# lexicographicallyPrecedes(_:by:) Returns a Boolean value indicating whether the sequence precedes another sequence in a lexicographical (dictionary) ordering, using the given predicate to compare elements. ``` func lexicographicallyPrecedes(_ other: OtherSequence, by areInIncreasingOrder: (Self.Element, Self.Element) throws -> Bool) rethrows -> Bool where OtherSequence : Sequence, Self.Element == OtherSequence.Element ``` ## Parameters `other` A sequence to compare to this sequence. `areInIncreasingOrder` A predicate that returns `true` if its first argument should be ordered before its second argument; otherwise, `false`. ## Return Value `true` if this sequence precedes `other` in a dictionary ordering as ordered by `areInIncreasingOrder`; otherwise, `false`. ## Discussion The predicate must be a *strict weak ordering* over the elements. That is, for any elements `a`, `b`, and `c`, the following conditions must hold: - `areInIncreasingOrder(a, a)` is always `false`. (Irreflexivity) - If `areInIncreasingOrder(a, b)` and `areInIncreasingOrder(b, c)` are both `true`, then `areInIncreasingOrder(a, c)` is also `true`. (Transitive comparability) - Two elements are *incomparable* if neither is ordered before the other according to the predicate. If `a` and `b` are incomparable, and `b` and `c` are incomparable, then `a` and `c` are also incomparable. (Transitive incomparability) > Note: This method implements the mathematical notion of lexicographical > ordering, which has no connection to Unicode. If you are sorting > strings to present to the end user, use `String` APIs that perform > localized comparison instead. > Complexity: O(*m*), where *m* is the lesser of the length of the > sequence and the length of `other`. --- Copyright © 2026 Apple Inc. All rights reserved. | [Terms of Use](https://www.apple.com/legal/internet-services/terms/site.html) | [Privacy Policy](https://www.apple.com/privacy/privacy-policy)