在两个或多个Numpy阵列的滚动窗口上同时迭代的更快方法?

本文介绍了在两个或多个Numpy阵列的滚动窗口上同时迭代的更快方法?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有两个numpy数组x和y.例如

I have two numpy arrays x and y. e.g.

x Out[1]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]) y Out[1]: array([100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119])

我想同时遍历x和y的n周期滚动窗口.

I want to simultaneously iterate over a n period rolling window over x and y.

我想尽快做到这一点，同时将滚动窗口保持为numpy数组.我毫不客气地从itertools文档中偷走了一些代码.然后制作了仅在np.array上运行的我自己的版本.但是想知道这里是否有人可以帮助我加快我的功能?

I would like to do this as quickly as possible, while keeping the rolling windows as numpy arrays. I shamelessly stole some code from the itertools documentation. And then made my own version that only operates on np.array's. But am wondering if anyone here could help me make my function faster?

我的源代码是:

from itertools import islice import numpy as np import time class Timer( object ): def __init__(self): pass def __enter__(self): self.start = time.time() return self def __exit__(self,a,b,c): print('ending') self.end = time.time() self.timetaken = self.end-self.start print( 'Took {0} seconds'.format( self.timetaken )) def window(seq, n=2): "Returns a sliding window (of width n) over data from the iterable" " s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ... " it = iter(seq) result = list(islice(it, n)) if len(result) == n: yield np.array( result ).T for elem in it: result = result[1:] + [elem,] yield np.array( result ).T def npwindow( seq, n=2): Zt = Z.T r = Zt[:n] for zt in Zt[n:]: r = np.roll( r, shift=-1, axis=0 ) r[-1] = zt yield r.T n = 100 N = 1000 x = np.arange(N) y = np.arange(100,N+100) Z = np.array( [x,y] ) def npwindow_test( Z,n ): window2 = npwindow( Z,n ) for w in window2: pass #print( 'npwindow: {0}'.format( w ) ) def window_test( Z, n ): window1 = window( zip(*Z),n ) for w in window1: pass #print( 'window: {0}'.format( w ) ) num_iter = 10 with Timer() as t0: for i in range(num_iter): npwindow_test( Z, n ) with Timer() as t1: for i in range(num_iter): window_test( Z, n ) print( ' ratio : {0}'.format( t0.timetaken / t1.timetaken ) )

推荐答案

滚动窗口可以通过np.lib.stride_tricks.as_strided()

参考链接: www.rigtorp.se /2011/01/01/rolling-statistics-numpy.html

这是代码:

import numpy as np n = 10 N = 20 x = np.arange(N) y = np.arange(100,N+100) Z = np.vstack((x, y)) def rolling_window(a, window): shape = a.shape[:-1] + (a.shape[-1] - window + 1, window) strides = a.strides + (a.strides[-1],) return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides) rZ = rolling_window(Z, n)

然后您可以使用rZ进行计算，而无需for循环.如果需要循环，请使用轴1进行循环:

Then you can do calculation with rZ without for loop. If you need for loop, do the loop with axis 1:

for i in xrange(rZ.shape[1]): print rZ[:, i, :]

输出:

[[ 0 1 2 3 4 5 6 7 8 9] [100 101 102 103 104 105 106 107 108 109]] [[ 1 2 3 4 5 6 7 8 9 10] [101 102 103 104 105 106 107 108 109 110]] [[ 2 3 4 5 6 7 8 9 10 11] [102 103 104 105 106 107 108 109 110 111]] [[ 3 4 5 6 7 8 9 10 11 12] [103 104 105 106 107 108 109 110 111 112]] [[ 4 5 6 7 8 9 10 11 12 13] [104 105 106 107 108 109 110 111 112 113]] [[ 5 6 7 8 9 10 11 12 13 14] [105 106 107 108 109 110 111 112 113 114]] [[ 6 7 8 9 10 11 12 13 14 15] [106 107 108 109 110 111 112 113 114 115]] [[ 7 8 9 10 11 12 13 14 15 16] [107 108 109 110 111 112 113 114 115 116]] [[ 8 9 10 11 12 13 14 15 16 17] [108 109 110 111 112 113 114 115 116 117]] [[ 9 10 11 12 13 14 15 16 17 18] [109 110 111 112 113 114 115 116 117 118]] [[ 10 11 12 13 14 15 16 17 18 19] [110 111 112 113 114 115 116 117 118 119]]