的顺序亚群说给出一个列表的列表
{"a", "b", "c", "d"}
有没有产生这样的顺序子集的列表更简单的方法(结果的顺序并不重要)
{
{"a"},
{"a b"},
{"a b c"},
{"a b c d"},
{"b"},
{"b c"},
{"b c d"},
{"c"},
{"c d"},
{"d"}
}
的顺序亚群说给出一个列表的列表
{"a", "b", "c", "d"}
有没有产生这样的顺序子集的列表更简单的方法(结果的顺序并不重要)
{
{"a"},
{"a b"},
{"a b c"},
{"a b c d"},
{"b"},
{"b c"},
{"b c d"},
{"c"},
{"c d"},
{"d"}
}
我想我喜欢这个最重要的是:
set = {"a", "b", "c", "d"};
ReplaceList[set, {___, x__, ___} :> {x}]
随着字符串加入:
ReplaceList[set, {___, x__, ___} :> "" <> Riffle[{x}, " "]]
与此类似,具体为字符串:
StringCases["abcd", __, Overlaps -> All]
由于纳赛尔说我作弊,这里是一个更手动的办法,也有大型成套更高的效率:
ClearAll[f, f2]
f[i_][x_] := NestList[i, x, [email protected] - 1]
f2[set_] := Join @@ (f[Most] /@ f[Rest][set])
f2[{"a", "b", "c", "d"}]
你可以像这样做:
a = {"a", "b", "c", "d"};
b = List[StringJoin[Riffle[#, " "]]] & /@
Flatten[Table[c = Drop[a, n];
Table[Take[c, i], {i, Length[c]}],
{n, 0, Length[a]}], 1]
输出将是这样的:
{{"a"}, {"a b"}, {"a b c"}, {"a b c d"}, {"b"}, {"b c"}, {"b c d"}, {"c"}, {"c d"}, {"d"}}
这里是一个可能的解决方案
a={"a","b","c","d"};
[email protected][#, " "] & /@
DeleteDuplicates[
LongestCommonSubsequence[a, #] & /@
DeleteCases[[email protected], {}]] // Column
结果
a
b
c
d
a b
b c
c d
a b c
b c d
a b c d
如何:
origset = {"a", "b", "c", "d"};
bdidxset = Subsets[Range[4], {1, 2}]
origset[[#[[1]] ;; #[[-1]]]] & /@ bdidxset
这给
{{"a"}, {"b"}, {"c"}, {"d"}, {"a", "b"}, {"a", "b", "c"}, {"a", "b",
"c", "d"}, {"b", "c"}, {"b", "c", "d"}, {"c", "d"}}
单程:
makeList[lst_] := Map[ Union[lst[[1 ;; #]]] &, [email protected][lst]]
r = Map[makeList[lst[[# ;; -1]]] &, [email protected][lst]];
Flatten[r, 1]
给出
{{"a"},
{"a", "b"},
{"a", "b", "c"},
{"a", "b", "c", "d"},
{"b"},
{"b", "c"},
{"b", "c", "d"},
{"c"},
{"c", "d"},
{"d"}}
Flatten[Partition[{a, b, c, d}, #, 1] & /@ {1, 2, 3, 4}, 1]
给出
{{A},{B},{C},{d},{一,b},{b,c},{c,d},{a,b,c},{b,c,d},{a, b,c,d}}
+1 ,比巫师先生复杂得多,但还是不错的。 – rcollyer 2012-01-16 02:50:13
我喜欢TomD的方法更好,但这是来到我的脑海里,SANS字符串处理:
set = {"a", "b", "c", "d"};
n = [email protected];
Join @@ Table[set~Take~{s, f}, {s, n}, {f, s, n}] // Column
为什么两个答案? – abcd 2012-01-14 15:36:49
@yoda更多代表,当然 – 2012-01-14 17:45:27
@yoda我认为答案是充分不同的性格。我看到一个综合性答案的问题:不清楚人们投票选择哪种方法,贬低投票的目的,选民可能会觉得如果他们想要投票,他们必须为不喜欢的方法投票(或彻底失败)为他们所做的一票投票。与彼得的断言相反,我不是在玩游戏票。 – 2012-01-15 00:02:30
+1,这是不公正的先生,你让Mathematica用这种方法做所有的辛苦工作。你应该做点什么! – Nasser 2012-01-14 13:31:13
这是神奇的,不是编程! – acl 2012-01-14 13:36:54
这太棒了!我认为这与Mathematica的核心概念非常吻合,而不是作弊:) – Silvia 2012-01-14 15:32:42