我试图在golang中实现一个简单的全文本搜索,但是我所有的实现都太慢而无法克服阈值。
任务如下:
文档是用空格分隔的小写单词的非空字符串
每个文档的隐式标识符等于其在输入数组中的索引
New()构造索引
Search():接受一个查询,该查询也是一串用空格分隔的小写单词,并返回一个包含文档唯一标识符的排序数组,该文档包含查询中所有单词的大小顺序
示例:
index := New([]string{
"this is the house that jack built", //: 0
"this is the rat that ate the malt", //: 1
})
index.Search("") // -> []
index.Search("in the house that jack built") // -> []
index.Search("malt rat") // -> [1]
index.Search("is this the") // -> [0, 1]
我已经尝试实现:
针对每个文档以及所有文档的二进制搜索树
每个文档和所有文档的特里树(前缀树)
倒排索引搜索
二进制搜索树(用于所有文档):
type Tree struct {
m map[int]bool
word string
left *Tree
right *Tree
}
type Index struct {
tree *Tree
}
二进制搜索树(每个文档的树):
type Tree struct {
word string
left *Tree
right *Tree
}
type Index struct {
tree *Tree
index int
next *Index
}
trie(对于所有文档):
type Trie struct {
m map[uint8]*Trie
end_node map[int]bool
}
type Index struct {
trie *Trie
}
trie(针对每个文档):
type Trie struct {
m map[uint8]*Trie
end_node bool
}
type Index struct {
trie *Trie
index int
next *Index
}
倒排索引:
type Index struct {
m map[string]map[int]bool
}
反向索引的新增和搜索实现:
// New creates a fulltext search index for the given documents
func New(docs []string) *Index {
m := make(map[string]map[int]bool)
for i := 0; i < len(docs); i++ {
words := strings.Fields(docs[i])
for j := 0; j < len(words); j++ {
if m[words[j]] == nil {
m[words[j]] = make(map[int]bool)
}
m[words[j]][i+1] = true
}
}
return &(Index{m})
}
// Search returns a slice of unique ids of documents that contain all words from the query.
func (idx *Index) Search(query string) []int {
if query == "" {
return []int{}
}
ret := make(map[int]bool)
arr := strings.Fields(query)
fl := 0
for i := range arr {
if idx.m[arr[i]] == nil {
return []int{}
}
if fl == 0 {
for value := range idx.m[arr[i]] {
ret[value] = true
}
fl = 1
} else {
tmp := make(map[int]bool)
for value := range ret {
if idx.m[arr[i]][value] == true {
tmp[value] = true
}
}
ret = tmp
}
}
ret_arr := []int{}
for value := range ret {
ret_arr = append(ret_arr, value-1)
}
sort.Ints(ret_arr)
return ret_arr
}
我做错了吗?或者在golang中有更好的搜索算法?
感谢您的帮助。
答案 0 :(得分:0)
对于语言的特定部分,我真的不能帮您,但是如果有帮助,这里是一个伪代码,描述了Trie实现以及以相当有效的方式解决当前问题的功能。
struct TrieNode{
map[char] children // maps character to children
set[int] contains // set of all ids of documents that contain the word
}
// classic search function in trie, except it returns a set of document ids instead of a simple boolean
function get_doc_ids(TrieNode node, string w, int depth){
if (depth == length(w)){
return node.contains
} else {
if (node.hasChild(w[depth]) {
return get_doc_ids(node.getChild(w[depth], w, depth+1)
} else {
return empty_set()
}
}
}
// the answering query function, as straight forward as it can be
function answer_query(TrieNode root, list_of_words L){
n = length(L)
result = get_docs_ids(root, L[0], 0)
for i from 1 to n-1 do {
result = intersection(result, get_docs_ids(root, L[i], 0)) // set intersection
if (result.is_empty()){
break // no documents contains them all, no need to check further
}
}
return result
}