在 Erlang 中转置二维矩阵

发布于 2024-10-25 01:54:57 字数 313 浏览 5 评论 0原文

给定如下所示的矩阵，将其变换 90 度，转换为下面的第二个矩阵。您将如何以最干净的方式做到这一点？首选简短/简洁/清晰的解决方案，其中要点易于掌握。

从

[[A1,A2,A3],
 [B1,B2,B3],
 [C1,C2,C3]]

到

[[A1,B1,C1],
 [A2,B2,C2],
 [A3,B3,C3]]

编辑：我意识到原来的问题并不清楚。我想知道如何在 Erlang 中执行此操作。

原文

Given a matrix like below, transform it, say, 90 degrees into the second matrix below. How would you go about doing this in the cleanest way possible? Short/succinct/clear solutions where the point is easy to grasp is preferred.

From

[[A1,A2,A3],
 [B1,B2,B3],
 [C1,C2,C3]]

[[A1,B1,C1],
 [A2,B2,C2],
 [A3,B3,C3]]

Edit: I realize it was not clear from original question. I'd like to know how to do this in Erlang.

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分开我的手 2024-11-01 01:54:57

简化已经给出的解决方案，您可以在最短的时间内完成：

-module(transp).

-export([transpose/1]).

transpose([[]|_]) -> [];
transpose(M) ->
  [lists:map(fun hd/1, M) | transpose(lists:map(fun tl/1, M))].

Simplifying the solutions already given, you can do it in as short as:

-module(transp).

-export([transpose/1]).

transpose([[]|_]) -> [];
transpose(M) ->
  [lists:map(fun hd/1, M) | transpose(lists:map(fun tl/1, M))].

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灯角 2024-11-01 01:54:57

这是我的示例解决方案：

-module(transp).

-export([transpose/1]).

transpose(L) ->
     transpose_do([], L).

transpose_do(Acc, [[]|_]) ->
     lists:reverse(Acc);
transpose_do(Acc, M) ->
     Row = lists:foldr(
          fun(Elem, FoldAcc) ->
                    [hd(Elem) | FoldAcc]
          end,
          [],
          M),
     transpose_do([Row|Acc], lists:map(fun(X) -> tl(X) end, M)).

测试：

1> M = [[a1,a2,a3],[b1,b2,b3],[c1,c2,c3]].
[[a1,a2,a3],[b1,b2,b3],[c1,c2,c3]]
2> transp:transpose(M).   
[[a1,b1,c1],[a2,b2,c2],[a3,b3,c3]]

Here's my sample solution:

-module(transp).

-export([transpose/1]).

transpose(L) ->
     transpose_do([], L).

transpose_do(Acc, [[]|_]) ->
     lists:reverse(Acc);
transpose_do(Acc, M) ->
     Row = lists:foldr(
          fun(Elem, FoldAcc) ->
                    [hd(Elem) | FoldAcc]
          end,
          [],
          M),
     transpose_do([Row|Acc], lists:map(fun(X) -> tl(X) end, M)).

Test:

1> M = [[a1,a2,a3],[b1,b2,b3],[c1,c2,c3]].
[[a1,a2,a3],[b1,b2,b3],[c1,c2,c3]]
2> transp:transpose(M).   
[[a1,b1,c1],[a2,b2,c2],[a3,b3,c3]]

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酒几许 2024-11-01 01:54:57

这是我认为我从 Haskell 标准库获得的一个实现：

%% Transpose rows and columns in a list of lists. Works even if sublists
%% are not of same length. Empty sublists are stripped.
transpose([[X | Xs] | Xss]) ->
    [[X | [H || [H | _] <- Xss]]
     | transpose([Xs | [T || [_ | T] <- Xss]])];
transpose([[] | Xss]) -> transpose(Xss);
transpose([]) -> [].

紧凑且有点令人费解。

Here's an implementation that I think I got from the Haskell standard library:

%% Transpose rows and columns in a list of lists. Works even if sublists
%% are not of same length. Empty sublists are stripped.
transpose([[X | Xs] | Xss]) ->
    [[X | [H || [H | _] <- Xss]]
     | transpose([Xs | [T || [_ | T] <- Xss]])];
transpose([[] | Xss]) -> transpose(Xss);
transpose([]) -> [].

Compact and slightly mind-bending.

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音栖息无 2024-11-01 01:54:57

在函数式编程语言中，矩阵转置的常用方法是使用unzip。

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你的心境我的脸 2024-11-01 01:54:57

您显示的不是矩阵旋转，而是矩阵转置。如果你调用第一个矩阵 A 和第二个 B ，那么你必须

A[i,j] = B[j,i]

从 A 到 B 你只需要两个嵌套循环，其中 i = 1 到 n 和 j = i+1 到 n ，并且在每次迭代时交换非对角线使用临时变量的条目。

What you are showing is not a matrix rotation, but rather matrix transposition. If you call the first matrix A and the second B then you have

A[i,j] = B[j,i]

To go from A to B you just need two nested loops with i = 1 to n and j = i+1 to n and at each iteration you swap the off-diagonal entries using a temporary variable.

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