Numpy where() 使用随数组中项目位置而变化的条件

Question

我正在尝试使用 numpy 构建一个网格世界。

网格为4*4，排列成正方形。
第一个和最后一个方格（即 1 和 16）是终端方格。
在每个时间步，您可以向任意方向移动一步：上、下、左或右。一旦您进入终端方块之一，就不可能再进行进一步的移动，游戏就会终止。

第一列和最后一列是正方形的左边缘和右边缘，而第一行和最后一行代表顶部边缘和底部边缘。如果您位于边缘（例如左边的边缘）并尝试向左移动，您不会向左移动，而是停留在开始所在的方格中。类似地，如果您尝试穿过任何其他边，您也会停留在同一个方格中。

虽然网格是一个正方形，但我将它实现为一个数组。

states_r 计算右移后状态的位置。1 和 16 保持原样，因为它们是终止状态（请注意，代码使用基于零的计数，因此 1 和 16 在代码中分别是 0 和 15）。
其余的方格都加一。states_r 的代码可以工作，但是右边缘上的那些方块，即 (4, 8, 12) 也应该保持在原来的位置，但 states_r 代码不会这样做。

State_l 是我尝试包含正方形左边缘的边缘条件。逻辑是相同的，终端状态 (1, 16) 不应移动，左侧边缘的那些方块 (5, 9, 13) 也不应移动。我认为一般逻辑是正确的，但它产生了错误。

states = np.arange(16)
states_r = states[np.where((states + 1 <= 15) & (states != 0), states + 1, states)]
states_l = states[np.where((max(1, (states // 4) * 4) <= states - 1) & (states != 15), states - 1, states)]

第一个示例 states_r 有效，它处理终端状态，但不处理边缘条件。第二个例子是我尝试包含边缘条件，但是它给了我以下错误：

“具有多个元素的数组的真值是不明确的。”

有人可以解释一下如何修复我的代码吗？
或者建议另一种解决方案，理想情况下我希望代码速度快（这样我就可以扩展它），所以我想尽可能避免 for 循环？

Answer 1

如果我理解正确的话，你需要数组来指示每个状态的下一个状态，具体取决于移动（右、左、上、下）。如果是这样，我猜你的退出执行state_r不正确。我建议切换到网格的 2D 表示，因为如果直接有 x 和 y （至少对我来说），您描述的很多事情会更容易、更直观地处理。import numpy as npn = 4states = np.arange(n*n).reshape(n, n)states_r, states_l, states_u, states_d = (states.copy(), states.copy(),&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; states.copy(), states.copy())states_r[:, :n-1] = states[:, 1:]states_l[:, 1:] = states[:, :n-1]states_u[1:, :] = states[:n-1, :]states_d[:n-1, :] = states[1:, :]#&nbsp; &nbsp; &nbsp; &nbsp; up&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[[ 0,&nbsp; 1,&nbsp; 2,&nbsp; 3],#&nbsp; left state right&nbsp; &nbsp; &nbsp; [ 0,&nbsp; 1,&nbsp; 2,&nbsp; 3],#&nbsp; &nbsp; &nbsp; &nbsp;down&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[ 4,&nbsp; 5,&nbsp; 6,&nbsp; 7],#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [ 8,&nbsp; 9, 10, 11]]##&nbsp; [[ 0,&nbsp; 0,&nbsp; 1,&nbsp; 2],&nbsp; &nbsp;[[ 0,&nbsp; 1,&nbsp; 2,&nbsp; 3],&nbsp; &nbsp;[[ 1,&nbsp; 2,&nbsp; 3,&nbsp; 3],#&nbsp; &nbsp;[ 4,&nbsp; 4,&nbsp; 5,&nbsp; 6],&nbsp; &nbsp; [ 4,&nbsp; 5,&nbsp; 6,&nbsp; 7],&nbsp; &nbsp; [ 5,&nbsp; 6,&nbsp; 7,&nbsp; 7],#&nbsp; &nbsp;[ 8,&nbsp; 8,&nbsp; 9, 10],&nbsp; &nbsp; [ 8,&nbsp; 9, 10, 11],&nbsp; &nbsp; [ 9, 10, 11, 11],#&nbsp; &nbsp;[12, 12, 13, 14]]&nbsp; &nbsp; [12, 13, 14, 15]]&nbsp; &nbsp; [13, 14, 15, 15]]##&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[[ 4,&nbsp; 5,&nbsp; 6,&nbsp; 7],#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [ 8,&nbsp; 9, 10, 11],#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [12, 13, 14, 15],#&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [12, 13, 14, 15]]如果你想排除终端状态，你可以这样做：terminal_states = np.zeros((n, n), dtype=bool)terminal_states[0, 0] = Trueterminal_states[-1, -1] = Truestates_r[terminal_states] = states[terminal_states]states_l[terminal_states] = states[terminal_states]states_u[terminal_states] = states[terminal_states]states_d[terminal_states] = states[terminal_states]如果您更喜欢一维方法：import numpy as npn = 4states = np.arange(n*n)valid_s = np.ones(n*n, dtype=bool)valid_s[0] = Falsevalid_s[-1] = Falsestates_r = np.where(np.logical_and(valid_s, states % n < n-1), states+1, states)states_l = np.where(np.logical_and(valid_s, states % n > 0),&nbsp; &nbsp;states-1, states)states_u = np.where(np.logical_and(valid_s, states > n-1),&nbsp; &nbsp; &nbsp;states-n, states)states_d = np.where(np.logical_and(valid_s, states < n**2-n),&nbsp; states+n, states)

Answer 2

另一种无需预分配数组的方法：states = np.arange(16).reshape(4,4)states_l = np.hstack((states[:,0][:,None],states[:,:-1],))states_r = np.hstack((states[:,1:],states[:,-1][:,None]))states_d = np.vstack((states[1:,:],states[-1,:]))states_u = np.vstack((states[0,:],states[:-1,:]))为了将它们全部变为一维，您始终可以使用flatten()/ravel()/reshape(-1)二维数组。&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [[ 0&nbsp; 1&nbsp; 2&nbsp; 3]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[ 0&nbsp; 1&nbsp; 2&nbsp; 3]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[ 4&nbsp; 5&nbsp; 6&nbsp; 7]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[ 8&nbsp; 9 10 11]][[ 0&nbsp; 0&nbsp; 1&nbsp; 2]&nbsp; &nbsp; [[ 0&nbsp; 1&nbsp; 2&nbsp; 3]&nbsp; &nbsp; [[ 1&nbsp; 2&nbsp; 3&nbsp; 3]&nbsp;[ 4&nbsp; 4&nbsp; 5&nbsp; 6]&nbsp; &nbsp; &nbsp;[ 4&nbsp; 5&nbsp; 6&nbsp; 7]&nbsp; &nbsp; &nbsp;[ 5&nbsp; 6&nbsp; 7&nbsp; 7]&nbsp;[ 8&nbsp; 8&nbsp; 9 10]&nbsp; &nbsp; &nbsp;[ 8&nbsp; 9 10 11]&nbsp; &nbsp; &nbsp;[ 9 10 11 11]&nbsp;[12 12 13 14]]&nbsp; &nbsp; [12 13 14 15]]&nbsp; &nbsp; [13 14 15 15]]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [[ 4&nbsp; 5&nbsp; 6&nbsp; 7]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[ 8&nbsp; 9 10 11]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[12 13 14 15]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;[12 13 14 15]]对于角落，你可以这样做：states_u[-1,-1] = 15states_l[-1,-1] = 15