将 Erlang 列表 X 拆分为所有 X 子列表的列表

发布于 2024-10-08 00:21:06 字数 347 浏览 14 评论 0原文

lists:sublist/2 和 lists:sublist/3 使从列表中提取单个子列表变得简单，但是是否有 BiF 或模块返回所有列表列表的子列表？

即

lists:awesome_sublist_function([1,2,3,4]) ->
  [[1], [2], [3], [4], [1,2], [1,3], [1,4], 
  [2,3], [2,4], [3,4], [1,2,3], [1,2,4], [1,3,4], [2,3,4], [1,2,3,4]]

可以构建我自己的，但想知道问题是否已经在任何地方得到解决？

原文

lists:sublist/2 and lists:sublist/3 make it simple to extract a single sublist from a list, but is there a BiF or module that returns a list of all of a list's sublists?

i.e.

lists:awesome_sublist_function([1,2,3,4]) ->
  [[1], [2], [3], [4], [1,2], [1,3], [1,4], 
  [2,3], [2,4], [3,4], [1,2,3], [1,2,4], [1,3,4], [2,3,4], [1,2,3,4]]

Can build my own, but wondered if the problem has been solved before anywhere?

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夏九 2024-10-15 00:21:06

我假设您的测试用例忘记了 [1,3,4]，但它可能看起来像这样：

-module(settheory).
-export([combinations/1]).

combinations([]) ->
    [];
combinations([H | T]) ->
    CT = combinations(T),
    [[H]] ++ [[H | L] || L <- CT] ++ CT.

-include_lib("eunit/include/eunit.hrl").
combinations_test() ->
    ?assertEqual(
       combinations([1,2,3,4]),
       lists:sort([[1], [2], [3], [4], [1,2], [1,3], [1,4],
                   [2,3], [2,4], [3,4], [1,2,3], [1,2,4], [1,3,4],
                   [2,3,4], [1,2,3,4]])),
    ok.

I assume your test case is forgetting [1,3,4], but it could look something like this:

-module(settheory).
-export([combinations/1]).

combinations([]) ->
    [];
combinations([H | T]) ->
    CT = combinations(T),
    [[H]] ++ [[H | L] || L <- CT] ++ CT.

-include_lib("eunit/include/eunit.hrl").
combinations_test() ->
    ?assertEqual(
       combinations([1,2,3,4]),
       lists:sort([[1], [2], [3], [4], [1,2], [1,3], [1,4],
                   [2,3], [2,4], [3,4], [1,2,3], [1,2,4], [1,3,4],
                   [2,3,4], [1,2,3,4]])),
    ok.

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