拆分字符串时，如何在输出中包含运算符？

Question

昨天我问了这个关于在 python 中拆分字符串的问题。从那以后，我决定用 Go 来做这个项目。我有以下几点：

input := "house-width + 3 - y ^ (5 * house length)"

s := regexp.MustCompile(" ([+-/*^]) ").Split(input, -1)

log.Println(s)  //  [house-width 3 y (5 house length)]

如何在此输出中包含运算符？例如，我想要以下输出：

['house-width', '+', '3', '-', 'y', '^', '(5', '*', 'house length)']

编辑：为了澄清我正在拆分以空格分隔的运算符，而不仅仅是运算符。运算符的两端必须有一个空格，以将其与破折号/连字符区分开来。如果需要，请参阅我链接到的原始 python 问题以进行澄清。

Answer 1

您可以使用regexp.Split()（就像您所做的那样）获取表达式的操作数，并且可以使用regexp.FindAllString().通过这样做，您将拥有 2 个单独的[]string切片，如果您希望将结果合并为一个切片，则可以合并这 2 个[]string切片。input := "house-width + 3 - y ^ (5 * house length)"r := regexp.MustCompile(`\s([+\-/*^])\s`)s1 := r.Split(input, -1)s2 := r.FindAllString(input, -1)fmt.Printf("%q\n", s1)fmt.Printf("%q\n", s2)all := make([]string, len(s1)+len(s2))for i := range s1 {&nbsp; &nbsp; all[i*2] = s1[i]&nbsp; &nbsp; if i < len(s2) {&nbsp; &nbsp; &nbsp; &nbsp; all[i*2+1] = s2[i]&nbsp; &nbsp; }}fmt.Printf("%q\n", all)输出（在Go Playground上试试）：["house-width" "3" "y" "(5" "house length)"][" + " " - " " ^ " " * "]["house-width" " + " "3" " - " "y" " ^ " "(5" " * " "house length)"]笔记：如果要修剪运算符中的空格，可以使用该strings.TrimSpace()函数：for i, v := range s2 {&nbsp; &nbsp; all[i*2+1] = strings.TrimSpace(v)}fmt.Printf("%q\n", all)输出：["house-width" "+" "3" "-" "y" "^" "(5" "*" "house length)"]

Answer 2

如果您打算在之后解析表达式，则必须进行一些更改：包括括号作为词素你不能让空格和破折号都是有效的标识符字符，因为例如在两者- y之间3并且^将是一个有效的标识符。完成后，您可以使用简单的线性迭代来对字符串进行词法分析：package mainimport (&nbsp; &nbsp; "bytes"&nbsp; &nbsp; "fmt")func main() {&nbsp; &nbsp; input := `house width + 3 - y ^ (5 * house length)`&nbsp; &nbsp; buffr := bytes.NewBuffer(nil)&nbsp; &nbsp; outpt := make([]string, 0)&nbsp; &nbsp; for _, r := range input {&nbsp; &nbsp; &nbsp; &nbsp; if r == '+' || r == '-' || r == '*' || r == '/' || r == '^' || r == '(' || r == ')' || (r >= '0' && r <= '9') {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; bs := bytes.TrimSpace(buffr.Bytes())&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if len(bs) > 0 {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; outpt = append(outpt, (string)(bs))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; outpt = append(outpt, (string)(r))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; buffr.Reset()&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; buffr.WriteRune(r)&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Printf("%#v\n", outpt)}词法分析后，使用 Dijkstra 的分流码算法构建 AST 或直接评估表达式。