3 stable releases
3.0.0 | Nov 1, 2024 |
---|---|
2.0.0 | Aug 9, 2024 |
1.0.0 | Aug 7, 2024 |
#462 in Visualization
14KB
219 lines
split-every
use split_every::prelude::*;
// This prints: [(0, 0), (0, 1)]
// [(0, 0)]
// [(0, 1), (0, 0)]
// [(0, 1)]
let mut splitter: SplitEvery<&[(u8, u8)], &[(u8, u8)]> = [
(0, 0), (0, 1), (0, 0),
(0, 0), (0, 0), (0, 1),
(0, 0), (0, 0), (0, 1),
].split_every_n_times(&[(0, 0)], 2);
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
// This prints: "Oh hi there"
// "I don't really"
// "know what to"
// "say".
let mut splitter: SplitEvery<&str, &str> =
"Oh hi there I don't really know what to say".split_every_n_times(" ", 3);
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
// This prints: ["This", "is", "you", "This"]
// ["me", "This", "is", "someone", "This"]
// ["them"]
let mut iter = [
["This", "is", "you"],
["This", "is", "me"],
["This", "is", "someone"],
["This", "is", "them"],
].iter().flatten().map(|val| *val);
let mut splitter: SplitEvery<Box<dyn FnMut() -> Option<&'static str>>, &str> =
SplitEvery::n_times_from_fn(Box::new(move || iter.next()), "is", 2);
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
println!("{:?}", splitter.next().unwrap());
✨ Split For Every N Occurrences Of A Pattern Iteratively
This crate helps you split data for every n
occurrences of a pattern
.
It contains an exclusive iterator
.
📄 Licensing
split-every
is licensed under the MIT LICENSE
; This is the summarization
.