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in Machine Learning by (19k points)

By processing a time series graph, I Would like to detect patterns that look similar to this:

enter image description here

Using a sample time series as an example, I would like to be able to detect the patterns as marked here:

enter image description here

What kind of AI algorithm (I am assuming machine learning techniques) do I need to use to achieve this? Is there any library (in C/C++) out there that I can use?

1 Answer

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by (33.1k points)

The following python code is the general approach to your problem. I have used ECG dataset from here.

Code:

import numpy as np

import numpy.random as rnd

import matplotlib.pyplot as plt 

import scipy.linalg as lin

import re

data=np.array(map(lambda l: map(float,filter(lambda x: len(x)>0,re.split('\\s+',l))),open('chfdb_chf01_275.txt'))).T

dK=230

pattern=data[1,:dK]

data=data[1,dK:]

def create_mats(dat):

    step=5 

    eps=.1

    dat=dat[::step]

    K=len(dat)+1

    A=np.zeros( (K,K) )

    A[0,1]=1.

    pA=np.zeros( (K,K) )

    pA[0,1]=1.

    for i in xrange(1,K-1):

        A[i,i]=(step-1.+eps)/(step+2*eps)

        A[i,i+1]=(1.+eps)/(step+2*eps)

        pA[i,i]=1.

        pA[i,i+1]=1.

    A[-1,-1]=(step-1.+eps)/(step+2*eps)

    A[-1,1]=(1.+eps)/(step+2*eps)

    pA[-1,-1]=1.

    pA[-1,1]=1.

    w=np.ones( (K,2) , dtype=np.float)

    w[0,1]=dat[0]

    w[1:-1,1]=(dat[:-1]-dat[1:])/step

    w[-1,1]=(dat[0]-dat[-1])/step

    return A,pA,w,K

A,pA,w,K=create_mats(pattern)

eta=10. #precision parameter for the autoregressive portion of the model 

lam=.1 #precision parameter for the weights prior 

N=1 

M=2 

T=len(data)

x=np.ones( (T+1,M) ) # sequence data (just one sequence)

x[0,1]=1

x[1:,0]=data

e=np.zeros( (T,K) )

#residuals

v=np.zeros( (T,K) )

#store the forward and backward recurrences

f=np.zeros( (T+1,K) )

fls=np.zeros( (T+1) )

f[0,0]=1

b=np.zeros( (T+1,K) )

bls=np.zeros( (T+1) )

b[-1,1:]=1./(K-1)

#hidden states

z=np.zeros( (T+1),dtype=np.int )

#expected hidden states

ex_k=np.zeros( (T,K) )

# expected pairs of hidden states

ex_kk=np.zeros( (K,K) )

nkk=np.zeros( (K,K) )

def fwd(xn):

    global f,e

    for t in xrange(T):

        f[t+1,:]=np.dot(f[t,:],A)*e[t,:]

        sm=np.sum(f[t+1,:])

        fls[t+1]=fls[t]+np.log(sm)

        f[t+1,:]/=sm

        assert f[t+1,0]==0

def bck(xn):

    global b,e

    for t in xrange(T-1,-1,-1):

        b[t,:]=np.dot(A,b[t+1,:]*e[t,:])

        sm=np.sum(b[t,:])

        bls[t]=bls[t+1]+np.log(sm)

        b[t,:]/=sm

def em_step(xn):

    global A,w,eta

    global f,b,e,v

    global ex_k,ex_kk,nkk

    x=xn[:-1] #current data vectors

    y=xn[1:,:1] #next data vectors predicted from current

    #compute residuals

    v=np.dot(x,w.T) # (N,K) <- (N,1) (N,K)

    v-=y

    e=np.exp(-eta/2*v**2,e)

    fwd(xn)

    bck(xn)

    # compute expected hidden states

    for t in xrange(len(e)):

        ex_k[t,:]=f[t+1,:]*b[t+1,:]

        ex_k[t,:]/=np.sum(ex_k[t,:])

    # compute expected pairs of hidden states    

    for t in xrange(len(f)-1):

        ex_kk=A*f[t,:][:,np.newaxis]*e[t,:]*b[t+1,:]

        ex_kk/=np.sum(ex_kk)

        nkk+=ex_kk

    A=pA+nkk

    A/=np.sum(A,1)[:,np.newaxis]

    # solve the weighted regression problem for emissions weights

    #  x and y are from above 

    for k in xrange(K):

        ex=ex_k[:,k][:,np.newaxis]

        dx=np.dot(x.T,ex*x)

        dy=np.dot(x.T,ex*y)

        dy.shape=(2)

        w[k,:]=lin.solve(dx+lam*np.eye(x.shape[1]), dy)

    #return the probability of the sequence (computed by the forward algorithm)

    return fls[-1]

if __name__=='__main__':

    #run the em algorithm

    for i in xrange(20):

        print em_step(x)

    r=np.arange(len(ex_k))[np.argmax(ex_k,1)<3]

    #plot

    plt.plot(range(T),x[1:,0])

    yr=[np.min(x[:,0]),np.max(x[:,0])]

    for i in r:

        plt.plot([i,i],yr,'-r')

    plt.show()

Hope this answer helps.

by (100 points)
This is great, but somewhat cryptic. A little bit of explanation of each function would be very much appreciated. Thanks a lot.

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