Itertools::tail: skip the starting part of the iterator

If the iterator is exact sized, then `.collect()` finishes the work.
More generally, if the size hint knows enough and `nth` is efficient, this might skip most of the iterator efficiently.

In the tests, `.filter(..)` is there to ensure that `tail` can't leverage a precise `size_hint` to entirely skip the iteration after initial `.collect()`.

Co-Authored-By: scottmcm <scottmcm@users.noreply.github.com>

Author: Philippe-Cholet

src/lib.rs

@@ -3147,6 +3147,8 @@ pub trait Itertools: Iterator {
     /// assert_equal(v.iter().tail(10), &v);
     ///
     /// assert_equal((0..100).tail(10), 90..100);
+    ///
+    /// assert_equal((0..100).filter(|x| x % 3 == 0).tail(10), (72..100).step_by(3));
     /// ```
     ///
     /// For double ended iterators without side-effects, you might prefer
@@ -3164,7 +3166,9 @@ pub trait Itertools: Iterator {
             }
             1 => self.last().into_iter().collect(),
             _ => {
-                let mut iter = self.fuse();
+                // Skip the starting part of the iterator if possible.
+                let (low, _) = self.size_hint();
+                let mut iter = self.fuse().skip(low.saturating_sub(n));
                 let mut data: Vec<_> = iter.by_ref().take(n).collect();
                 // Update `data` cyclically.
                 let idx = iter.fold(0, |i, val| {

tests/quick.rs

@@ -1952,6 +1952,8 @@ quickcheck! {
 
     fn tail(v: Vec<i32>, n: u8) -> bool {
         let n = n as usize;
-        itertools::equal(v.iter().tail(n), &v[v.len().saturating_sub(n)..])
+        let result = &v[v.len().saturating_sub(n)..];
+        itertools::equal(v.iter().tail(n), result)
+            && itertools::equal(v.iter().filter(|_| true).tail(n), result)
     }
 }