Python中重叠窗口的平均值 [英] Average on overlapping windows in Python

问题描述

我正在尝试计算移动平均线，但是在每个平均线之间设置了步长.例如，如果我正在计算每2个元素4个元素窗口的平均值:

I'm trying to compute a moving average but with a set step size between each average. For example, if I was computing the average of a 4 element window every 2 elements:

data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

这应该产生[1、2、3、4]，[3、4、5、6]，[5、6、7、8]，[7、8、9、10]的平均值.

This should produce the average of [1, 2, 3, 4], [3, 4, 5, 6], [5, 6, 7, 8], [7, 8, 9, 10].

window_avg = [2.5, 4.5, 6.5, 8.5]

我的数据使得结尾将在处理之前被截断，因此长度相对于窗口大小没有问题.

My data is such that the ending will be truncated before processing so there is no problem with the length with respect to window size.

我已经阅读了一些有关如何在Python中进行移动平均的知识，并且itertools的用法似乎很多；但是，迭代器一次只包含一个元素，我无法弄清楚在每次平均值计算之间如何具有步长. (如何在Python 3中计算移动平均值? )

I've read a bit about how to do moving averages in Python and there seems to be a lot of usage of itertools; however, the iterators go one element at a time and I can't figure out how to have a step size between each calculation of the average. (How to calculate moving average in Python 3?)

在MATLAB之前，我还能够通过创建重叠的索引矩阵，然后为数据向量建立索引并执行按列的均值(

I have also been able to do this before in MATLAB by creating a matrix of indices which are overlapping and then indexing the data vector and performing a column wise mean (Create matrix by repeatedly overlapping a vector). However, since this vector is rather large (~70 000 elements, window of 450 samples, average every 30 samples), the computation would probably require too much memory.

任何帮助将不胜感激.我正在使用Python 2.7.

Any help would be greatly appreciated. I am using Python 2.7.

推荐答案

在Python中计算列表中滑动窗口平均值的一种方法是使用列表理解.您可以使用

One way to compute the average of a sliding window across a list in Python is to use a list comprehension. You can use

>>> range(0, len(data), 2)
[0, 2, 4, 6, 8]

获取每个窗口的起始索引，然后numpy的mean函数获取每个窗口的平均值.请参见下面的演示:

to get the starting indices of each window, and then numpy's mean function to take the average of each window. See the demo below:

>>> import numpy as np
>>> window_size = 4
>>> stride = 2
>>> window_avg = [ np.mean(data[i:i+window_size]) for i in range(0, len(data), stride)
                   if i+window_size <= len(data) ]
>>> window_avg
[2.5, 4.5, 6.5, 8.5]

请注意，列表理解确实具有确保仅计算全窗口"或具有完全window_size个元素的子列表的平均值的条件.

Note that the list comprehension does have a condition to ensure that it only computes the average of "full windows", or sublists with exactly window_size elements.

在OP中讨论的大小的数据集上运行时，此方法将在我的MBA上花费200毫秒以上的时间:

When run on a dataset of the size discussed in the OP, this method computes on my MBA in a little over 200 ms:

In [5]: window_size = 450
In [6]: data = range(70000)
In [7]: stride = 30
In [8]: timeit [ np.mean(data[i:i+window_size]) for i in range(0, len(data), stride)
                 if i+window_size <= len(data) ]
1 loops, best of 3: 220 ms per loop

在我的机器上，它的速度是@Abhijit提出的itertools方法的两倍:

It is about twice as fast on my machine to the itertools approach presented by @Abhijit:

In [9]: timeit map(np.mean, izip(*(islice(it, i, None, stride) for i, it in enumerate(tee(data, window_size)))))
1 loops, best of 3: 436 ms per loop

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