python中的重复图 [英] Recurrence plot in python
问题描述
我正在尝试按时间顺序对模式进行聚类
I am trying to clusterize paterns in time series as I ask in
我尝试使用递归图技术来解决我的问题，因此我在python中编写了一些代码来重现这些图.我想知道我的代码是否正常，我使用声音时序序列进行了尝试，并且根据距离参数值得到了这种结果:
I try using to solve my problem using the recurrence plots technique, so I make some code in python to reproduce these plots. I want o know if my code is ok, I tried it with a sound temporal series and I am getting this kind of result depending on the distance parameter value:
http://ceciliajarne.web.unq.edu.ar/envelope-问题/
我也包括数据集.我正在使用ch2.这是我的代码:
Also I include the data set. I am using ch2. This is my code:
import numpy as np
import scipy
import os
from scipy.io import wavfile
import wave, struct
import matplotlib.pyplot as pp
from pylab import *
import scipy.signal.signaltools as sigtool
import scipy, pylab
from scipy.io import wavfile
import wave, struct
import scipy.signal as signal
from scipy.fftpack import fft
#Data set input
data=np.random.rand(44000*3)
#random secuence to compare with almost 3 seconds of data, cold be other
print 'data:', data
#set size
sissse=data.size
print 'size: ',sissse
print '---------------'
#empty vectors
x_filt_all_p=[]
y_filt_all_p=[]
los_p_filt_all_p=[]
#creating the list to fill
dif=[]
dif_abs=[]
p=1
#for each i-element of data vector for each p
for p in range(1,sissse,4400):
for i in enumerate(data):
#print i
j=i[0]
#print 'j: ',j
if (j<sissse-p):
dif_aux=data[j+p]-data[j]
#print 'dif=',dif_aux
dif.append(dif_aux)
dif_abs.append(abs(data[j+p]-data[j]))
#print'.........'
print'.........'
#print 'dif=',dif
print'.........'
#print 'Absolute difference=',dif_abs
print'.........'
#vector with index and diferences in absolute value
pepe= np.vstack([np.arange(len(dif_abs)),dif_abs])
print 'pepe0: ', pepe[0]
xx=pepe[0]
print 'pepe1: ', pepe[1]
yy=pepe[1]
#filtering the elements with diference<delta
delta= 0.001
# Now let's extract only the part of the data we're interested in...
los_p = np.empty(len(pepe[1]))#dif_abs
los_p.fill(p)
x_filt = xx[yy<delta]
y_filt = yy[yy<delta]
los_p_filt= los_p[yy<delta]
print 'value of coordinate i', x_filt
print 'absolute difference', y_filt
print 'value of coordinate p', los_p_filt
print '------------------------'
if (p==1):
x_filt_all_p=x_filt
y_filt_all_p=y_filt
los_p_filt_all_p=los_p_filt
else:
x_filt_all_p=np.concatenate((x_filt_all_p,x_filt))
y_filt_all_p=np.concatenate((y_filt_all_p,y_filt))
los_p_filt_all_p=np.concatenate((los_p_filt_all_p,los_p_filt))
print 'full value of coordinate i: ', x_filt_all_p
print 'full absolute difference', y_filt_all_p
print 'full value of coordinate p: ', los_p_filt_all_p
#trying to plot the "recurrence plots" together with the envelope.
pp.subplot(211)
pp.plot(arange(data.size),data, color='c',label='Time Signal 2')
pp.legend(fontsize= 'small')
pp.grid(True)
pp.xlabel('Time (s)')
pp.ylabel('Amplitude')
#pp.xlim([0,3])
pp.subplot(212)
base='test_plot'
pp.title('Recurrence plot delta=')
markerline2, stemlines2, baseline2 = stem(x_filt_all_p*float(1)/float(w[0]), los_p_filt_all_p*float(1)/float(w[0]),'b',linefmt=" ",)
pp.matplotlib.markers.MarkerStyle('.')
setp(markerline2,'markerfacecolor','b',label='points')
pp.legend(fontsize= 'small')
pp.grid(True)
pp.xlabel('Time i [s]')
pp.ylabel('Time p [s]')
#pp.xlim(0,3)
#pp.ylim(0,3)
pp.show()
#pp.savefig('plots/%s.jpg' %(str(base))
pp.close()
但我不确定我的代码是否 100% 工作正常.有人可以看看我的代码给我一些建议，以测试它吗?我不想使用 matlab 和 mathematica.这个想法是用python创建一个独立的代码.另外我还有另一个较小的问题，我无法更改绘图中的点大小.最后，我已经尝试将交叉检查与 http://recurrence-plot.tk/online 一起使用/index.php?state= 对接我无法使其正常工作.非常欢迎对我的代码或可能的交叉检查提出任何建议.预先感谢
But I am not sure 100% that my code I working ok. Could someone take a look to my code to give me some advise of how to test it? I don't want to use neither matlab nor mathematica. The idea was to create an independent code in python. Also I have another smaller problem, I could not change the dot size in my plot. Finally I alrady try to use crosscheck with http://recurrence-plot.tk/online/index.php?state= butt I couldn't make it work. Any suggestion on my code or possible crosscheck wold be very welcome. Thanks in advance
推荐答案
我知道这个问题已经很老了，但是将来有人会偶然发现这个问题.
I understand this question is quite old, but maybe someone will stumble on this in future.
既然你已经在使用 NumPy，那么让我推荐这个片段:
Since you are already using NumPy let me suggest this snippet:
import numpy as np
def rec_plot(s, eps=0.1, steps=10):
N = s.size
S = np.repeat(s[None,:], N, axis=0)
Z = np.floor(np.abs(S-S.T)/eps)
Z[Z>steps] = steps
return Z
最初会创建(N，N)大小的方形空数组.然后它通过 S-S.T 减去所有可能的点组合，这隐式等效于进行矩阵减法，其中一个矩阵具有所有行 S，另一个矩阵具有所有列 S.
It initially creates square empty array of (N, N) size. Then it subtract all possible combinations of points via S-S.T, which is implicitly equivalent to having matrix subtraction where one matrix has all rows S and the other all columns S.
除以 eps 和 flooring 是一种简写，用于询问这些点之间有多少 eps 差异.然后 Z [Z> steps] 是有界的，因此，只要某点大于 steps 乘以该点的 eps ，则该最大值为只需绘制具有相同值的图形即可.
Dividing by eps and flooring is for a short hand for asking of how many eps difference is between those points. Then Z[Z>steps] is bounding, so that whenever something is more than steps times eps from the point, then it's max and will be simply plot with same value.
此解决方案不是最优的，因为它首先创建了两个NxN矩阵，对于大的N而言，这太大了.对于 N>10000 这绝对不好.由于您使用的是SciPy，因此我们可以使用其 distance 库.下面是更优化的实现:
This solution is suboptimal as it first creates two NxN matrices, which for large N is too much. For N>10000 this is definitely not good. Since you are using SciPy we can use its distance library. Below is more optimal implementation:
import numpy as np
from scipy.spatial.distance import pdist, squareform
def rec_plot(s, eps=0.1, steps=10):
d = pdist(s[:,None])
d = np.floor(d/eps)
d[d>steps] = steps
Z = squareform(d)
return Z
您可以找到使用示例https://laszukdawid.com/tag/recurrence-plot/ 或 https://github.com/laszukdawid/recurrence-plot.
Examples of usage you can find https://laszukdawid.com/tag/recurrence-plot/ or https://github.com/laszukdawid/recurrence-plot.
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