I am completely new to python multiprocessing and a bit overwhelmbed by the vast amount of online resources, so I kinda wanna a more clear-cut approach from here. My code looks like the following: the two functions, forward and backward are very computaionally expensive. On my input dataset, each takes about 13mins. I want to compute the two matrices (forward and backward, see the 3rd and 4th line of code in decode() function) simultaneouly. I looked into some online tutorials and I was thinking I could use multiprocessing.process to do this. However, I am not sure how to retrieve the numpy array. I know there are things like Queue, Array, but these seem quite restrictive in their usage and doesn't seem to be suitable here. Thanks in advance! '''
def forward(self, emis):
# Given the observed haplotype, compute its forward matrix
f = np.full((self.n1+self.n2, self.numSNP), np.nan)
# initialization
f[:,0] = (-math.log(self.n1+self.n2) + emis[0]).flatten()
# fill in forward matrix
for j in range(1, self.numSNP):
T = self.transition(self.D[j])
# using axis=1, logsumexp sum over each column of the transition matrix
f[:, j] = emis[j] + logsumexp(f[:,j-1][:,np.newaxis] + T, axis=0)
return f
#@profile
def backward(self, emis):
# Given the observed haplotype, compute its backward matrix
b = np.full((self.n1+self.n2, self.numSNP), np.nan)
# initialization
b[:, self.numSNP-1] = np.full(self.n1+self.n2, 0)
for j in range(self.numSNP-2, -1, -1):
T = self.transition(self.D[j+1])
b[:,j] = logsumexp(T + emis[j+1] + b[:,j+1], axis=1)
return b
#@profile
def decode(self, obs):
# infer hidden state of each SNP sites in the given haplotype
# state[j] = 0 means site j was most likely copied from population 1
# and state[j] = 1 means site j was most likely copies from population 2
start = time.time()
emis = self.emissionALL(obs)
f = self.forward(emis)
b = self.backward(emis)
end= time.time()
print(f'uncached version takes time {end-start}')
print(f'forward probability:{logsumexp(f[:,-1])}')
print(f'backward probability:{logsumexp(-math.log(self.n1+self.n2)+emis[0]+b[:,0])}')
return 0
'''