通过平均调整大小或重新绑定一个numpy的二维数组

问题描述

我正在尝试在python中重新实现IDL函数:

I am trying to reimplement in python an IDL function:

http://star.pst.qub.ac.uk/idl/REBIN.html

通过平均将二维数组缩小为整数因子.

which downsizes by an integer factor a 2d array by averaging.

例如:

>>> a=np.arange(24).reshape((4,6))
>>> a
array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23]])

我想通过取相关样本的平均值将其调整为(2,3)，预期输出为:

I would like to resize it to (2,3) by taking the mean of the relevant samples, the expected output would be:

>>> b = rebin(a, (2, 3))
>>> b
array([[  3.5,   5.5,  7.5],
       [ 15.5, 17.5,  19.5]])

即b[0,0] = np.mean(a[:2,:2]), b[0,1] = np.mean(a[:2,2:4])，依此类推.

我相信我应该将其整形为4维数组，然后在正确的切片上取均值，但无法找出算法.你有什么提示吗?

I believe I should reshape to a 4 dimensional array and then take the mean on the correct slice, but could not figure out the algorithm. Would you have any hint?

推荐答案

以下是基于(为清楚起见):

Here's an example based on the answer you've linked (for clarity):

>>> import numpy as np
>>> a = np.arange(24).reshape((4,6))
>>> a
array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23]])
>>> a.reshape((2,a.shape[0]//2,3,-1)).mean(axis=3).mean(1)
array([[  3.5,   5.5,   7.5],
       [ 15.5,  17.5,  19.5]])

功能:

def rebin(a, shape):
    sh = shape[0],a.shape[0]//shape[0],shape[1],a.shape[1]//shape[1]
    return a.reshape(sh).mean(-1).mean(1)

这篇关于通过平均调整大小或重新绑定一个numpy的二维数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！