假设我有一个字符串
String test = "This is a test string and I have some stopwords in here";
我希望看到下面数组中的单词与我的字符串匹配的次数
psudocode
array = a,and,the,them,they,I
所以答案是“3”
只是好奇在java中最有效的方法是什么？
答案 0 :(得分：5)
我可能会将输入中的单词存储到HashSet中，然后遍历数组，看看数组中的每个单词是否都是.contains。
这里是代码......输入是“Around the world in 80 days”。
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
public class Main
{
public static void main(final String[] argv)
throws FileNotFoundException
{
final File file;
final String[] wordsToFind;
file = new File(argv[0]);
wordsToFind = getWordsToFind(file);
a(file, wordsToFind);
b(file, wordsToFind);
c(file, wordsToFind);
d(file, wordsToFind);
}
// this just reads the file into the disk cache
private static String[] getWordsToFind(final File file)
throws FileNotFoundException
{
final Scanner scanner;
final Set<String> words;
scanner = new Scanner(file);
words = new HashSet<String>();
while(scanner.hasNext())
{
final String word;
word = scanner.next();
words.add(word);
}
return (words.toArray(new String[words.size()]));
}
// bad way, read intpo a list and then iterate over the list until you find a match
private static void a(final File file,
final String[] wordsToFind)
throws FileNotFoundException
{
final long start;
final long end;
final long total;
final Scanner scanner;
final List<String> words;
int matches;
scanner = new Scanner(file);
words = new ArrayList<String>();
while(scanner.hasNext())
{
final String word;
word = scanner.next();
words.add(word);
}
start = System.nanoTime();
{
matches = 0;
for(final String wordToFind : wordsToFind)
{
for(final String word : words)
{
if(word.equals(wordToFind))
{
matches++;
break;
}
}
}
System.out.println(matches);
}
end = System.nanoTime();
total = end - start;
System.out.println("a: " + total);
}
// slightly better way, read intpo a list and then iterate over the set (which reduces the number of things you progbably
// have to read until you find a match), until you find a match
private static void b(final File file,
final String[] wordsToFind)
throws FileNotFoundException
{
final long start;
final long end;
final long total;
final Scanner scanner;
final Set<String> words;
int matches;
scanner = new Scanner(file);
words = new HashSet<String>();
while(scanner.hasNext())
{
final String word;
word = scanner.next();
words.add(word);
}
start = System.nanoTime();
{
matches = 0;
for(final String wordToFind : wordsToFind)
{
for(final String word : words)
{
if(word.equals(wordToFind))
{
matches++;
break;
}
}
}
System.out.println(matches);
}
end = System.nanoTime();
total = end - start;
System.out.println("b: " + total);
}
// my way
private static void c(final File file,
final String[] wordsToFind)
throws FileNotFoundException
{
final long start;
final long end;
final long total;
final Scanner scanner;
final Set<String> words;
int matches;
scanner = new Scanner(file);
words = new HashSet<String>();
while(scanner.hasNext())
{
final String word;
word = scanner.next();
words.add(word);
}
start = System.nanoTime();
{
matches = 0;
for(final String wordToFind : wordsToFind)
{
if(words.contains(wordToFind))
{
matches++;
}
}
System.out.println(matches);
}
end = System.nanoTime();
total = end - start;
System.out.println("c: " + total);
}
// Nikita Rybak way
private static void d(final File file,
final String[] wordsToFind)
throws FileNotFoundException
{
final long start;
final long end;
final long total;
final Scanner scanner;
final Set<String> words;
int matches;
scanner = new Scanner(file);
words = new HashSet<String>();
while(scanner.hasNext())
{
final String word;
word = scanner.next();
words.add(word);
}
start = System.nanoTime();
{
words.retainAll(new HashSet<String>(Arrays.asList(wordsToFind)));
matches = words.size();
System.out.println(matches);
}
end = System.nanoTime();
total = end - start;
System.out.println("d: " + total);
}
}
结果（经过几次运行后，每次运行几乎都是相同的）：
12596
a: 2440699000
12596
b: 2531635000
12596
c: 4507000
12596
d: 5597000
如果你通过在getWordsToFind的每个单词中添加“XXX”来修改它（所以没有找到单词），你得到：
0
a: 7415291000
0
b: 4688973000
0
c: 2849000
0
d: 7981000
而且，为了完整起见，我试着搜索单词“I”，结果如下：
1
a: 235000
1
b: 351000
1
c: 75000
1
d: 10725000
答案 1 :(得分：5)
这样的东西？不确定“最有效”，但很简单。
Set<String> s1 = new HashSet<String>(Arrays.asList("This is a test string and I have some stopwords in here".split("\\s")));
Set<String> s2 = new HashSet<String>(Arrays.asList("a", "and", "the", "them", "they", "I"));
s1.retainAll(s2);
System.out.println(s1.size());
两组词的交集。
答案 2 :(得分：3)
最有效的方法是对'test'和'array'进行排序，然后迭代两者：n.log（n）+ n
测试 - ＆gt; ['a'，'和'，'有'，'这里'，in，is，...，'This'] 数组 - ＆gt; ['a'，'和'，'''，'他们'，'他们'，'我']
阵列测试匹配 'a''a'1 'a''和'1 '和''和'2 '和''拥有'2 ''''这里'2 '''''2' '''是'2' ...
答案 3 :(得分：0)
Nikita答案的一个小变化（尼基塔增加1）。如果对s1使用List，则会得到出现次数（如果单词在句子中多次出现）。
List<String> s1 = new ArrayList<String>(Arrays.asList("This is a test string and I have some stopwords in here".split("\\s")));
Set<String> s2 = new HashSet<String>(Arrays.asList("a", "and", "the", "them", "they", "I"));
s1.retainAll(s2);
System.out.println(s1.size());
答案 4 :(得分：0)
将您的字符串存储在哈希表（HashMap of（String and Integer））中，然后对文本进行迭代，并增加哈希表中匹配单词的整数值。然后在哈希表上迭代，并求和所有整数值。