Try to find first index of no_pad more intelligently.

Author: groutr

toolz/itertoolz.py

@@ -720,12 +720,22 @@ def partition_all(n, seq):
         yield prev
         prev = item
     if prev[-1] is no_pad:
-        # Get first index of no_pad without using .index()
-        # https://github.com/pytoolz/toolz/issues/387
-        for ind, item in enumerate(reversed(prev)):
-            if item is not no_pad:
-                yield prev[:len(prev)-ind]
-                break
+        try:
+            # If seq defines __len__, then we can quickly calculate where no_pad starts
+            yield prev[:len(seq) % n]
+        except TypeError:
+            # Get first index of no_pad without using .index()
+            # https://github.com/pytoolz/toolz/issues/387
+            # We can employ modified binary search here to speed things up from O(n) to O(log n)
+            # Binary search from CPython's bisect module, modified for identity testing.
+            lo, hi = 0, len(prev)
+            while lo < hi:
+                mid = (lo + hi) // 2
+                if prev[mid] is no_pad:
+                    hi = mid
+                else:
+                    lo = mid + 1
+            yield prev[:lo]
     else:
         yield prev