让我们为python的numpy做一个N维像素分组/分组的参考实现

Question

I frequently want to pixel bin/pixel bucket a numpy array, meaning, replace groups of N consecutive pixels with a single pixel which is the sum of the N replaced pixels. For example, start with the values:

x = np.array([1, 3, 7, 3, 2, 9])

with a bucket size of 2, this transforms into:

bucket(x, bucket_size=2) 
= [1+3, 7+3, 2+9]
= [4, 10, 11]

As far as I know, there's no numpy function that specifically does this (please correct me if I'm wrong!), so I frequently roll my own. For 1d numpy arrays, this isn't bad:

import numpy as np

def bucket(x, bucket_size):
    return x.reshape(x.size // bucket_size, bucket_size).sum(axis=1)

bucket_me = np.array([3, 4, 5, 5, 1, 3, 2, 3])
print(bucket(bucket_me, bucket_size=2)) #[ 7 10  4  5]

...however, I get confused easily for the multidimensional case, and I end up rolling my own buggy, half-assed solution to this "easy" problem over and over again. I'd love it if we could establish a nice N-dimensional reference implementation.

• Preferably the function call would allow different bin sizes along different axes (perhaps something like bucket(x, bucket_size=(2, 2, 3)) )

• Preferably the solution would be reasonably efficient (reshape and sum are fairly quick in numpy)

• Bonus points for handling edge effects when the array doesn't divide nicely into an integer number of buckets.

• Bonus points for allowing the user to choose the initial bin edge offset.

As suggested by Divakar, here's my desired behavior in a sample 2-D case:

x = np.array([[1, 2, 3, 4],
              [2, 3, 7, 9],
              [8, 9, 1, 0],
              [0, 0, 3, 4]])

bucket(x, bucket_size=(2, 2))
= [[1 + 2 + 2 + 3, 3 + 4 + 7 + 9],
   [8 + 9 + 0 + 0, 1 + 0 + 3 + 4]]
= [[8, 23],
   [17, 8]]

...hopefully I did my arithmetic correctly ;)

Answer 1

I think you can do most of the fiddly work with skimage's view_as_blocks . This function is implemented using as_strided so it is very efficient (it just changes the stride information to reshape the array). Because it's written in Python/NumPy, you can always copy the code if you don't have skimage installed.

After applying that function, you just need to sum the N trailing axes of the reshaped array (where N is the length of the bucket_size tuple). Here's a new bucket() function:

from skimage.util import view_as_blocks

def bucket(x, bucket_size):
    blocks = view_as_blocks(x, bucket_size)
    tup = tuple(range(-len(bucket_size), 0))
    return blocks.sum(axis=tup)

Then for example:

>>> x = np.array([1, 3, 7, 3, 2, 9])
>>> bucket(x, bucket_size=(2,))
array([ 4, 10, 11])

>>> x = np.array([[1, 2, 3, 4],
                  [2, 3, 7, 9],
                  [8, 9, 1, 0],
                  [0, 0, 3, 4]])

>>> bucket(x, bucket_size=(2, 2))
array([[ 8, 23],
       [17,  8]])

>>> y = np.arange(6*6*6).reshape(6,6,6)
>>> bucket(y, bucket_size=(2, 2, 3))
array([[[ 264,  300],
        [ 408,  444],
        [ 552,  588]],

       [[1128, 1164],
        [1272, 1308],
        [1416, 1452]],

       [[1992, 2028],
        [2136, 2172],
        [2280, 2316]]])

Answer 2

To specify different bin sizes along each axis for ndarray cases, you can use iteratively use np.add.reduceat along each axis of it, like so -

def bucket(x, bin_size):
    ndims = x.ndim
    out = x.copy()
    for i in range(ndims):
        idx = np.append(0,np.cumsum(bin_size[i][:-1]))
        out = np.add.reduceat(out,idx,axis=i)
    return out

Sample run -

In [126]: x
Out[126]: 
array([[165, 107, 133,  82, 199],
       [ 35, 138,  91, 100, 207],
       [ 75,  99,  40, 240, 208],
       [166, 171,  78,   7, 141]])

In [127]: bucket(x, bin_size = [[2, 2],[3, 2]])
Out[127]: 
array([[669, 588],
       [629, 596]])

#  [2, 2] are the bin sizes along axis=0
#  [3, 2] are the bin sizes along axis=1

# array([[165, 107, 133, | 82, 199],
#        [ 35, 138,  91, | 100, 207],
# -------------------------------------
#        [ 75,  99, 40,  | 240, 208],
#        [166, 171, 78,  | 7, 141]])

In [128]: x[:2,:3].sum()
Out[128]: 669

In [129]: x[:2,3:].sum()
Out[129]: 588

In [130]: x[2:,:3].sum()
Out[130]: 629

In [131]: x[2:,3:].sum()
Out[131]: 596

Answer 3

Natively from as_strided :

x = array([[1, 2, 3, 4],
           [2, 3, 7, 9],
           [8, 9, 1, 0],
           [0, 0, 3, 4]])

from numpy.lib.stride_tricks import as_strided     
def bucket(x,bucket_size):
      x=np.ascontiguousarray(x)
      oldshape=array(x.shape)
      newshape=concatenate((oldshape//bucket_size,bucket_size))
      oldstrides=array(x.strides)
      newstrides=concatenate((oldstrides*bucket_size,oldstrides))
      axis=tuple(range(x.ndim,2*x.ndim))
      return as_strided (x,newshape,newstrides).sum(axis)

if a dimension not divide evenly into the corresponding dimension of x, remaining elements are lost.

verification :

In [9]: bucket(x,(2,2))
Out[9]: 
array([[ 8, 23],
       [17,  8]])