Java中用于分割复数字符串的正则表达式
在Java中寻找正则表达式来分隔表示复数的字符串。一个代码示例就好了。 输入字符串的格式为：Java中用于分割复数字符串的正则表达式,java,regex,Java,Regex,在Java中寻找正则表达式来分隔表示复数的字符串。一个代码示例就好了。 输入字符串的格式为： "a+bi" Example: "300+400i", "4+2i", "5000+324i". 我需要从字符串中检索300和400。” 我知道我们可以用这种方法粗略地做 str.substring(0, str.indexOf('+')); str.substring(str.indexOf('+')+1,str.indexOf("i")); 我需要从字符串中检索300和400 使用String.
"a+bi"
Example: "300+400i", "4+2i", "5000+324i".
str.substring(0, str.indexOf('+'));
str.substring(str.indexOf('+')+1,str.indexOf("i"));
String.split（regex）String s[] = "300-400i".split("[\\Q+-\\Ei]");
System.out.println(s[0]+" "+s[1]); //prints 300 400
String.split（regex）String s[] = "300-400i".split("[\\Q+-\\Ei]");
System.out.println(s[0]+" "+s[1]); //prints 300 400
[0-9]{1,}
[0-9]{1,}
希望有帮助。与此匹配的正则表达式是：
/[0-9]{1，}[+-][0-9]{1，}i/Pattern complexNumberPattern = Pattern.compile("[0-9]{1,}");
Matcher complexNumberMatcher = complexNumberPattern.matcher(myString);
并使用
complexNumberMatcherfindgroupmyString/[0-9]{1，}[+-][0-9]{1，}i/Pattern complexNumberPattern = Pattern.compile("[0-9]{1,}");
Matcher complexNumberMatcher = complexNumberPattern.matcher(myString);
然后使用
complexNumberMatcherfindgroupmyStringpublic static void main(String[] args) {
String[] attempts = new String[]{"300+400i", "4i+2", "5000-324i", "555", "2i", "+400", "-i"};
for (String s : attempts) {
System.out.println("Parsing\t" + s);
printComplex(s);
}
}
static void printComplex(String in) {
String[] parts = in.split("[+-]");
int re = 0, im = 0, pos = -1;
for (String s : parts) {
if (pos != -1) {
s = in.charAt(pos) + s;
} else {
pos = 0;
if ("".equals(s)) {
continue;
}
}
pos += s.length();
if (s.lastIndexOf('i') == -1) {
if (!"+".equals(s) && !"-".equals(s)) {
re += Integer.parseInt(s);
}
} else {
s = s.replace("i", "");
if ("+".equals(s)) {
im++;
} else if ("-".equals(s)) {
im--;
} else {
im += Integer.parseInt(s);
}
}
}
System.out.println("Re:\t" + re + "\nIm:\t" + im);
}
Parsing 300+400i
Re: 300
Im: 400
Parsing 4i+2
Re: 2
Im: 4
Parsing 5000-324i
Re: 5000
Im: -324
Parsing 555
Re: 555
Im: 0
Parsing 2i
Re: 0
Im: 2
试试这个。对我来说，它是有效的
public static void main(String[] args) {
String[] attempts = new String[]{"300+400i", "4i+2", "5000-324i", "555", "2i", "+400", "-i"};
for (String s : attempts) {
System.out.println("Parsing\t" + s);
printComplex(s);
}
}
static void printComplex(String in) {
String[] parts = in.split("[+-]");
int re = 0, im = 0, pos = -1;
for (String s : parts) {
if (pos != -1) {
s = in.charAt(pos) + s;
} else {
pos = 0;
if ("".equals(s)) {
continue;
}
}
pos += s.length();
if (s.lastIndexOf('i') == -1) {
if (!"+".equals(s) && !"-".equals(s)) {
re += Integer.parseInt(s);
}
} else {
s = s.replace("i", "");
if ("+".equals(s)) {
im++;
} else if ("-".equals(s)) {
im--;
} else {
im += Integer.parseInt(s);
}
}
}
System.out.println("Re:\t" + re + "\nIm:\t" + im);
}
Parsing 300+400i
Re: 300
Im: 400
Parsing 4i+2
Re: 2
Im: 4
Parsing 5000-324i
Re: 5000
Im: -324
Parsing 555
Re: 555
Im: 0
Parsing 2i
Re: 0
Im: 2
http://rubular.com/r/FfOAt1zk0v
string regexPattern =
// Match any float, negative or positive, group it
@"([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)" +
// ... possibly following that with whitespace
@"\s*" +
// ... followed by a plus
@"\+" +
// and possibly more whitespace:
@"\s*" +
// Match any other float, and save it
@"([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)" +
// ... followed by 'i'
@"i";
Regex regex = new Regex(regexPattern);
Console.WriteLine("Regex used: " + regex);
while (true)
{
Console.WriteLine("Write a number: ");
string imgNumber = Console.ReadLine();
Match match = regex.Match(imgNumber);
double real = double.Parse(match.Groups[1].Value, CultureInfo.InvariantCulture);
double img = double.Parse(match.Groups[2].Value, CultureInfo.InvariantCulture);
Console.WriteLine("RealPart={0};Imaginary part={1}", real, img);
}
http://rubular.com/r/FfOAt1zk0v
string regexPattern =
// Match any float, negative or positive, group it
@"([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)" +
// ... possibly following that with whitespace
@"\s*" +
// ... followed by a plus
@"\+" +
// and possibly more whitespace:
@"\s*" +
// Match any other float, and save it
@"([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)" +
// ... followed by 'i'
@"i";
Regex regex = new Regex(regexPattern);
Console.WriteLine("Regex used: " + regex);
while (true)
{
Console.WriteLine("Write a number: ");
string imgNumber = Console.ReadLine();
Match match = regex.Match(imgNumber);
double real = double.Parse(match.Groups[1].Value, CultureInfo.InvariantCulture);
double img = double.Parse(match.Groups[2].Value, CultureInfo.InvariantCulture);
Console.WriteLine("RealPart={0};Imaginary part={1}", real, img);
}
理论上，你可以用这样的方法：
Pattern complexNumberPattern = Pattern.compile("(.*)+(.*)");
Matcher complexNumberMatcher = complexNumberPattern.matcher(myString);
if (complexNumberMatcher.matches()) {
String prePlus = complexNumberMatcher.group(1);
String postPlus = complexNumberMatcher.group(2);
}
编辑：只是注意到你不想要这些字母，所以没关系，但这会让你在其他字母出现的情况下对其进行更多的控制。理论上，你可以使用这样的方式：
Pattern complexNumberPattern = Pattern.compile("(.*)+(.*)");
Matcher complexNumberMatcher = complexNumberPattern.matcher(myString);
if (complexNumberMatcher.matches()) {
String prePlus = complexNumberMatcher.group(1);
String postPlus = complexNumberMatcher.group(2);
}
编辑：只是注意到你不想要这些信，所以没关系，但是这会给你更多的控制权，以防其他字母出现在其中。为什么你更喜欢正则表达式而不是那几行非常简洁的代码？只是想看看它是否可能，语法：）为什么你更喜欢正则表达式而不是那几行非常简洁的代码？只是想看看它是否可能，语法：）嘿，MJafar，你的代码对我有用，但我选择Sage的答案是因为简单和缺乏输入。但是我喜欢这种方法：）我也喜欢他的答案，非常好的方法，但是记住我的代码，把它保存在某个地方，这是一种在更大的字符串中获得与特定模式匹配的子字符串的方法。当输入有额外的空格时，这种方法也有效，如
300+400itrim（）300+400itrim（）a-bia-bi