical order, not by the magnitude of the count. IF sorting the entries by count is important, there is a technique where you can "reverse" the map and make the values the keys and the keys to values. However, since two or more words could have the same count, you will need to create a new map of <Integer, Set>, so that you can group together words with the same count.
Iterate over your list of words
At this point, you should have a list of words and a map structure to collect the count. Using a lambda expression, you should be able to perform a count() or your words very easily. But, if you are not familiarized or comfortable with Lambda expressions, you can use a regular looping structure to iterate over your list, do a containsKey() check to see if the word was encountered before, get() the value if the map already contains the word, and then add "1" to the previous value. Lastly, put() the new count in the map.
Display results
Again, you can use a Lambda Expression to print out the EntrySet key value pairs or simply iterate over the entry set to display the results.
Based on all of the above points, a potential solution should look like this (not using Lambda for the OPs sake)
public static void main(String[] args) {
String text = "Over time, computer engineers take advantage of each other's work and invent algorithms for new things. Algorithms combine with other algorithms to utilize the results of other algorithms, in turn producing results for even more algorithms.";
text = text.replaceAll("\\p{P}", ""); // replace all punctuations
text = text.toLowerCase(); // turn all words into lowercase
String[] wordArr = text.split(" "); // create list of words
Map<String, Integer> wordCount = new TreeMap<>();
// Collect the word count
for (String word : wordArr) {
if(!wordCount.containsKey(word)){
wordCount.put(word, 1);
} else {
int count = wordCount.get(word);
wordCount.put(word, count + 1);
}
}
Iterator<Entry<String, Integer>> iter = wordCount.entrySet().iterator();
System.out.println("Output: ");
while(iter.hasNext()) {
Entry<String, Integer> entry = iter.next();
System.out.println(entry.getKey() + ": " + entry.getValue());
}
}
This produces the following output
Output:
advantage: 1
algorithms: 5
and: 1
combine: 1
computer: 1
each: 1
engineers: 1
even: 1
for: 2
in: 1
invent: 1
more: 1
new: 1
of: 2
other: 2
others: 1
over: 1
producing: 1
results: 2
take: 1
the: 1
things: 1
time: 1
to: 1
turn: 1
utilize: 1
with: 1
work: 1
Why did I break down the problem like this for such mundane task? Simple. I believe each of those discrete steps should be extracted into functions to improve code reusability. Yes, it is cool to use a Lambda expression to do everything at once and make your code look much simplified. But what if you need to some intermediate step over and over? Most of the time, code is duplicated to accomplish this. In reality, often a better solution is to break these tasks into methods. Some of these tasks, like transforming the input text, can be done in a single method since that activity seems to be related in nature. (There is such a thing as a method doing "too little.")
public String[] createWordList(String text) {
return text.replaceAll("\\p{P}", "").toLowerCase().split(" ");
}
public Map<String, Integer> createWordCountMap(String[] wordArr) {
Map<String, Integer> wordCountMap = new TreeMap<>();
for (String word : wordArr) {
if(!wordCountMap.containsKey(word)){
wordCountMap.put(word, 1);
} else {
int count = wordCountMap.get(word);
wordCountMap.put(word, count + 1);
}
}
return wordCountMap;
}
String void displayCount(Map<String, Integer> wordCountMap) {
Iterator<Entry<String, Integer>> iter = wordCountMap.entrySet().iterator();
while(iter.hasNext()) {
Entry<String, Integer> entry = iter.next();
System.out.println(entry.getKey() + ": " + entry.getValue());
}
}
Now, after doing that, your main method looks more readable and your code is more reusable.
public static void main(String[] args) {
WordCount wc = new WordCount();
String text = "...";
String[] wordArr = wc.createWordList(text);
Map<String, Integer> wordCountMap = wc.createWordCountMap(wordArr);
wc.displayCount(wordCountMap);
}
UPDATE:
One small detail I forgot to mention is that, if instead of a TreeMap a HashMap is used, the output will come sorted by count value in descending order. This is because the hashing function will use value of the entry as the hash. Therefore, you won't need to "reverse" the map for this purpose. So, after switching to HashMap, the output should be as follows:
Output:
algorithms: 5
other: 2
for: 2
turn: 1
computer: 1
producing: 1
...

my suggestion is to use regexp and split and stream with grouping example 3
EX1 this solution does not use a collection LIST/MAP only array for me it is not optimal
#Test
public void testApp2() {
final String text = "Over time, computer engineers take advantage of each other's work and invent algorithms for new things. Algorithms combine with other algorithms to utilize the results of other algorithms, in turn producing results for even more algorithms.";
final String lowerText = text.toLowerCase();
final String[] split = lowerText.split("\\W+");
System.out.println("Output: ");
for (String s : split) {
if (s == null) {
continue;
}
int count = 0;
for (int i = 0; i < split.length; i++) {
final boolean sameWorld = s.equals(split[i]);
if (sameWorld) {
count = count + 1;
split[i] = null;
}
}
System.out.println(s + " " + count);
}
}
EX2 I think that's what you mean, but I'm not sure if I used too much for the list
#Test
public void testApp() {
final String text = "Over time, computer engineers take advantage of each other's work and invent algorithms for new things. Algorithms combine with other algorithms to utilize the results of other algorithms, in turn producing results for even more algorithms.";
final String[] split = text.split("\\W+");
final List<String> list = new ArrayList<>();
System.out.println("Output: ");
for (String s : split) {
if(!list.contains(s)){
list.add(s.toUpperCase());
final long count = Arrays.stream(split).filter(s::equalsIgnoreCase).count();
System.out.println(s+" "+count);
}
}
}
EX3 below is a test for your example but use MAP
#Test
public void test() {
final String text = "Over time, computer engineers take advantage of each other's work and invent algorithms for new things. Algorithms combine with other algorithms to utilize the results of other algorithms, in turn producing results for even more algorithms.";
Map<String, Long> result = Arrays.stream(text.split("\\W+")).collect(Collectors.groupingBy(String::toLowerCase, Collectors.counting()));
assertEquals(result.get("algorithms"), new Long(5));
System.out.println("Output: ");
result.entrySet().stream().forEach(x -> System.out.println(x.getKey() + " " + x.getValue()));
}

Looping through an ArrayList with another Arraylist in Java

I have a large array list of sentences and another array list of words.
My program loops through the array list and removes an element from that array list if the sentence contains any of the words from the other.
The sentences array list can be very large and I coded a quick and dirty nested for loop. While this works for when there are not many sentences, in cases where their are, the time it takes to finish this operation is ridiculously long.
for (int i = 0; i < SENTENCES.size(); i++) {
for (int k = 0; k < WORDS.size(); k++) {
if (SENTENCES.get(i).contains(" " + WORDS.get(k) + " ") == true) {
//Do something
}
}
}
Is there a more efficient way of doing this then a nested for loop?

There's a few inefficiencies in your code, but at the end of the day, if you've got to search for sentences containing words then there's no getting away from loops.
That said, there are couple of things to try.
First, make WORDS a HashSet, the contains method will be far quicker than for an ArrayList because it's doing a hash look-up to get the value.
Second, switch the logic about a bit like this:
Iterator<String> sentenceIterator = SENTENCES.iterator();
sentenceLoop:
while (sentenceIterator.hasNext())
{
String sentence = sentenceIterator.next();
for (String word : sentence.replaceAll("\\p{P}", " ").toLowerCase().split("\\s+"))
{
if (WORDS.contains(word))
{
sentenceIterator.remove();
continue sentenceLoop;
}
}
}
This code (which assumes you're trying to remove sentences that contain certain words) uses Iterators and avoids the string concatenation and parsing logic you had in your original code (replacing it with a single regex) both of which should be quicker.
But bear in mind, as with all things performance you'll need to test these changes to see they improve the situation.

I̶ ̶w̶o̶u̶l̶d̶ ̶s̶a̶y̶ ̶n̶o̶,̶ ̶b̶u̶t̶ what you must change is the way you handle the removal of the data. This is noted by this part of the explanation of your problem:
The sentences array list can be very large (...). While this works for when there are not many sentences, in cases where their are, the time it takes to finish this operation is ridiculously long.
The cause of this is that removal time in ArrayList takes O(N), and since you're doing this inside a loop, then it will take at least O(N^2).
I recommend using LinkedList rather than ArrayList to store the sentences, and use Iterator rather than your naive List#get since it already offers Iterator#remove in time O(1) for LinkedList.
In case you cannot change the design to LinkedList, I recommend storing the sentences that are valid in a new List, and in the end replace the contents of your original List with this new List, thus saving lot of time.
Apart from this big improvement, you can improve the algorithm even more by using a Set to store the words to lookup rather than using another List since the lookup in a Set is O(1).

What you could do is put all your words into a HashSet. This allows you to check if a word is in the set very quickly. See https://docs.oracle.com/javase/8/docs/api/java/util/HashSet.html for documentation.
HashSet<String> wordSet = new HashSet();
for (String word : WORDS) {
wordSet.add(word);
}
Then it's just a matter of splitting each sentence into the words that make it up, and checking if any of those words are in the set.
for (String sentence : SENTENCES) {
String[] sentenceWords = sentence.split(" "); // You probably want to use a regex here instead of just splitting on a " ", but this is just an example.
for (String word : sentenceWords) {
if (wordSet.contains(word)) {
// The sentence contains one of the special words.
// DO SOMETHING
break;
}
}
}

I will create a set of words from second ArrayList:
Set<String> listOfWords = new HashSet<String>();
listOfWords.add("one");
listOfWords.add("two");
I will then iterate over the set and the first ArrayList and use Contains:
for (String word : listOfWords) {
for(String sentence : Sentences) {
if (sentence.contains(word)) {
// do something
}
}
}
Also, if you are free to use any open source jar, check this out:
searching string in another string

First, your program has a bug: it would not count words at the beginning and at the end of a sentence.
Your current program has runtime complexity of O(s*w), where s is the length, in characters, of all sentences, and w is the length of all words, also in characters.
If words is relatively small (a few hundred items or so) you could use regex to speed things up considerably: construct a pattern like this, and use it in a loop:
StringBuilder regex = new StringBuilder();
boolean first = true;
// Let's say WORDS={"quick", "brown", "fox"}
regex.append("\\b(?:");
for (String w : WORDS) {
if (!first) {
regex.append('|');
} else {
first = false;
}
regex.append(w);
}
regex.append(")\\b");
// Now regex is "\b(?:quick|brown|fox)\b", i.e. your list of words
// separated by OR signs, enclosed in non-capturing groups
// anchored to word boundaries by '\b's on both sides.
Pattern p = Pattern.compile(regex.toString());
for (int i = 0; i < SENTENCES.size(); i++) {
if (p.matcher(SENTENCES.get(i)).find()) {
// Do something
}
}
Since regex gets pre-compiled into a structure more suitable for fast searches, your program would run in O(s*max(w)), where s is the length, in characters, of all sentences, and w is the length of the longest word. Given that the number of words in your collection is about 200 or 300, this could give you an order of magnitude decrease in running time.

If you have enough memory you can tokenize SENTENCES and put them in a Set. Then it would be better in performance and also more correct than current implementation.

Well, looking at your code I would suggest two things that will improve the performance from each iteration:
Remove " == true". The contains operation already returns a boolean, so it is enough for the if, comparing it with true adds one extra operation for each iteration that is not needed.
Do not concatenate Strings inside a loop (" " + WORDS.get(k) + " ") as it is a quite expensive operation because + operator creates new objects. Better use a string buffer / builder and clear it after each iteration with stringBuffer.setLength(0);.
Besides that, for this case I do not know any other approach, maybe you can use regular expressions if you can abstract a pattern out of those words you want to remove and have then only one loop.
Hope it helps!

If you concern about the efficiency, I think that the most effective way to do this is to use Aho-Corasick's algorithm. While you have 2 nested loops here and a contains() method (that I think takes at the best length of sentence + length of word time), Aho-Corasick gives you one loop over sentences and for checking of containing words it takes length of sentence, which is length of word times faster (+ a preprocessing time for creation of finite state machine, which is relatively small).

I'll approach this in more theoretical view.. If you don't have memory limitation, you can try to mimic the logic in counting sort
say M1 = sentences.size, M2 = number of word per sentences, and N = word.size
Assume all sentences has the same number of words just for simplicity
your current approach's complexity is O(M1.M2.N)
We can create a mapping of words - position in sentences.
Loop through your arraylist of sentences, and change them into two dimensional jagged array of words. Loop through the new array, create a HashMap where key,value = words, arraylist of word position (say with length X). That's O(2M1.M2.X) = O(M1.M2.X)
Then loop through your words arraylist, access your word hashmap, loop through the list of word position. remove each one. That's O(N.X)
Say you're need to give the result in arraylist of string, we need another loop and concat everything. That's O(M1.M2)
Total complexity is O(M1.M2.X) + O(N.X) + O(M1.M2)
assumming X is way smaller than N, you'll probably get better performance

Java - Search keywords list in another string list

I have a list of keywords in a List and I have data coming from some source which will be a list too.
I would like to find if any of keywords exists in the data list, if yes add those keywords to another target list.
E.g.
Keywords list = FIRSTNAME, LASTNAME, CURRENCY & FUND
Data list = HUSBANDFIRSTNAME, HUSBANDLASTNAME, WIFEFIRSTNAME, SOURCECURRENCY & CURRENCYRATE.
From above example, I would like to make a target list with keywords FIRSTNAME, LASTNAME & CURRENCY, however FUND should not come as it doesn't exists in the data list.
I have a solution below that works by using two for loops (one inside another) and check with String contains method, but I would like to avoid two loops, especially one inside another.
for (int i=0; i<dataList.size();i++) {
for (int j=0; j<keywordsList.size();j++) {
if (dataList.get(i).contains(keywordsList.get(j))) {
targetSet.add(keywordsList.get(j));
break;
}
}
}
Is there any other alternate solution for my problem?

Here's a one loop approach using regex. You construct a pattern using your keywords, and then iterate through your dataList and see if you can find a match.
public static void main(String[] args) throws Exception {
List<String> keywords = new ArrayList(Arrays.asList("FIRSTNAME", "LASTNAME", "CURRENCY", "FUND"));
List<String> dataList = new ArrayList(Arrays.asList("HUSBANDFIRSTNAME", "HUSBANDLASTNAME", "WIFEFIRSTNAME", "SOURCECURRENCY", "CURRENCYRATE"));
Set<String> targetSet = new HashSet();
String pattern = String.join("|", keywords);
for (String data : dataList) {
Matcher matcher = Pattern.compile(pattern).matcher(data);
if (matcher.find()) {
targetSet.add(matcher.group());
}
}
System.out.println(targetSet);
}
Results:
[CURRENCY, LASTNAME, FIRSTNAME]

Try Aho–Corasick algorithm. This algorithm can get the count of appearance of every keyword in the data (You just need whether it appeared or not).
The Complexity is O(Sum(Length(Keyword)) + Length(Data) + Count(number of match)).
Here is the wiki-page:
In computer science, the Aho–Corasick algorithm is a string searching
algorithm invented by Alfred V. Aho and Margaret J. Corasick. It is
a kind of dictionary-matching algorithm that locates elements of a
finite set of strings (the "dictionary") within an input text. It
matches all patterns simultaneously. The complexity of the algorithm
is linear in the length of the patterns plus the length of the
searched text plus the number of output matches.
I implemented it(about 200 lines) years ago for similar case, and it works well.
If you just care keyword appeared or not, you can modify that algorithm for your case with a better complexity:
O(Sum(Length(Keyword)) + Length(Data)).
You can find implementation of that algorithm from internet everywhere but I think it's good for you to understand that algorithm and implement it by yourself.
EDIT:
I think you want to eliminate two-loops, so we need find all keywords in one loop. We call it Set Match Problem that a set of patterns(keywords) to match a text(data). You want to solve Set Match Problem, then you should choose Aho–Corasick algorithm which is particularly designed for that case. In that way, we will get one loop solution:
for (int i=0; i < dataList.size(); i++) {
targetSet.addAll(Ac.run(keywordsList));
}
You can find a implementation from here.

Java - Search performantly for subset of String in String list

I want to search through a list of Strings and return the values, which contains which contain the search string.
The list could look like this (can have up to 1000 entries). Although it is not guranteed that it is always letters and then a digit. It could be digits only, words only or even both mixed up:
entry 1
entry 2
entry 3
entry 4
test 1
test 2
test 3
tst 4
If the user does search for 1, these should be returned:
entry 1
test 1
The situation is that the user has a search bar and can enter a search string. This search string is used to search through the list.
How can this be done performantly?
Currently, I have got:
for (String s : strings) {
if (s.contains(searchedText)) result.add(s);
}
It is O(N) and really slow. Especially if the user types many characters at a time.

Maybe I don't understand your question, but as you know n Java, String objects are immutable, but also can represent collection(array) of chars. So one thing what you can do is to perform search with better algorithms as binary_search, Aho-Corasick, Rabin–Karp, Boyer–Moore string search, StringSearch or one of these. Also you may consider some usage of Abstract_data_types with better performance (hashing, trees etc.).

This is very simple if you use streams:
final List<String> items = Arrays.asList("entry 1", "entry 2", "entry 3", "test 1", "test 2", "test 3");
final String searchString = "1";
final List<String> results = items.parallelStream() // work in parallel
.filter(s -> s.contains(searchString)) // pick out items that match
.collect(Collectors.toList()); // and turn those into a result list
results.forEach(System.out::println);
Notice the parallelStream() which will cause the list to be filtered and traversed using all available CPUs.
In your case you can use the results when the user expands the search term (while typing) to reduce the amount of items that need to be filtered, because if 's' matches all items in result, all those that match 'se' will be a sub-list of result.

If you don't use any additional structures, you cannot perform faster, than look though your data. That takes O(N).
If you can do some preparations, like building text index, you can increase performance of search. General information: http://en.wikipedia.org/wiki/Full_text_search. If you can make some assumptions about your data (like the last symbol is number and you are going to search only by it), it'll be easy to create such index.

Depending on the upper limit of the number in the string and if you have no concerns about space, use an Array of ArrayLists where the array index is the number of the string:
ArrayList<String>[] data = new ArrayList<String>[1000];
for ( int i = 0; i < 1000; i++ )
data[i] = new ArrayList<String>();
//inserting data
int num = Integer.parseInt(datastring.substring(datastring.length-1));
data[i].add(datastring);
//getting all data that has a 1
for ( String s: data[1] )
result.add(s);
Using a Hashmap would overwrite previous mapped values when trying to put new values into it.
i.e. if 1 maps to entry, then you try to add 1 mapping to test, the entry would get replaced with test.
As another idea, you could just keep a count of the number of strings with each number, so when you're searching, you know how many to look for, so as soon as you find all of them, you stop searching:
int[] str_count = new int[1000];
for ( int i = 0; i < 1000; i++ )
str_count[i] = 0;
//when storing data into the list:
int num = Integer.parseInt(datastring.substring(datastring.length-1));
str_count[i]++;
//when searching the list for 1s:
int count = str_count[1];
for (String s : strings) {
if (s.contains(searchedText))
result.add(s);
if (result.size() == count)
break;
}
While the first idea would be much faster, it would take up more space. Yet, the second idea takes up less space, the worst case scenario would search O(N) still.

How can i extract specific terms from string lines in Java?

I have a serious problem with extracting terms from each string line. To be more specific, I have one csv formatted file which is actually not csv format (it saves all terms into line[0] only)
So, here's just example string line among thousands of string lines:
(split() doesn't work.!!! )
test.csv
"31451 CID005319044 　　15939353　　 C8H14O3S2 　　　beta-lipoic acid　　 C1C[S#](=O)S[C##H]1CCCCC(=O)O "
"12232 COD05374044 23439353　　C924O3S2 　　　saponin　　 CCCC(=O)O "
"9048 　 CTD042032　23241　　C3HO4O3S2　Berberine　 [C##H]1CCCCC(=O)O "
I want to extract "beta-lipoic acid" ,"saponin" and "Berberine" only which is located in 5th position.
You can see there are big spaces between terms, so that's why I said 5th position.
In this case, how can I extract terms located in 5th position for each line?
One more thing: the length of whitespace between each of the six terms is not always equal. the length could be one, two, three, four, or five, or something like that.
Because the length of whitespace is random, I can not use the .split() function.
For example, in the first line I would get "beta-lipoic" instead "beta-lipoic acid.**

Here is a solution for your problem using the string split and index of,
import java.util.ArrayList;
public class StringSplit {
public static void main(String[] args) {
String[] seperatedStr = null;
int fourthStrIndex = 0;
String modifiedStr = null, finalStr = null;
ArrayList<String> strList = new ArrayList<String>();
strList.add("31451 CID005319044 　　15939353　　 C8H14O3S2 beta-lipoic acid C1C[S#](=O)S[C##H]1CCCCC(=O)O ");
strList.add("12232 COD05374044 23439353 C924O3S2 saponin CCCC(=O)O ");
strList.add("9048 CTD042032 23241 C3HO4O3S2 Berberine [C##H]1CCCCC(=O)O ");
for (String item: strList) {
seperatedStr = item.split("\\s+");
fourthStrIndex = item.indexOf(seperatedStr[3]) + seperatedStr[3].length();
modifiedStr = item.substring(fourthStrIndex, item.length());
finalStr = modifiedStr.substring(0, modifiedStr.indexOf(seperatedStr[seperatedStr.length - 1]));
System.out.println(finalStr.trim());
}
}
}
Output:
beta-lipoic acid
saponin
Berberine

Option 1 : Use spring.split and check for multiple consecutive spaces. Like the code below:
String s[] = str.split("\\s\\s+");
for (String string : s) {
System.out.println(string);
}
Option 2 : Implement your own string split logic by browsing through all the characters. Sample code below (This code is just to give an idea. I didnot test this code.)
public static List<String> getData(String str) {
List<String> list = new ArrayList<>();
String s="";
int count=0;
for(char c : str.toCharArray()){
System.out.println(c);
if (c==' '){
count++;
}else {
s = s+c;
}
if(count>1&&!s.equalsIgnoreCase("")){
list.add(s);
count=0;
s="";
}
}
return list;
}

This would be a relatively easy fix if it weren't for beta-lipoic acid...
Assuming that only spaces/tabs/other whitespace separate terms, you could split on whitespace.
Pattern whitespace = Pattern.compile("\\s+");
String[] terms = whitespace.split(line); // Not 100% sure of syntax here...
// Your desired term should be index 4 of the terms array
While this would work for the majority of your terms, this would also result in you losing the "acid" in "beta-lipoic acid"...
Another hacky solution would be to add in a check for the 6th spot in the array produced by the above code and see if it matches English letters. If so, you can be reasonably confident that the 6th spot is actually part of the same term as the 5th spot, so you can then concatenate those together. This falls apart pretty quickly though if you have terms with >= 3 words. So something like
Pattern possibleEnglishWord = Pattern.compile([[a-zA-Z]*); // Can add dashes and such as needed
if (possibleEnglishWord.matches(line[5])) {
// return line[4].append(line[5]) or something like that
}
Another thing you can try is to replace all groups of spaces with a single space, and then remove everything that isn't made up of just english letters/dashes
line = whitespace.matcher(line).replaceAll("");
Pattern notEnglishWord = Pattern.compile("^[a-zA-Z]*"); // The syntax on this is almost certainly wrong
notEnglishWord.matcher(line).replaceAll("");
Then hopefully the only thing that is left would be the term you're looking for.
Hopefully this helps, but I do admit it's rather convoluted. One of the issues is that it appears that non-term words may have only one space between them, which would fool Option 1 as presented by Hirak... If that weren't the case that option should work.
Oh by the way, if you do end up doing this, put the Pattern declarations outside of any loops. They only need to be created once.