從數學方程中提取變量

Question

我有一個像

a +（b * 6）<=貓* 45 &&貓=狗

我試圖提取變量a, b, cat, dog 。 下面是我的代碼。

        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());
        }

我得到的輸出是：

ab
cat
dog

要求的輸出：

a
b
cat
dog

在這種情況下，變量可能會或可能不會由空格分隔。 有空格時，輸出良好。 但是如果變量沒有用空格隔開，則輸出錯誤。 有人可以建議我正確的Pattern嗎？

Answer 1

為什么要使用正則表達式find()循環提取單詞，然后將它們全部串聯成一個字符串以再次拆分該字符串？

只需使用正則表達式找到的單詞即可。

好吧，就是說，從表達式中刪除空格（ \\\\s ）並使其與整個單詞（ + ）匹配之后，當然。

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

Answer 2

如果變量只是字母字符串，則可以使用簡單的正則表達式像這樣簡單地搜索它們。

正則表達式： [A-Za-z]+

Regex101演示

Answer 3

此正則表達式應該有效（ variable name can start with uppercase or lowercase and can then contain digit(s), underscore, uppercase and lowercase ）

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

正則表達式演示

Java代碼

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演示

Answer 4

我相信您應該用“ [A-Za-z] +”代替您的正則表達式。 我只是用Python模擬

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

接下來，將結果列表放入集合中：

>>> 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

完整的工作代碼

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());
    }
}