Question

我试图通过取平均值来将numpy数组分组为更小的数组。例如，在100x100阵列中取平均foreach 5x5子阵列以创建20x20大小的阵列。由于我需要操作大量数据，这是一种有效的方法吗？

Answer 1

我已经尝试过更小的数组，所以请用你的测试：

import numpy as np

nbig = 100
nsmall = 20
big = np.arange(nbig * nbig).reshape([nbig, nbig]) # 100x100

small = big.reshape([nsmall, nbig//nsmall, nsmall, nbig//nsmall]).mean(3).mean(1)

6x6的示例 - ＆gt; 3×3：

nbig = 6
nsmall = 3
big = np.arange(36).reshape([6,6])
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],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

small = big.reshape([nsmall, nbig//nsmall, nsmall, nbig//nsmall]).mean(3).mean(1)

array([[  3.5,   5.5,   7.5],
       [ 15.5,  17.5,  19.5],
       [ 27.5,  29.5,  31.5]])

Answer 2

这很简单，虽然我觉得它可能会更快：

from __future__ import division
import numpy as np
Norig = 100
Ndown = 20
step = Norig//Ndown
assert step == Norig/Ndown # ensure Ndown is an integer factor of Norig
x = np.arange(Norig*Norig).reshape((Norig,Norig)) #for testing
y = np.empty((Ndown,Ndown)) # for testing
for yr,xr in enumerate(np.arange(0,Norig,step)):
    for yc,xc in enumerate(np.arange(0,Norig,step)):
        y[yr,yc] = np.mean(x[xr:xr+step,xc:xc+step])

您可能还会发现scipy.signal.decimate很有趣。在对数据进行下采样之前，它应用比简单平均更复杂的低通滤波器，尽管你必须抽取一个轴，然后另一个轴。

Answer 3

在大小为NxN的子阵列上平均2D数组：

height, width = data.shape
data = average(split(average(split(data, width // N, axis=1), axis=-1), height // N, axis=1), axis=-1)

Answer 4

请注意，eumiro's approach不适用于屏蔽数组，因为.mean(3).mean(1)假设沿轴3的每个平均值都是根据相同数量的值计算的。如果数组中有掩码元素，则此假设不再适用。在这种情况下，您必须跟踪用于计算.mean(3)的值的数量，并用加权平均值替换.mean(1)。权重是用于计算.mean(3)的标准化值。

以下是一个例子：

import numpy as np


def gridbox_mean_masked(data, Nbig, Nsmall):
    # Reshape data
    rshp = data.reshape([Nsmall, Nbig//Nsmall, Nsmall, Nbig//Nsmall])

    # Compute mean along axis 3 and remember the number of values each mean
    # was computed from
    mean3 = rshp.mean(3)
    count3 = rshp.count(3)

    # Compute weighted mean along axis 1
    mean1 = (count3*mean3).sum(1)/count3.sum(1)
    return mean1


# Define test data
big = np.ma.array([[1, 1, 2],
                   [1, 1, 1],
                   [1, 1, 1]])
big.mask = [[0, 0, 0],
            [0, 0, 1],
            [0, 0, 0]]
Nbig = 3
Nsmall = 1

# Compute gridbox mean
print gridbox_mean_masked(big, Nbig, Nsmall)

平均分组2D numpy数组

4 个答案: