从数学方程中提取变量

Question

我有一个类似

的字符串

a+(b * 6)

我正在尝试提取变量a, b, cat, dog。下面是我的代码。

        Set<String> varList = null; 
        StringBuilder sb = null; 
        String expression = "a+(b * 6) <= cat*45 && cat = dog";
        if (expression!=null)
        {
            sb = new StringBuilder(); 

            //list that will contain encountered words,numbers, and white space
            varList = new HashSet<String>();

            Pattern p = Pattern.compile("[A-Za-z\\s]");
            Matcher m = p.matcher(expression);

            //while matches are found 
            while (m.find())
            {
                //add words/variables found in the expression 
                sb.append(m.group());
            }//end while 

            //split the expression based on white space 
            String [] splitExpression = sb.toString().split("\\s");
            for (int i=0; i<splitExpression.length; i++)
            {
                varList.add(splitExpression[i]);
            }
        }

        Iterator iter = varList.iterator();
        while (iter.hasNext()) {
            System.out.println(iter.next());
        }

我得到的输出是：

ab
cat
dog

需要的输出：

a
b
cat
dog

在这种情况下，变量可能会也可能不会被空格分隔。当有空白时，输出是好的。但如果变量没有用空格分隔，我得到错误的输出。有人可以建议我正确的Pattern 吗？

Answer 1

为什么要使用正则表达式 find() 循环来提取单词，然后将它们全部连接成一个字符串来再次拆分该字符串？

只需使用正则表达式找到的单词。

当然，也就是说，在从表达式中删除空格 (\\s) 并使其匹配整个单词 (+) 之后。

Pattern p = Pattern.compile("[A-Za-z]+");
Matcher m = p.matcher(expression);
while (m.find())
{
    varList.add(m.group());
}

Answer 2

如果你的变量只是一串字母，你可以简单地使用像这样的简单正则表达式来搜索它们。

正则表达式： [A-Za-z]+

Regex101 Demo

Answer 3

这个正则表达式应该可以工作 (variable name can start with uppercase or lowercase and can then contain digit(s), underscore, uppercase and lowercase)

\b[A-Za-z]\w*\b

Regex Demo

Java 代码

Set<String> set = new HashSet<String>();
String line = "a+(b * 6) <= cat*45 && cat = dog";
String pattern = "\\b([A-Za-z]\\w*)\\b";

Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(line);

while (m.find()) {
    set.add(m.group());
}
System.out.println(set);

Ideone Demo

Answer 4

我相信你应该用“[A-Za-z]+”替换你的正则表达式。 我只是用 Python 模拟的

>>> re.findall('[A-Za-z]+', 'a+(b * 6) <= cat*45 && cat = dog')
['a', 'b', 'cat', 'cat', 'dog']
>>>

那么接下来，将结果列表放入一个集合中：

>>> rs = set(re.findall('[A-Za-z]+', 'a+(b * 6) <= cat*45 && cat = dog'))
>>> for w in rs:
...     print w,
...
a b dog cat
>>>

Answer 5

完整的工作代码

public static void main(String[] args) {
    Set<String> varList = null; 
    StringBuilder sb = null; 
    String expression = "a+(b * 6) <= cat*45 && cat = dog";
    if (expression!=null)
    {
        sb = new StringBuilder(); 

        //list that will contain encountered words,numbers, and white space
        varList = new HashSet<String>();

        Pattern p = Pattern.compile("[A-Za-z\\s]+");
        Matcher m = p.matcher(expression);

        //while matches are found 
        while (m.find())
        {
            //add words/variables found in the expression 
            sb.append(m.group());
            sb.append(",");
        }//end while 

        //split the expression based on white space 
        String [] splitExpression = sb.toString().split(",");
        for (int i=0; i<splitExpression.length; i++)
        {
            if(!splitExpression[i].isEmpty() && !splitExpression[i].equals(" "))
                varList.add(splitExpression[i].trim());
        }
    }

    Iterator iter = varList.iterator();
    while (iter.hasNext()) {
        System.out.println(iter.next());
    }
}