[英]Count number of sentence in a java string
嗨,我想計算一個字符串中的句子數,到目前為止我正在使用這個:
int count = str.split("[!?.:]+").length;
但我的字符串包含“。” 例如在名字和單詞之間
“他的名字是 Walton DC,他去年剛剛完成了他的 B.Tech。”
現在使用上面的行作為示例 count 將返回 4 個句子,但只有一個。
那么如何應對這些情況呢?
您可以使用BreakIterator ,並檢測不同類型的文本邊界
在你的情況下句子:
private static void markBoundaries(String target, BreakIterator iterator) {
StringBuffer markers = new StringBuffer();
markers.setLength(target.length() + 1);
for (int k = 0; k < markers.length(); k++) {
markers.setCharAt(k, ' ');
}
int count = 0;
iterator.setText(target);
int boundary = iterator.first();
while (boundary != BreakIterator.DONE) {
markers.setCharAt(boundary, '^');
++count;
boundary = iterator.next();
}
System.out.println(target);
System.out.println(markers);
System.out.println("Number of Boundaries: " + count);
System.out.println("Number of Sentences: " + (count-1));
}
public static void main(String[] args) {
Locale currentLocale = new Locale("en", "US");
BreakIterator sentenceIterator
= BreakIterator.getSentenceInstance(currentLocale);
String someText = "He name is Walton D.C. and he just completed his B.Tech last year.";
markBoundaries(someText, sentenceIterator);
someText = "This order was placed for QT3000! MK?";
markBoundaries(someText, sentenceIterator);
}
輸出將是:
He name is Walton D.C. and he just completed his B.Tech last year.
^ ^
Number of Boundaries: 2
Number of Sentences: 1
This order was placed for QT3000! MK?
^ ^ ^
Number of Boundaries: 3
Number of Sentences: 2
解決方案可能是在點的情況下,您可以檢查后面是否有空格和大寫字母。
“[點][空格][大寫字母]”
這將是對判決的肯定
更新相同的代碼:
public static void main( String args[] ) {
// String to be scanned to find the pattern.
String line = "This order was placed for QT3000! MK? \n Thats amazing. \n But I am not sure.";
String pattern = "([.!?])([\\s\\n])([A-Z]*)";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
int count=0;
while (m.find( )) {
count++;
}
count++; //for the last line, which will not get included here.
System.out.println("COUNT=="+count);
}
如果前面有一個或多個大寫字母,則一種解決方案是跳過點。 在這種情況下,名稱(如果它們是大寫的)。 實現這一點,你將只有一個句子。
另一種解決方案:改進這里的一個答案可能是:[小寫]([點]或[?]或[!])[空格][大寫]
但是就像我說的,如果沒有確切的規則,那幾乎是不可能的。
簡單的方法
公共類計數線{
public static void main(String[] args) {
// TODO Auto-generated method stub
String s="Find the number Sentence";
int count=0;
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i)==' ') {
count++;
}
}
count=count+1;
System.out.println(count);
}
}
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