java 使用正则表达式拆分简单的数学表达式

Splitting a simple maths expression with regex

提问by sutoL

I am trying to have a regular expression split on equations like 1.5+4.2*(5+2) with operators - + * / so the output would be input into a array so I can parse individually

我正在尝试使用运算符 - + * / 在 1.5+4.2*(5+2) 等等式上拆分正则表达式，以便将输出输入到数组中，以便我可以单独解析

[0]1.5
[1]+
[2]4.2
[3]*
[4](
[5]5
[6]+
[7]2
[8]) 

I have found out that the \bwill work on 1+2+3however if I were to have decimal points it would not split.

我已经发现，在\b将工作于1+2+3但是如果我有它不会分裂小数点。

I have tried splitting with \b(\.\d{1,2})however it does not split on the decimal point

我试过拆分，\b(\.\d{1,2})但它不会在小数点上拆分

回答by polygenelubricants

You can use zero-width matching lookahead and lookbehind combo as alternates.

您可以使用零宽度匹配前瞻和后视组合作为替代。

    String equation = "1.5+4.2*(5+2)";

    String regex = "(?<=op)|(?=op)".replace("op", "[-+*/()]");

    // actual regex becomes (?<=[-+*/()])|(?=[-+*/()])

    System.out.println(java.util.Arrays.toString(
        equation.split(regex)
    ));
    //  ___  _  ___  _  _  _  _  _  _
    // [1.5, +, 4.2, *, (, 5, +, 2, )]

Explanation

解释

• […]is a character class definition
• (?<=…)is a lookbehind; it asserts that we can match …to the left
• (?=…)is a lookahead; it asserts that we can match …to the right
• this|thatis alternation
• Thus, (?<=op)|(?=op)matches everywhere after or before op
  • ... where opis replaced by [-+*/()], i.e. a character class that matches operators
    • Note that -is first here so that it doesn't become a range definition meta character

• […]是字符类定义
• (?<=…)是回顾；它断言我们可以匹配…到左边
• (?=…)是一个前瞻；它断言我们可以匹配…到右边
• this|that是交替
• 因此，(?<=op)|(?=op)匹配之后或之前的任何地方op
  • ... whereop被替换为[-+*/()]，即匹配运算符的字符类
    • 请注意，-这里是第一个，这样它就不会成为范围定义元字符

References

参考

• regular-expressions.info/Lookaroundsand Alternation with Vertical Barand Character Class

• 正则表达式.info/Lookarounds和垂直条和字符类的交替

Related questions

相关问题

• How does the regular expression (?<=#)[^#]+(?=#)work?
• Regex: why doesn't [01-12] range work as expected?

• 正则表达式是如何(?<=#)[^#]+(?=#)工作的？
• 正则表达式：为什么 [01-12] 范围没有按预期工作？

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More examples of zero-width matching regex for splitting

用于拆分的零宽度匹配正则表达式的更多示例

Here are more examples of splitting on zero-width matching constructs; this can be used to split a string but also keep delimiters.

以下是在零宽度匹配构造上拆分的更多示例；这可用于拆分字符串，但也可以保留分隔符。

Simple sentence splitting, keeping punctuation marks:

简单的分句，保留标点符号：

    String str = "Really?Wow!This.Is.Awesome!";
    System.out.println(java.util.Arrays.toString(
        str.split("(?<=[.!?])")
    )); // prints "[Really?, Wow!, This., Is., Awesome!]"

Splitting a long string into fixed-length parts, using \G

将长字符串拆分为固定长度的部分，使用 \G

    String str = "012345678901234567890";
    System.out.println(java.util.Arrays.toString(
        str.split("(?<=\G.{4})")
    )); // prints "[0123, 4567, 8901, 2345, 6789, 0]"

Split before capital letters (except the first!)

在大写字母前拆分（第一个除外！）

    System.out.println(java.util.Arrays.toString(
        "OhMyGod".split("(?=(?!^)[A-Z])")
    )); // prints "[Oh, My, God]"

A variety of examples is provided in related questions below.

下面的相关问题提供了各种示例。

Related questions

相关问题

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  • "abc<def>ghi<x><x>" -> "abc", "<def>", "ghi", "<x>", "<x>"
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  • C# version: is there a elegant way to parse a word and add spaces before capital letters
• Java split is eating my characters
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• Regex split string but keep separators

• 您可以在字符串拆分中使用零宽度匹配正则表达式吗？
  • "abc<def>ghi<x><x>" -> "abc", "<def>", "ghi", "<x>", "<x>"
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  • "AnXMLAndXSLT2.0Tool" -> "An XML And XSLT 2.0 Tool"
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• Java 分裂正在吞噬我的角色
• 有没有办法用 String.split() 拆分字符串并包含分隔符？
• 正则表达式拆分字符串但保留分隔符

回答by Marimuthu Madasamy

Pattern pattern = Pattern.compile("((\d*\.\d+)|(\d+)|([\+\-\*/\(\)]))");
Matcher m = pattern.matcher("1.5+4.2*(5+2)/10-4");
while(m.find()) {
    System.out.printf("%s ", m.group());
}

output: 1.5 + 4.2 * ( 5 + 2 ) / 10 - 4

You can also use ?: to avoid capturing groups. I left it to make it simple.

您还可以使用 ?: 来避免捕获组。我把它留下来让它变得简单。

回答by bcosca

Use match, instead of split:

使用匹配而不是拆分：

(?:\d+\.)?\d*(?:e[+\-]?\d+)?|[\s\-\/()+*%=]

This regex will also accept valid floats like: 1.2e+3 * 2which should equal 2400. the regexes given by the other respondents will fail.

此正则表达式还将接受有效的浮点数，例如：1.2e+3 * 2which should equal 2400。其他受访者给出的正则表达式将失败。

回答by Allain Lalonde

Split the string using [+-/*()].

使用 拆分字符串[+-/*()]。