英文:
golang drop a dimension of array
问题
我可以帮你翻译这段代码。这是一个使用golang编写的代码。
func GetIndexes(body string) ([]int, error) {
indexPattern, err := regexp.Compile(`<div class="post article" id="([0-9]+)">`)
res := indexPattern.FindAllStringSubmatch(body, -1)
fmt.Printf("%v\n", res)
//Just for debug
return make([]int, 5), err
}
例如，结果可能是这样的：
[
[<div class="post article" id="55987"> 55987]
[<div class="post article" id="6717024"> 6717024]
[<div class="post article" id="6440542"> 6440542]
[<div class="post article" id="6800745"> 6800745]
[<div class="post article" id="449954"> 449954]
[<div class="post article" id="427586"> 427586]
[<div class="post article" id="5418445"> 5418445]
[<div class="post article" id="559225"> 559225]
...
]
我想要的是只获取一个类似这样的数组：
[55987, 6717024, 6717024, ...]
我可以遍历数组并复制我想要的值，但我不确定这是否是最好的方法。因此，我想知道是否有可能删除数组的列，或者是否可以创建带有lambda函数或其他函数的切片。
谢谢。
英文:
I've a code like that in golang
func GetIndexes(body string) ([]int, error) {
indexPattern, err := regexp.Compile(`<div class="post article" id="([0-9]+)">`)
res := indexPattern.FindAllStringSubmatch(body, -1)
fmt.Printf("%v\n", res)
//Just for debug
return make([]int, 5), err
}
for exemple the result is like :
[
[<div class="post article" id="55987"> 55987]
[<div class="post article" id="6717024"> 6717024]
[<div class="post article" id="6440542"> 6440542]
[<div class="post article" id="6800745"> 6800745]
[<div class="post article" id="449954"> 449954]
[<div class="post article" id="427586"> 427586]
[<div class="post article" id="5418445"> 5418445]
[<div class="post article" id="559225"> 559225]
...
]
And I'm looking for a way to get just an array like
[55987, 6717024, 6717024, ...]
I could range the array and copy value which I looking for, but i'm not sure it's the better way.
It's why I ask myself if it's possible to drop column of this array, or why not create kind of slices with lambdas functions or other...
Thank you
答案1
得分: 0
这更多是一个正则表达式引擎的问题，因为引擎将以以下格式输出结果：
res[0] // 将是第一个匹配的“整个字符串”出现
res[0][0] // 将是整个字符串的匹配
res[0][1] // 将是第一个分组匹配 -- 括号中的内容
res[0]['some_name'] // 将是一个命名分组匹配
你需要做的是迭代res[i]并检索res[i][1]。
由于你尝试匹配的正则表达式可能非常复杂 -- 可能有许多分组匹配、许多命名分组匹配等 -- 结果变量也可能相当复杂。
由于结果变量的（可能）复杂性，正则表达式库没有必要为你提供完全符合你所描述需求的函数，因为这些函数的用途非常有限。
此外，编写这样的代码片段或函数是一个简单的任务，所以你必须根据你非常特定的需求自己混合和匹配。
英文:
This is more of a RegEx engine issue because the engine will output the results in the following format:
res[0] // will be the first matched "whole string" occurrence
res[0][0] // will be the whole string match
res[0][1] // will be the first grouped match -- the things in parenthesis
res[0]['some_name'] // will be a named group match
You have to do is iterate over res[i] and retrieve res[i][1].
Since the RegEx you're trying to match may be very complex -- it may have many grouped matches, many named grouped matches etc. -- the result variable might also be fairly complex.
Because of the (possible) complexity of the result variable there would be no point for a RegEx library to provide you with functions that do exactly what you described because those functions would be of very limited use.
Also writing such a snippet of code or function is trivial task so you have to mix and match your own according to your very particular set of needs.
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