Part of my program requires me to be able to randomly shuffle list elements. I need a function such that when i give it a list, it will pseudo-randomly re-arrange the elements in the list.
A change in arrangement Must be visible at each call with the same list.
My implementation seems to work just fine but i feel that its rather long and is increasing my code base and also, i have a feeling that it ain't the best solution for doing this. So i need a much shorter implementation. Here is my implementation:
-module(shuffle). -export([list/1]). -define(RAND(X),random:uniform(X)). -define(TUPLE(Y,Z,E),erlang:make_tuple(Y,Z,E)). list(L)-> Len = length(L), Nums = lists:seq(1,Len), tuple_to_list(?TUPLE(Len,[],shuffle(Nums,L,[]))). shuffle([],_,Buffer)-> Buffer; shuffle(Nums,[Head|Items],Buffer)-> {Pos,NewNums} = pick_position(Nums), shuffle(NewNums,Items,[{Pos,Head}|Buffer]). pick_position([N])-> {N,[]}; pick_position(Nos)-> T = lists:max(Nos), pick(Nos,T). pick(From,Max)-> random:seed(begin (case random:seed(now()) of undefined -> NN = element(3,now()), {?RAND(NN),?RAND(NN),?RAND(NN)}; Any -> Any end) end ), T2 = random:uniform(Max), case lists:member(T2,From) of false -> pick(From,Max); true -> {T2,From -- [T2]} end.
On running it in shell:
F:\> erl Eshell V5.8.4 (abort with ^G) 1> c(shuffle). {ok,shuffle} 2> shuffle:list([a,b,c,d,e]). [c,b,a,e,d] 3> shuffle:list([a,b,c,d,e]). [e,c,b,d,a] 4> shuffle:list([a,b,c,d,e]). [a,b,c,e,d] 5> shuffle:list([a,b,c,d,e]). [b,c,a,d,e] 6> shuffle:list([a,b,c,d,e]). [c,e,d,b,a]I am motivated by the fact that in the STDLIB there is no such function. Somewhere in my game, i need to shuffle things up and also i need to find the best efficient solution to the problem, not just one that works.
Python Random shuffle() Method The shuffle() method takes a sequence, like a list, and reorganize the order of the items. Note: This method changes the original list, it does not return a new list.
The random. sample() function returns the random list with the sample size you passed to it. For example, sample(myList, 3) will return a list of 3 random items from a list. If we pass the sample size the same as the original list's size, it will return us the new shuffled list.
shuffle() function in Python. The shuffle() is an inbuilt method of the random module. It is used to shuffle a sequence (list).
Python3. This is most recommended method to shuffle a list. Python in its random library provides this inbuilt function which in-place shuffles the list.
1> L = lists:seq(1,10).
[1,2,3,4,5,6,7,8,9,10]
Associate a random number R with each element X in L by making a list of tuples {R, X}. Sort this list and unpack the tuples to get a shuffled version of L.
1> [X||{_,X} <- lists:sort([ {random:uniform(), N} || N <- L])].
[1,6,2,10,5,7,9,3,8,4]
2>
Please note that karl's answer is much more concise and simple.
Here's a fairly simple solution, although not necessarily the most efficient:
-module(shuffle).
-export([list/1]).
list([]) -> [];
list([Elem]) -> [Elem];
list(List) -> list(List, length(List), []).
list([], 0, Result) ->
Result;
list(List, Len, Result) ->
{Elem, Rest} = nth_rest(random:uniform(Len), List),
list(Rest, Len - 1, [Elem|Result]).
nth_rest(N, List) -> nth_rest(N, List, []).
nth_rest(1, [E|List], Prefix) -> {E, Prefix ++ List};
nth_rest(N, [E|List], Prefix) -> nth_rest(N - 1, List, [E|Prefix]).
For example, one could probably do away with the ++
operation in nth_rest/3
. You don't need to seed the random algorithm in every call to random
. Seed it initially when you start your program, like so: random:seed(now())
. If you seed it for every call to uniform/1
your results become skewed (try with [shuffle:list([1,2,3]) || _ <- lists:seq(1, 100)]
).
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