如何在prolog中转置矩阵

时间:2010-11-25 21:34:21

标签: list matrix prolog transpose

如何将[[1,2,3][4,5,6][6,7,8]]之类的列表转置为[[1,4,6],[2,7,8],[3,6,9]]

描绘它:我想将矩阵向左翻转90度。我怎么能这样做?

8 个答案:

答案 0 :(得分:7)

不确定你的例子是否正确,但我明白了。

如果使用SWI-PROLOG,您可以使用CLPFD module,如下所示:

:- use_module(library(clpfd)).

允许您使用transpose/2谓词,如下所示:

1 ?- transpose([[1,2,3],[4,5,6],[6,7,8]], X).
X = [[1, 4, 6], [2, 5, 7], [3, 6, 8]].

否则(如果没有SWI-PROLOG),你可以简单地使用这个实现(在SWI的clpfd中碰巧是旧实现):

transpose([], []).
transpose([F|Fs], Ts) :-
    transpose(F, [F|Fs], Ts).

transpose([], _, []).
transpose([_|Rs], Ms, [Ts|Tss]) :-
        lists_firsts_rests(Ms, Ts, Ms1),
        transpose(Rs, Ms1, Tss).

lists_firsts_rests([], [], []).
lists_firsts_rests([[F|Os]|Rest], [F|Fs], [Os|Oss]) :-
        lists_firsts_rests(Rest, Fs, Oss).

有关使用foldl和maplist内置插件的更新版本,请参阅clpfd.pl

答案 1 :(得分:5)

这是我能提出的最小解决方案。

代码

transpose([[]|_], []).
transpose(Matrix, [Row|Rows]) :- transpose_1st_col(Matrix, Row, RestMatrix),
                                 transpose(RestMatrix, Rows).
transpose_1st_col([], [], []).
transpose_1st_col([[H|T]|Rows], [H|Hs], [T|Ts]) :- transpose_1st_col(Rows, Hs, Ts).

测试

:- transpose([[1,2,3],
              [4,5,6],
              [7,8,9]], R),
   print(R).

打印:

[[1,4,7],
 [2,5,8],
 [3,6,9]]

解释

它的工作方式是transpose递归调用transpose_1st_col,它提取并转置矩阵的第一列。例如:

:- transpose_1st_col([[1,2,3],
                      [4,5,6],
                      [7,8,9]], Row, RestMatrix),
   print(Row),
   print(RestMatrix).

将打印

[1,4,7]

[[2,3],
 [5,6],
 [8,9]]

重复此操作直到输入矩阵为空,此时所有列都已转置。然后将转置的列连接到转置矩阵中。

答案 2 :(得分:2)

这是a fragment of a larger answer

% transposed(+A, ?B) iff matrix B is transposed matrix A
transposed(A, B) :- transposed(A, [], B).
transposed(M, X, X) :- empty(M), !.
transposed(M, A, X) :- columns(M, Hs, Ts), transposed(Ts, [Hs|A], X).

% empty(+A) iff A is empty list or a list of empty lists
empty([[]|A]) :- empty(A).
empty([]).

% columns(+M, ?Hs, ?Ts) iff Hs is the first column
%   of matrix M and Ts is the rest of matrix M
columns([[Rh|Rt]|Rs], [Rh|Hs], [Rt|Ts]) :- columns(Rs, Hs, Ts).
columns([[]], [], []).
columns([], [], []).

答案 3 :(得分:1)

另一种简单的方法:

transpose(M0, M) :-
    nonvar(M0),
    findall(L, maplist(nth1(_), M0, L), M).

?- transpose([[1,2,3],[4,5,6],[7,8,9]], M). 
M = [[1, 4, 7], [2, 5, 8], [3, 6, 9]]. `

答案 4 :(得分:0)

迭代方法:

trans([H|R],[H1|R1]):-trans2([H|R],[H|R],[],[H1|R1],0),!.
trans2([A|_],_,_,[],N):-length(A,N).
trans2(M,[],H1,[H1|R1],N):-N1 is N+1, trans2(M,M,[],R1,N1).
trans2(M,[H|R],L,[H1|R1],N):-nth0(N,H,X),
   append(L,[X],L1),trans2(M,R,L1,[H1|R1],N).

答案 5 :(得分:0)

我的解决方案全名,以便更好地理解:

% emptyMatrix(Line, EmptyMatrix)
emptyMatrix([],[]).
emptyMatrix([_|T1],[[]|T2]):-emptyMatrix(T1,T2).
% only length of parameter 'Line' is interesting. It ignores its content.    

% appendElement(Element, InputList, OutputList)
appendElement(E,[],[E]).
appendElement(E,[H|T],[H|L]):-appendElement(E,T,L).

% appendTransposed(NestedList, InputMatrix, OutputMatrix)
appendTransposed([],[],[]).
appendTransposed([X|T1],[],[[X]|T3]):-appendTransposed(T1,[],T3).
appendTransposed([X|T1],[R|T2],[C|T3]):-appendElement(X,R,C),appendTransposed(T1,T2,T3).

% transposeMatrix(InputMatrix, TransposedMatrix)
transposeMatrix([L|M],T):-emptyMatrix(L,A),transpose([L|M],T,A).
transpose([],T,T).
transpose([L|M],T,A):-appendTransposed(L,A,B),transpose(M,T,B).

'line'可以是col或row。

这个想法在于将元素附加到空矩阵的列表中。 (例如,第一行的所有元素=所有cols的第一个元素 =>第一行第i行的所有元素=所有列的第i个元素

它可以在我的机器上运行,因为此会话协议向我显示:

5 ?- transposeMatrix([[1,2],[3,4]],T).
T = [[1, 3], [2, 4]] ;
false.

6 ?- transposeMatrix([[1],[2]],T).
T = [[1, 2]] ;
false.

7 ?- transposeMatrix([[1,2,3],[4,5,6]],T).
T = [[1, 4], [2, 5], [3, 6]] ;
false.

8 ?- transposeMatrix([[1]],T).
T = [[1]] ;
false.

答案 6 :(得分:0)

更简单的方法:

trans(M, [P|T]):- first(M, P, A), trans(A, T).
trans(Empty, []):- empty(Empty).

empty([[]|T]):- empty(T).
empty([[]]).

first([[P|A]|R], [P|Ps], [A|As]):- first(R, Ps, As).
first([], [], []).

效率也很高

[debug] 36 ?- time(trans([[1,2,3],[4,5,6],[7,8,9]],A)).
% 21 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips)
A = [[1,4,7],[2,5,8],[3,6,9]] ;
% 12 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips)
false.

答案 7 :(得分:0)

另一种方法:

delete_one_list([], []).
delete_one_list([[_|L]|LLs], [L|Ls]) :-
  delete_one_list(LLs, Ls).

transpose_helper([], []).
transpose_helper([[X|_]|Xs], [X|Ys]) :-
  transpose_helper(Xs, Ys).

transpose([[]|_], []).
transpose(List, [L|Ls]) :-
  transpose_helper(List, L),
  delete_one_list(List, NewList),
  transpose(NewList, Ls).