零填充数据，直到它的长度等于2的幂

Question

我正试图写一个函数，可以零填充一些数据，直到它的长度等于最接近的更高的2次幂。我也想成为能够做一定数量的迭代。现在这是我的函数：

def pad_to_next_2n(array, iterations = 1): 
    n = 1 
    while n <= iterations: 
     l = len(array) 
     padding = l +1 #incase array is already length equal to 2^n 
     while np.log2(padding) % 1 != 0: 
      padding += 1 
     if l % 2 == 0: 
      total_padding = padding - l 
      array = np.pad(array, (total_padding/2,), 'constant') 
      n += 1 
     else: 
      total_padding = padding - l 
      left_padding = (total_padding - 1)/2 
      right_padding = total_padding - left_padding 
      print total_padding 
      print left_padding 
      print right_padding 
      array = np.pad(array, (left_padding, right_padding), 'constant') 
      n += 1 
    return array 

这并不工作，但它真的很慢的迭代大于5。我在想，如果有人可以帮助提高这个速度或看到一个更好的方式来做到这一点。我相信最大的问题来自于

while np.log2(padding) % 1 != 0: 
       padding += 1 

部分，但我不知道如何使效率更高。

Answer 1

你不需要做很多你正在做的事情 - 你只是想找出下一个你需要将数组的长度扩展到2的力量，然后你可以调用pad阵列将它填充到你需要的长度。

这使用shift_bit_length来自另一个关于在Python中获得2的下一个幂的最快方法。

import numpy as np 

def shift_bit_length(x): 
    return 1<<(x-1).bit_length() 

def padpad(data, iterations = 1): 
    narray = data 
    for i in xrange(iterations): 
     length = len(narray) 
     diff = shift_bit_length(length + 1) - length 
     if length % 2 == 0: 
      pad_width = diff/2 
     else: 
      # need an uneven padding for odd-number lengths 
      left_pad = diff/2 
      right_pad = diff - left_pad 
      pad_width = (left_pad, right_pad) 
     narray = np.pad(narray, pad_width, 'constant') 
    return narray 

一些测试：

>> arr = np.array([1, 2]) 

>> padpad(arr, 1) 
Out[2]: array([0, 1, 2, 0]) 

>> len(padpad(arr, 1) 
Out[3]: 4 

>> padpad(arr, 5) 
Out[4]: array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
     0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
     0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]) 

>> len(padpad(arr, 5)) 
Out[5]: 32 

>> padpad(arr, 8) 
Out[6]: array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0]) 

Answer 2

您不必通过每次增加一个填充来迭代。只需计算你需要的下一个尺寸。下一个尺寸是2**int(log2(current_size)+1)。

然后你可以减去current_size并知道你需要多少填充。