使用正则表达式从句子中找到[a-zA-Z]的单词-findingwordswith[a-zA-Z]fromasentenceusingRegex

6  

Assuming your regex engine supports lookahead/lookbehind assertions, you can use something like the following:

假设您的正则表达式引擎支持前瞻/后瞻断言,您可以使用以下内容:

(^|(?<= )[a-zA-Z]+($|(?= ))

Here's a quick description of what each component does:

以下是每个组件的功能的简要说明:

(^|(?<= )): This says "if a word starts here, we're interested". Specifically,
  ^: Match the beginning of the line, or
  (?<= ): Match any point that is preceded by a space, without actually consuming the space itself. This is called a positive lookbehind assertion.

(^ |(?<=)):这说“如果一个词从这里开始,我们就会感兴趣”。具体来说,^:匹配行的开头,或(?<=):匹配任何以空格开头的点,而不实际占用空间本身。这被称为积极的后视断言。

[a-zA-Z]+: This should be obvious, but it matches any run of sequential ASCII alphabetic characters.

[a-zA-Z] +:这应该是显而易见的,但它匹配任何连续的ASCII字母字符。

($|(?= )): This says "if the word is finished here, we're done". Specifically,
  $: Match the end of the line, or
  (?= ): Match any point that is followed by a space, without actually consuming the space itself. This is called a positive lookahead assertion.

($ |(?=)):这说“如果这个词在这里完成,我们就完成了”。具体来说,$:匹配行的结尾,或(?=):匹配任何后跟空格的点,而不实际占用空间本身。这被称为积极的先行断言。

Note that this particular regex doesn't count a word as a word if it's followed by punctuation. This may actually not be what you want, but you described checking for spaces so that's what the regex does. If you want to support words that are followed by simple punctuation you might amend that last atom to be

请注意,如果单词后跟标点符号,则此特定正则表达式不会将单词计为单词。这实际上可能不是你想要的,但你描述了检查空格,这就是正则表达式所做的。如果你想支持简单标点符号后面的单词,你可以修改最后一个原子

($|(?=[ .,!?]))

which will match the word if it's followed by a space, period, comma, exclamation mark, or question mark. You can be more elaborate too if you want.

如果后跟空格,句号,逗号,感叹号或问号,则会匹配该单词。如果你愿意,你也可以更精细。

2  

Could you use a simpler pattern like \b[A-Za-z]+\b instead? (The \b metacharacter separates word characters (e.g., letters) from nonword characters (e.g., spaces and punctuation.))

您可以使用更简单的模式,例如\ b [A-Za-z] + \ b吗? (\ b元字符将单词字符(例如,字母)与非单词字符(例如,空格和标点符号)分开。))

The code

Pattern p = Pattern.compile("\\b[A-Za-z]+\\b");
Matcher m = p.matcher("i am good");
while(m.find()) System.out.println(m.group());

Produces {"i", "am", "good"} .

产生{“i”,“am”,“good”}。

Edit As mathematical.coffee commented, the above fails. The expression

编辑如math.coffee评论,上述内容失败。表达方式

(?<=^|\s)[A-Za-z]+(?=\W*(?:\s*$|\s))

may work better. For the string I a1m a b*y boy am is!! or, matching produces "I", "a", "boy", "am", "is", "or".

可能会更好。对于字符串我a1m a b * y boy am is !!或者,匹配产生“I”,“a”,“boy”,“am”,“is”,“or”。

If in the previous expression "is!!" should be ignored, the expression (?<=^|\s)[A-Za-z]+(?=$|\s) can be used instead. In the previous example, it does not return "is" but returns the other words (I, a, boy, am, or).

如果在前一个表达式中“是!!”应该忽略,可以使用表达式(?<= ^ | \ s)[A-Za-z] +(?= $ | \ s)代替。在前面的示例中,它不返回“is”但返回其他单词(I,a,boy,am或)。

0  

This is just a note, if you didn't want to use something like Kevin Ballard has suggested. You can break the string into tokens and from there you can check each token to make sure it only contains only [a-zA-Z].

如果您不想使用Kevin Ballard建议的内容,这只是一个注释。您可以将字符串分解为标记,然后可以检查每个标记以确保它仅包含[a-zA-Z]。

To break it into tokens, do something like this:

要将其分解为令牌,请执行以下操作:

String message="The text of the message to be scanned.";
StringTokenizer st=new StringTokenizer(message);
while (st.hasMoreTokens())
    {
      checkWord(st.nextToken()); 
       idx++;
    }

And then you would write a function to check if that token is composed of [a-zA-Z]. Since there will be no white space to deal with, I think you'll find it much easier to deal with these tokens rather than the full string.

然后你会编写一个函数来检查该标记是否由[a-zA-Z]组成。由于没有空白处理空间,我认为你会发现处理这些令牌而不是完整的字符串要容易得多。

Best of luck.

祝你好运。