Numpy將立方體拆分成立方體

Question

有一個功能np.split()可以沿1軸拆分數組。例如，我想知道是否有多軸版本可以沿軸（0,1,2）分割。

Answer 1

假設cube的形狀爲(W, H, D)並且您希望將其分解爲N形狀的小立方體(w, h, d)。由於NumPy陣列具有固定長度的軸，因此w必須均勻分配W，並且類似地h和d。

然後有一種方法可以將形狀爲(W, H, D)的立方體重新塑造成形狀爲(N, w, h, d)的新陣列。

例如，如果arr = np.arange(4*4*4).reshape(4,4,4)（所以(W,H,D) = (4,4,4)），我們希望把它分解成形狀(2,2,2)的立方體，然後我們可以使用

In [283]: arr.reshape(2,2,2,2,2,2).transpose(0,2,4,1,3,5).reshape(-1,2,2,2) 
Out[283]: 
array([[[[ 0, 1], 
     [ 4, 5]], 

     [[16, 17], 
     [20, 21]]], 

... 
     [[[42, 43], 
     [46, 47]], 

     [[58, 59], 
     [62, 63]]]]) 

這裏的想法是多餘的軸添加到陣列排序充當位置標記的：

number of repeats act as placemarkers 
o---o---o 
| | | 
v v v 
(2,2,2,2,2,2) 
^^^
    | | | 
    o---o---o 
    newshape 

然後，我們可以重新排序的軸（使用transpose），使得重複的次數至上，和newshape自帶底：

arr.reshape(2,2,2,2,2,2).transpose(0,2,4,1,3,5) 

最後，請撥reshape(-1, w, h, d)將所有地標軸壓扁到一個軸上。這產生了一個形狀數組(N, w, h, d)其中N是小立方體的數量。

---

以上使用的思想是將this idea推廣到3維。它可以進一步推廣到任何尺寸的ndarrays：

import numpy as np 
def cubify(arr, newshape): 
    oldshape = np.array(arr.shape) 
    repeats = (oldshape/newshape).astype(int) 
    tmpshape = np.column_stack([repeats, newshape]).ravel() 
    order = np.arange(len(tmpshape)) 
    order = np.concatenate([order[::2], order[1::2]]) 
    # newshape must divide oldshape evenly or else ValueError will be raised 
    return arr.reshape(tmpshape).transpose(order).reshape(-1, *newshape) 

print(cubify(np.arange(4*6*16).reshape(4,6,16), (2,3,4)).shape) 
print(cubify(np.arange(8*8*8*8).reshape(8,8,8,8), (2,2,2,2)).shape) 

產生形狀的新的陣列

(16, 2, 3, 4) 
(256, 2, 2, 2, 2) 

---

到 「uncubify」 陣列：

def uncubify(arr, oldshape): 
    N, newshape = arr.shape[0], arr.shape[1:] 
    oldshape = np.array(oldshape)  
    repeats = (oldshape/newshape).astype(int) 
    tmpshape = np.concatenate([repeats, newshape]) 
    order = np.arange(len(tmpshape)).reshape(2, -1).ravel(order='F') 
    return arr.reshape(tmpshape).transpose(order).reshape(oldshape) 

---

這裏有一些測試代碼來檢查cubify和uncubify是反轉。

import numpy as np 
def cubify(arr, newshape): 
    oldshape = np.array(arr.shape) 
    repeats = (oldshape/newshape).astype(int) 
    tmpshape = np.column_stack([repeats, newshape]).ravel() 
    order = np.arange(len(tmpshape)) 
    order = np.concatenate([order[::2], order[1::2]]) 
    # newshape must divide oldshape evenly or else ValueError will be raised 
    return arr.reshape(tmpshape).transpose(order).reshape(-1, *newshape) 

def uncubify(arr, oldshape): 
    N, newshape = arr.shape[0], arr.shape[1:] 
    oldshape = np.array(oldshape)  
    repeats = (oldshape/newshape).astype(int) 
    tmpshape = np.concatenate([repeats, newshape]) 
    order = np.arange(len(tmpshape)).reshape(2, -1).ravel(order='F') 
    return arr.reshape(tmpshape).transpose(order).reshape(oldshape) 

tests = [[np.arange(4*6*16), (4,6,16), (2,3,4)], 
     [np.arange(8*8*8*8), (8,8,8,8), (2,2,2,2)]] 

for arr, oldshape, newshape in tests: 
    arr = arr.reshape(oldshape) 
    assert np.allclose(uncubify(cubify(arr, newshape), oldshape), arr) 
    # cuber = Cubify(oldshape,newshape) 
    # assert np.allclose(cuber.uncubify(cuber.cubify(arr)), arr) 

Answer 2

我不認爲有一個多軸版本，你可以沿某些給定的軸分割。但是你可以一次把它分成一個維度。例如像這樣：

def split2(arys, sections, axis=[0, 1]): 
    if not isinstance(arys, list): 
     arys = [arys] 
    for ax in axis: 
     arys = [np.split(ary, sections, axis=ax) for ary in arys] 
     arys = [ary for aa in arys for ary in aa] # Flatten 
    return arys 

它可以像這樣使用：

In [1]: a = np.array(range(100)).reshape(10, 10) 
In [2]: split2(a, 2, axis=[0, 1]) 
Out[2]: 
[array([[ 0, 1, 2, 3, 4], 
     [10, 11, 12, 13, 14], 
     [20, 21, 22, 23, 24], 
     [30, 31, 32, 33, 34], 
     [40, 41, 42, 43, 44]]), 
array([[ 5, 6, 7, 8, 9], 
     [15, 16, 17, 18, 19], 
     [25, 26, 27, 28, 29], 
     [35, 36, 37, 38, 39], 
     [45, 46, 47, 48, 49]]), 
array([[50, 51, 52, 53, 54], 
     [60, 61, 62, 63, 64], 
     [70, 71, 72, 73, 74], 
     [80, 81, 82, 83, 84], 
     [90, 91, 92, 93, 94]]), 
array([[55, 56, 57, 58, 59], 
     [65, 66, 67, 68, 69], 
     [75, 76, 77, 78, 79], 
     [85, 86, 87, 88, 89], 
     [95, 96, 97, 98, 99]])] 

Answer 3

除了我的額外問題@ unutbu的答案，我想我已經得到了相反的工作（在你想要的情況下將立方體拆分爲立方體，對每個立方體應用一個功能，然後將它們組合起來）。

import numpy as np 
import pdb 
np.set_printoptions(precision=3,linewidth=300) 

class Cubify(): 
    def __init__(self,oldshape,newshape): 
     self.newshape = np.array(newshape) 
     self.oldshape = np.array(oldshape) 
     self.repeats = (oldshape/newshape).astype(int) 
     self.tmpshape = np.column_stack([self.repeats, newshape]).ravel() 
     order = np.arange(len(self.tmpshape)) 
     self.order = np.concatenate([order[::2], order[1::2]]) 
     self.reverseOrder = self.order.copy() 
     self.reverseOrder = np.arange(len(self.tmpshape)).reshape(2, -1).ravel(order='F') 
     self.reverseReshape = np.concatenate([self.repeats,self.newshape]) 

    def cubify(self,arr): 
     # newshape must divide oldshape evenly or else ValueError will be raised 
     return arr.reshape(self.tmpshape).transpose(self.order).reshape(-1, *self.newshape) 

    def uncubify(self,arr): 
     return arr.reshape(self.reverseReshape).transpose(self.reverseOrder).reshape(self.oldshape) 

if __name__ == "__main__": 
    N = 9 
    x = np.arange(N**3).reshape(N,N,N) 
    oldshape = x.shape 
    newshape = np.array([3,3,3]) 
    cuber = Cubify(oldshape,newshape) 
    out = cuber.cubify(x) 
    back = cuber.uncubify(out)