正则表达式-捕获空格分隔的单词

Question

I have a string:

2001 970451 4 l 97 0451 iver b y c 3 0 1 8 4 1 4 hundred 2001 970451 nama 4 l 97 0451 iver hundred blah

I need an appropriate regular expression to capture series of characters and spaces such as b y c 3 0 1 8 4 1 4?

I have tried:

(\b[a-z0-9]{1}\s{1})+ - I get l

EDIT:

To further explain what I need, I need to be able to capture similar series of text where a single alphanum character is continuously/repeatedly followed by single space character to a point where this is no longer true.

Answer 1

Is regexp a hard requirement?

It would be far simpler for you, in the long term, to just use something like strings.Fields and filter the resulting array by length (you can apply any other requirements too).

Example:
(Give it a try on the playground! https://play.golang.org/p/Ue2wO5d-Te)

package main

import (
    "fmt"
    "strings"
)

func CaptureGroups(input string) (output [][]string) {
    fields := strings.Fields(input)

    var group []string
    for _, field := range fields {
        if len(field) == 1 {
            group = append(group, field)
        } else {
            if len(group) > 0 {
                output = append(output, group)
                group = make([]string, 0)
            }
        }
    }

    return
}

func main() {
    input := "2001 970451 4 l 97 0451 iver b y c 3 0 1 8 4 1 4 hundred 2001 970451 nama  4 l 97 0451 iver hundred blah"
    output := CaptureGroups(input)

    fmt.Printf("Groups: %q", output)
}