将带有双精度和字符的字符串转换为Java中的双精度的最佳方法

Question

请原谅我，如果我已经问过一个已经回答的问题，但我不确定这是一个初学者的最好方法。

我有一个像下面这样的字符串，什么是转换此字符串的最佳方式，以便我只有双打3.4134388041,0.63117288等我丢弃其余的？

谢谢！

Shading:Building:Detailed,
    39,                      !- Name
    ,                        !- Transmittance Schedule Name
    4,                       !- Number of Vertices
    3.4134388041,            !- Vertex 1 X-coordinate {m}
    0.63117288,              !- Vertex 1 Y-coordinate {m}
    2.2012378517,            !- Vertex 1 Z-coordinate {m}
    3.4134388041,            !- Vertex 2 X-coordinate {m}
    10.01517288,             !- Vertex 2 Y-coordinate {m}
    2.2012378517,            !- Vertex 2 Z-coordinate {m}
    2.9134388041,            !- Vertex 3 X-coordinate {m}
    10.01517288,             !- Vertex 3 Y-coordinate {m}
    2.2012378517,            !- Vertex 3 Z-coordinate {m}
    2.9134388041,            !- Vertex 4 X-coordinate {m}
    0.63117288,              !- Vertex 4 Y-coordinate {m}
    2.2012378517;            !- Vertex 4 Z-coordinate {m}

Answer 1

以下代码应隔离字符串中不能完全被1整除的每个数字，分别打印出来：

public static void main(String[] args)
{
    String text = ""; // string was too long to fit here for tidiness
    Matcher search = Pattern.compile("(?!=\\d\\.\\d\\.)([\\d.]+)").matcher(text);
    while (search.find())
    {
        double doub = Double.parseDouble(search.group(1));
        if (doub mod 1 != 0)
        {
            System.out.println(doub);
        }
    }
}