有效地將Numpy數組重新排列/重新排列爲正確排序的Tile（圖像）

Question

我希望能夠以某種方式對numpy數組重新排序以便高效處理tile。

我得到了什麼：

>>> A = np.array([[1,2],[3,4]]).repeat(2,0).repeat(2,1) 
>>> A # image like array 
array([[[1, 1, 2, 2], 
     [1, 1, 2, 2]], 

     [[3, 3, 4, 4], 
     [3, 3, 4, 4]]]) 

>>> A.reshape(2,2,4) 
array([[[1, 1, 2, 2], 
     [1, 1, 2, 2]], 

     [[3, 3, 4, 4], 
     [3, 3, 4, 4]]]) 

我想要什麼：如何獲得X：

>>> X 
array([[[1, 1, 1, 1], 
     [2, 2, 2, 2]], 

     [[3, 3, 3, 3], 
     [4, 4, 4, 4]]]) 

這是可能沒有一個緩慢的蟒蛇循環？

獎勵：轉換從X到後面有一個

巴戈指出了一個類似的問題，我已經錯過了：Creating a 4D view on 2D array to divide it into cells of fixed size

---

爲什麼所有的麻煩呢？爲了能夠做這樣的事情：

>>> X[X.sum(2)>12] -= 1 
>>> X 
array([[[1, 1, 1, 1], 
     [2, 2, 2, 2]], 

     [[3, 3, 3, 3], 
     [3, 3, 3, 3]]]) 

編輯：解：

# MIT - License 

import numpy as np 

def tile_reorder(A, t): 
    """reshape and transpose quadratic array for easy access to quadratic tiles of size t""" 
    l = A.shape[0]/t 
    X = A.reshape((l, t, l, t)) 
    X = X.transpose([0, 2, 1, 3]) 
    return X 

def tile_reorder_reverse(X): 
    l = X.shape[0] * X.shape[2] 
    A = X.transpose([0, 2, 1, 3]) 
    A = A.reshape((l, l)) 
    return A 

if __name__ == "__main__": 
    A = np.array([[1,2,3],[3,4,5],[6,7,8]]).repeat(4,0).repeat(4,1) 

    print "A:\n", A, "\n\n" 

    print "A_tiled:\n", tile_reorder(A,4), "\n\n" 

    print "A_tiled_reversed:\n", tile_reorder_reverse(tile_reorder(A,4)), "\n\n" 

    X = tile_reorder(A,4) 
    X[X.sum((2)).sum(2)>63,:,:] += 10 
    B = tile_reorder_reverse(X) 

    print "B_processed:\n", B, "\n\n" 

Answer 1

你想要的東西，如：

>>> X = A.reshape(2,2,2,2) 
>>> X = X.transpose([0,2,1,3]) 
>>> X = X.reshape((2,2,-1)) 
>>> X 
array([[[1, 1, 1, 1], 
     [2, 2, 2, 2]], 

     [[3, 3, 3, 3], 
     [4, 4, 4, 4]]]) 
>>> B = X.reshape((2,2,2,2)) 
>>> B = B.transpose([0,2,1,3]) 
>>> B = B.reshape((2,2,-1)) 
>>> B 
array([[[1, 1, 2, 2], 
     [1, 1, 2, 2]], 

     [[3, 3, 4, 4], 
     [3, 3, 4, 4]]]) 

Answer 2

我覺得你只是看你的陣列的錯片。以下應該做你想要的。

>>> X[(X.sum(2)>12),:] -= 1 
>>> X 
array([[[1, 1, 1, 1], 
     [2, 2, 2, 2]], 

     [[3, 3, 3, 3], 
     [3, 3, 3, 3]]]) 

回到原來的順序有點複雜。我相信你可以將下面的代碼修改成一個名爲get_back()的函數或者其他東西，所以你不必一直重複使用它。

import numpy as np 
X = np.asarray([[[1,1,1,1],[2,2,2,2]],[[3,3,3,3],[4,4,4,4]]]) 
print X 

# Make the change to the array 
# Note that I make the change to +=1 so we can indeed see that 
# the reshaping worked by looking for the 5's. 
X[(X.sum(2)>12),:] += 1 

(M,N,P) = X.shape;  # Number of tiles is MxN 
orig_i = 2; orig_j = 2; # (Original rows/cols in a tile). 
orig = (orig_i, orig_j) # Define as a tuple for convenience. 
hstacks = [] 
for ii in range(M): 
    hstacks.append(X[ii,0,:].reshape(orig)) 
    for jj in range(1,N): 
     hstacks[-1] = np.hstack((hstacks[-1], X[ii,jj,:].reshape(orig))) 

orig_array = hstacks[0] 
for kk in range(1,len(hstacks)): 
    orig_array = np.vstack((orig_array, hstacks[kk])) 

# Print to confirm that it worked. 
print orig_array 

我得到如下：

>>> [[[1 1 1 1] 
     [2 2 2 2]] 

    [[3 3 3 3] 
     [4 4 4 4]]] 

>>> [[1 1 2 2] 
    [1 1 2 2] 
    [3 3 5 5] 
    [3 3 5 5]] 

，並測試，我與瓷磚的大格局嘗試它，以及它給出正確的結果。請注意，我使用函數get_back()來封裝上面的代碼，並且我只是在平鋪尺寸參數中對函數進行了硬編碼。真的，他們應該是輸入參數。

>>> X = np.asarray([[[1,1,1,1],[2,2,2,2],[3,3,3,3]], 
        [[4,4,4,4],[5,5,5,5],[6,6,6,6]]]) 
>>> X 
array([[[1, 1, 1, 1], 
     [2, 2, 2, 2], 
     [3, 3, 3, 3]], 

     [[4, 4, 4, 4], 
     [5, 5, 5, 5], 
     [6, 6, 6, 6]]]) 

>>> get_back(X) 
array([[1 1 2 2 3 3] 
     [1 1 2 2 3 3] 
     [4 4 5 5 6 6] 
     [4 4 5 5 6 6]])