python - 在python中计算重叠滑动窗口中的值

Question

给定一个排序值数组 , 和一个范围数组 ,a计算每个范围 ,中有多少个值的最有效方法是什么？ binsarngbins

目前我正在做以下事情：

def sliding_count(a, end, window, start=0, step=1):
    bins = [(x, x + window) for x in range(start, (end + 1) - window, step)]
    counts = np.zeros(len(bins))
    for i, rng in enumerate(bins):
        count = len(a[np.where(np.logical_and(a>=rng[0], a<=rng[1]))])
        counts[i] = count
    return counts

a = np.array([1, 5, 8, 11, 14, 19])
end = 20
window = 10
sliding_count(a, end, window)

返回预期的数组

array([3., 4., 3., 3., 4., 4., 3., 3., 3., 3., 3.])

但我觉得必须有一种更有效的方法来做到这一点？

标签： pythonalgorithmnumpy

Answer 1

import numpy as np

def alt(a, end, window, start=0, step=1):
    bin_starts = np.arange(start, end+1-window, step)
    bin_ends = bin_starts + window
    last_index = np.searchsorted(a, bin_ends, side='right')
    first_index = np.searchsorted(a, bin_starts, side='left')
    return  last_index - first_index

def sliding_count(a, end, window, start=0, step=1):
    bins = [(x, x + window) for x in range(start, (end + 1) - window, step)]
    counts = np.zeros(len(bins))
    for i, rng in enumerate(bins):
        count = len(a[np.where(np.logical_and(a>=rng[0], a<=rng[1]))])
        counts[i] = count
    return counts

a = np.array([1, 5, 8, 11, 14, 19])
end = 20
window = 10

print(sliding_count(a, end, window))
# [3. 4. 3. 3. 4. 4. 3. 3. 3. 3. 3.]

print(alt(a, end, window))
# [3 4 3 3 4 4 3 3 3 3 3]

---

alt 的工作原理：

生成 bin 的起始值和结束值：

In [73]: bin_starts = np.arange(start, end+1-window, step); bin_starts
Out[73]: array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10])

In [74]: bin_ends = bin_starts + window; bin_ends
Out[74]: array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20])

由于a按排序顺序，您可以使用np.searchsorted查找第一个和最后一个索引bin_starts以及bin_ends每个值a适合的位置：

In [75]: last_index = np.searchsorted(a, bin_ends, side='right'); last_index
Out[75]: array([3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6])

In [76]: first_index = np.searchsorted(a, bin_starts, side='left'); first_index
Out[76]: array([0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3])

这count只是指数的差异：

In [77]: last_index - first_index
Out[77]: array([3, 4, 3, 3, 4, 4, 3, 3, 3, 3, 3])

---

这是一个perfplot比较 与 的性能alt作为sliding_count长度的函数a：

import perfplot

def make_array(N):
    a = np.random.randint(10, size=N)
    a = a.cumsum()
    return a

def using_sliding(a):
    return sliding_count(a, end, window)

def using_alt(a):
    return alt(a, end, window)

perfplot.show(
    setup=make_array,
    kernels=[using_sliding, using_alt],
    n_range=[2**k for k in range(22)],
    logx=True,
    logy=True,
    xlabel='len(a)')

Perfplot 还检查返回的值是否using_sliding等于返回的值using_alt。

Matt Timmermans 的想法“从那个箱子的计数中减去position_in_a”触发了这个解决方案。