**Map** -base

*Note:* You should use Data.Map.Strict instead of this module if:
* You will eventually need all the values stored.
* The stored values don't represent large virtual data structures to be lazily computed.
An efficient implementation of ordered maps from keys to values (dictionaries).
These modules are intended to be imported qualified, to avoid name clashes with Prelude functions, e.g.
> import qualified Data.Map as Map
The implementation of Map is based on *size balanced* binary trees (or trees of *bounded balance*) as described by:
* Stephen Adams, "*Efficient sets: a balancing act*", Journal of Functional Programming 3(4):553-562, October 1993, http://www.swiss.ai.mit.edu/~adams/BB/.
* J. Nievergelt and E.M. Reingold, "*Binary search trees of bounded balance*", SIAM journal of computing 2(1), March 1973.
Note that the implementation is *left-biased* -- the elements of a first argument are always preferred to the second, for example in union or insert.
Operation comments contain the operation time complexity in the Big-O notation (http://en.wikipedia.org/wiki/Big_O_notation).
A Map from keys k to values a.

*O(n)* map f xs is the ByteString obtained by applying f to each element of xs
*O(n)* map f t is the Text obtained by applying f to each element of t. Subject to fusion. Performs replacement on invalid scalar values.
*O(n*min(n,W))*. map f s is the set obtained by applying f to each element of s.
It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y
*O(n)* map f xs is the ByteString obtained by applying f to each element of xs.
*O(n)* map f xs is the ByteString obtained by applying f to each element of xs. This function is subject to array fusion.
*O(n)*. Map a function over all values in the map.
> map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]

*O(n)*. Map a function over all values in the map.
> map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]

*O(n*log n)*. map f s is the set obtained by applying f to each element of s.
It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y
Show more results