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Make split_best_frequency returns references instead of owned data

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This commit is contained in:
ManyTheFish 2022-04-07 17:05:44 +02:00
parent fa7d3a37c0
commit b1905dfa24
1 changed files with 8 additions and 5 deletions
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13
milli/src/search/query_tree.rs
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@ -257,7 +257,10 @@ impl<'a> QueryTreeBuilder<'a> {
}
/// Split the word depending on the frequency of subwords in the database documents.
fn split_best_frequency(ctx: &impl Context, word: &str) -> heed::Result<Option<(String, String)>> {
fn split_best_frequency<'a>(
ctx: &impl Context,
word: &'a str,
) -> heed::Result<Option<(&'a str, &'a str)>> {
let chars = word.char_indices().skip(1);
let mut best = None;
@ -273,7 +276,7 @@ fn split_best_frequency(ctx: &impl Context, word: &str) -> heed::Result<Option<(
}
}
Ok(best.map(|(_, left, right)| (left.to_string(), right.to_string())))
Ok(best.map(|(_, left, right)| (left, right)))
}
#[derive(Clone)]
@ -343,7 +346,7 @@ fn create_query_tree(
PrimitiveQueryPart::Word(word, prefix) => {
let mut children = synonyms(ctx, &[&word])?.unwrap_or_default();
if let Some((left, right)) = split_best_frequency(ctx, &word)? {
children.push(Operation::Phrase(vec![left, right]));
children.push(Operation::Phrase(vec![left.to_string(), right.to_string()]));
}
let (word_len_one_typo, word_len_two_typo) = ctx.min_word_len_for_typo()?;
let exact_words = ctx.exact_words()?;
@ -499,8 +502,8 @@ fn create_matching_words(
}
if let Some((left, right)) = split_best_frequency(ctx, &word)? {
let left = MatchingWord::new(left, 0, false);
let right = MatchingWord::new(right, 0, false);
let left = MatchingWord::new(left.to_string(), 0, false);
let right = MatchingWord::new(right.to_string(), 0, false);
matching_words.push((vec![left, right], vec![id]));
}