You should use n-grams
for that matter so you just count the number of times a sequence of contiguous n
words appear. Because you don't know how many words will be repeating you can try several n
for n-grams
, ie. from 2 to 6.
Java ngrams example tested on JDK 1.8.0
:
import java.util.*;
public class NGramExample{
public static HashMap<String, Integer> ngrams(String text, int n) {
ArrayList<String> words = new ArrayList<String>();
for(String word : text.split(" ")) {
words.add(word);
}
HashMap<String, Integer> map = new HashMap<String, Integer>();
int c = words.size();
for(int i = 0; i < c; i++) {
if((i + n - 1) < c) {
int stop = i + n;
String ngramWords = words.get(i);
for(int j = i + 1; j < stop; j++) {
ngramWords +=" "+ words.get(j);
}
map.merge(ngramWords, 1, Integer::sum);
}
}
return map;
}
public static void main(String []args){
System.out.println("Ngrams: ");
HashMap<String, Integer> res = ngrams("Patient name xyz phone no 12345 emailid xyz@abc.com. Patient name abc address some us address", 2);
for (Map.Entry<String, Integer> entry : res.entrySet()) {
System.out.println(entry.getKey() + ":" + entry.getValue().toString());
}
}
}
The output:
Ngrams:
name abc:1
xyz@abc.com. Patient:1
emailid xyz@abc.com.:1
phone no:1
12345 emailid:1
Patient name:2
xyz phone:1
address some:1
us address:1
name xyz:1
some us:1
no 12345:1
abc address:1
So you see how 'Patient name' has the max count, 2 times. You could use this function with several n
values and retrieve the max occurrences.
Edit: I will leave this Python code here for historic reasons.
A simple Python (using nltk
) working example to show you what I mean:
from nltk import ngrams
from collections import Counter
paragraph = 'Patient name xyz phone no 12345 emailid xyz@abc.com. Patient name abc address some us address'
n = 2
words = paragraph.split(' ') # of course you should split sentences in a better way
bigrams = ngrams(words, n)
c = Counter(bigrams)
c.most_common()[0]
This gives you the output:
>> (('Patient', 'name'), 2)