Question

我使用了voronoi binning：我有每个bin的中心坐标，我想找到每个bin中包含的所有像素的平均值。我无法理解如何切割numpy数组来进行矢量化。

这是我目前的代码：X和Y是一维数组，每个数据库的中心有x和y坐标; f在2d图像中：

import numpy as np
from scipy.spatial import KDTree

def rebin(f, X, Y):
    s = f.shape
    x_grid = np.arange(s[0])
    y_grid = np.arange(s[1])
    x_grid, y_grid = np.meshgrid(x_grid,y_grid)
    x_grid, y_grid = x_grid.flatten(),  y_grid.flatten()

    tree = KDTree(zip(X,Y))
    _, b = tree.query(zip(x_grid, y_grid))
    out = X*np.nan

    for i in range(max(b)):
        out[i] = np.nanmean(f[x_grid[b==i], y_grid[b==i]])
    return out

for循环目前是一个巨大的瓶颈。可能有一个非常简单的答案 - 我现在无法看到它！

Answer 1

out = X*np.nan
for i in range(max(b)):
    out[i] = np.nanmean(f[x_grid[b==i], y_grid[b==i]])

可以替换为np.bincount的两次调用：

total = np.bincount(b, weights=f[x_grid, y_grid], minlength=len(X))
count = np.bincount(b, minlength=len(X))
out = total/count

或拨打stats.binned_statistic：

out, bin_edges, binnumber = stats.binned_statistic(
    x=b, values=f[x_grid, y_grid], statistic='mean', bins=np.arange(len(X)+1))

例如，

import numpy as np
from scipy.spatial import KDTree
import scipy.stats as stats
np.random.seed(2017)

def rebin(f, X, Y):
    s = f.shape
    x_grid = np.arange(s[0])
    y_grid = np.arange(s[1])
    x_grid, y_grid = np.meshgrid(x_grid,y_grid)
    x_grid, y_grid = x_grid.flatten(),  y_grid.flatten()

    tree = KDTree(np.column_stack((X,Y)))
    _, b = tree.query(np.column_stack((x_grid, y_grid)))

    out, bin_edges, binnumber = stats.binned_statistic(
        x=b, values=f[x_grid, y_grid], statistic='mean', bins=np.arange(len(X)+1))
    # total = np.bincount(b, weights=f[x_grid, y_grid], minlength=len(X))
    # count = np.bincount(b, minlength=len(X))
    # out = total/count
    return out

def orig(f, X, Y):
    s = f.shape
    x_grid = np.arange(s[0])
    y_grid = np.arange(s[1])
    x_grid, y_grid = np.meshgrid(x_grid,y_grid)
    x_grid, y_grid = x_grid.flatten(),  y_grid.flatten()

    tree = KDTree(np.column_stack((X,Y)))
    _, b = tree.query(np.column_stack((x_grid, y_grid)))

    out = X*np.nan
    for i in range(len(X)):
        out[i] = np.nanmean(f[x_grid[b==i], y_grid[b==i]])
    return out

N = 100
X, Y = np.random.random((2, N))
f = np.random.random((N, N))

expected = orig(f, X, Y)
result = rebin(f, X, Y)
print(np.allclose(expected, result, equal_nan=True))
# True

Answer 2

我刚刚发现有一个名为cKDTree的KDTree的cython功能完整版本，速度要快得多。

将数据分箱到非基于网格的箱柜中

2 个答案: