2017-02-17 40 views

回答

3

假设cube的形状为(W, H, D)并且您希望将其分解为N形状的小立方体(w, h, d)。由于NumPy阵列具有固定长度的轴,因此w必须均匀分配W,并且类似地hd

然后有一种方法可以将形状为(W, H, D)的立方体重新塑造成形状为(N, w, h, d)的新阵列。

例如,如果arr = np.arange(4*4*4).reshape(4,4,4)(所以(W,H,D) = (4,4,4)),我们希望把它分解成形状(2,2,2)的立方体,然后我们可以使用

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]]]]) 

这里的想法是多余的轴添加到阵列排序充当位置标记的:

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

然后,我们可以重新排序的轴(使用transpose),使得重复的次数至上,和newshape自带底:

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

最后,请拨reshape(-1, w, h, d)将所有地标轴压扁到一个轴上。这产生了一个形状数组(N, w, h, d)其中N是小立方体的数量。


以上使用的思想是将this idea推广到3维。它可以进一步推广到任何尺寸的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) 

产生形状的新的阵列

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

到 “uncubify” 阵列:

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) 

这里有一些测试代码来检查cubifyuncubify是反转。

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) 
+0

如果我想解开它什么是转置重塑组合我需要回到大立方体? – mattdns

+1

@mattdns:我在上面添加了一个'uncubify'函数。我们对'reverseOrder'的定义有所不同(或者我在'uncubify'函数中称为'order')。它对高维数组有所帮助。将定义更改为'self.reverseOrder = np.arange(len(self.tmpshape))。reshape(2,-1).ravel(order ='F')'将修复您的代码。 – unutbu

0

我不认为有一个多轴版本,你可以沿某些给定的轴分割。但是你可以一次把它分成一个维度。例如像这样:

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 

它可以像这样使用:

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]])] 
0

除了我的额外问题@ unutbu的答案,我想我已经得到了相反的工作(在你想要的情况下将立方体拆分为立方体,对每个立方体应用一个功能,然后将它们组合起来)。

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)