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python-3.x - 同时迭代两个或多个 Numpy 数组的滚动窗口的更快方法?

转载 作者:行者123 更新时间:2023-12-05 03:15:05 29 4
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我有两个 numpy 数组 x 和 y。例如

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])

我想在 x 和 y 上同时迭代 n 个周期的滚动窗口。

我想尽快完成此操作,同时将滚动窗口保持为 numpy 数组。我无耻地从 itertools 文档中窃取了一些代码。然后制作了我自己的版本,它只在 np.array 上运行。但是想知道这里是否有人可以帮助我加快我的功能?

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 ) )

最佳答案

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

引用链接:http://www.rigtorp.se/2011/01/01/rolling-statistics-numpy.html

代码如下:

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)

然后你就可以不用for循环用rZ进行计算了。如果需要 for 循环,请使用轴 1 执行循环:

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]]

关于python-3.x - 同时迭代两个或多个 Numpy 数组的滚动窗口的更快方法?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21204206/

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