英文:
How do I include the operators in my output when I split my string?
问题
昨天我问了一个关于在Python中拆分字符串的问题。后来我决定用Go来完成这个项目。我有以下代码：
input := "house-width + 3 - y ^ (5 * house length)"s := regexp.MustCompile(" ([+-/*^]) ").Split(input, -1)log.Println(s) // [house-width 3 y (5 house length)]
如何在输出中包含运算符？例如，我希望得到以下输出：
['house-width', '+', '3', '-', 'y', '^', '(5', '*', 'house length)']
编辑：
为了澄清，我是在以空格分隔的运算符上进行拆分，而不仅仅是运算符本身。运算符两边必须有一个空格，以便将其与破折号/连字符区分开来。如果需要，可以参考我链接的原始Python问题以获得澄清。
英文:
Yesterday I asked this question about splitting a string in python. I've since decided to do this project in Go instead. I have the following:
input := "house-width + 3 - y ^ (5 * house length)"s := regexp.MustCompile(" ([+-/*^]) ").Split(input, -1)log.Println(s) // [house-width 3 y (5 house length)]
How do I include the operators in this output? e.g. I'd like the following output:
['house-width', '+', '3', '-', 'y', '^', '(5', '*', 'house length)']
EDIT:
To clarify I am splitting on the space-separated operators and not just the operator. The operator must have a space on both ends to differentiate it from a dash/hyphen. Please refer to my original python question I linked to for clarification if needed.
答案1
得分: 1
你可以使用regexp.Split()来获取表达式的操作数（就像你之前做的那样），并且可以使用regexp.FindAllString()来获取运算符（分隔符）。
通过这样做，你将得到两个独立的[]string切片，如果你想要将结果合并到一个[]string切片中，你可以合并这两个切片。
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 {all[i*2] = s1[i]if i < len(s2) {all[i*2+1] = s2[i]}}fmt.Printf("%q\n", all)
输出结果（在Go Playground上尝试）：
["house-width" "3" "y" "(5" "house length)"][" + " " - " " ^ " " * "]["house-width" " + " "3" " - " "y" " ^ " "(5" " * " "house length)"]
注意：
如果你想要去除运算符周围的空格，你可以使用strings.TrimSpace()函数：
for i, v := range s2 {all[i*2+1] = strings.TrimSpace(v)}fmt.Printf("%q\n", all)
输出结果：
["house-width" "+" "3" "-" "y" "^" "(5" "*" "house length)"]
英文:
You can get the operands of your expression using regexp.Split() (just as you did) and you can get the operators (the separators) using regexp.FindAllString().
By doing this you will have 2 separate []string slices, you can merge these 2 slices if you want the result in one []string slice.
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 {all[i*2] = s1[i]if i < len(s2) {all[i*2+1] = s2[i]}}fmt.Printf("%q\n", all)
Output (try it on the Go Playground):
["house-width" "3" "y" "(5" "house length)"][" + " " - " " ^ " " * "]["house-width" " + " "3" " - " "y" " ^ " "(5" " * " "house length)"]
Note:
If you want to trim the spaces from the operators, you can use the strings.TrimSpace() function for that:
for i, v := range s2 {all[i*2+1] = strings.TrimSpace(v)}fmt.Printf("%q\n", all)
Output:
["house-width" "+" "3" "-" "y" "^" "(5" "*" "house length)"]
答案2
得分: 1
如果你计划在之后解析这个表达式，你需要做一些改动：
- 将括号包含在词素中
- 不能同时将空格和破折号作为有效的标识符字符，因为例如
- y在3和^之间将成为一个有效的标识符。
完成这些之后，你可以使用简单的线性迭代来对字符串进行词法分析：
package mainimport ("bytes""fmt")func main() {input := `house width + 3 - y ^ (5 * house length)`buffr := bytes.NewBuffer(nil)outpt := make([]string, 0)for _, r := range input {if r == '+' || r == '-' || r == '*' || r == '/' || r == '^' || r == '(' || r == ')' || (r >= '0' && r <= '9') {bs := bytes.TrimSpace(buffr.Bytes())if len(bs) > 0 {outpt = append(outpt, string(bs))}outpt = append(outpt, string(r))buffr.Reset()} else {buffr.WriteRune(r)}}fmt.Printf("%#v\n", outpt)}
词法分析完成后，使用Dijkstra的逆波兰算法构建一个AST或直接计算表达式。
英文:
If you're planing to parse the expression afterwards you'll have to make some changes:
- Include parentheses as lexemes
- You can't have both spaces and dashes be valid identifier characters because e.g.
- yinbetween3and^would be a valid identifier.
After that's done, you can use use a simple linear iteration to lex your string:
package mainimport ("bytes""fmt")func main() {input := `house width + 3 - y ^ (5 * house length)`buffr := bytes.NewBuffer(nil)outpt := make([]string, 0)for _, r := range input {if r == '+' || r == '-' || r == '*' || r == '/' || r == '^' || r == '(' || r == ')' || (r >= '0' && r <= '9') {bs := bytes.TrimSpace(buffr.Bytes())if len(bs) > 0 {outpt = append(outpt, (string)(bs))}outpt = append(outpt, (string)(r))buffr.Reset()} else {buffr.WriteRune(r)}}fmt.Printf("%#v\n", outpt)}
Once lexed, use Dijkstra's shunting-yard algorithm to build an AST or directly evaluate the expression.
答案3
得分: 0
我认为FindAll()可能是一个可行的方法。
扩展的正则表达式：
\s* # 去除前导空格( # (1 开始)，运算符/非运算符字符(?: # 集群组\w - # 单词破折号| - \w # 或者，破折号单词| [^+\-/*^] # 或者，非运算符字符)+ # 结束集群，出现1次或多次| # 或者，[+\-/*^] # 单个简单的数学运算符) # (1 结束)\s* # 去除尾随空格
Go代码片段：
http://play.golang.org/p/bHZ21B6Tzi
package mainimport ("log""regexp")func main() {in := []byte("house-width + 3 - y ^ (5 * house length)")rr := regexp.MustCompile("\\s*((?:\\w-|-\\w|[^+\\-/*^])+|[+\\-/*^])\\s*")s := r.FindAll( in, -1 )for _, ss:=range s{log.Println(string(ss))}}
输出：
house-width+3-y^(5*house length)
英文:
I think FindAll() may be the way to go.
Expanded regex:
\s* # Trim preceding whitespace( # (1 start), Operator/Non-Operator chars(?: # Cluster group\w - # word dash| - \w # or, dash word| [^+\-/*^] # or, a non-operator char)+ # End cluster, do 1 to many times| # or,[+\-/*^] # A single simple math operator) # (1 end)\s* # Trim trailing whitespace
Go code snippet:
http://play.golang.org/p/bHZ21B6Tzi
package mainimport ("log""regexp")func main() {in := []byte("house-width + 3 - y ^ (5 * house length)")rr := regexp.MustCompile("\\s*((?:\\w-|-\\w|[^+\\-/*^])+|[+\\-/*^])\\s*")s := r.FindAll( in, -1 )for _, ss:=range s{log.Println(string(ss))}}
Output:
house-width+3-y^(5*house length)
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