Dandandan commented on code in PR #5444:
URL: https://github.com/apache/arrow-rs/pull/5444#discussion_r1514241520
##########
arrow-cast/src/parse.rs:
##########
@@ -556,65 +556,30 @@ fn parse_date(string: &str) -> Option<NaiveDate> {
.map(|dt| dt.date_naive())
.ok();
};
- let mut digits = [0; 10];
- let mut mask = 0;
-
- // Treating all bytes the same way, helps LLVM vectorise this correctly
- for (idx, (o, i)) in digits.iter_mut().zip(string.bytes()).enumerate() {
- *o = i.wrapping_sub(b'0');
- mask |= ((*o < 10) as u16) << idx
- }
-
- const HYPHEN: u8 = b'-'.wrapping_sub(b'0');
-
- // refer to https://www.rfc-editor.org/rfc/rfc3339#section-3
- if digits[4] != HYPHEN {
- let (year, month, day) = match (mask, string.len()) {
- (0b11111111, 8) => (
- digits[0] as u16 * 1000
- + digits[1] as u16 * 100
- + digits[2] as u16 * 10
- + digits[3] as u16,
- digits[4] * 10 + digits[5],
- digits[6] * 10 + digits[7],
- ),
- _ => return None,
- };
- return NaiveDate::from_ymd_opt(year as _, month as _, day as _);
- }
-
- let (month, day) = match mask {
- 0b1101101111 => {
- if digits[7] != HYPHEN {
- return None;
- }
- (digits[5] * 10 + digits[6], digits[8] * 10 + digits[9])
- }
- 0b101101111 => {
- if digits[7] != HYPHEN {
- return None;
- }
- (digits[5] * 10 + digits[6], digits[8])
- }
- 0b110101111 => {
- if digits[6] != HYPHEN {
- return None;
- }
- (digits[5], digits[7] * 10 + digits[8])
+ use regex::Regex;
Review Comment:
You'll want to precompile / reuse the regex. That might return most of the
performance (although manual implementation probably would be still more
efficient).
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