Question on Debugging a code line
subhabangalore at gmail.com
subhabangalore at gmail.com
Sun May 11 02:20:32 EDT 2014
On Sunday, May 11, 2014 12:57:34 AM UTC+5:30, subhaba... at gmail.com wrote:
> Dear Room,
>
>
>
> I was trying to go through a code given in http://en.wikipedia.org/wiki/Forward%E2%80%93backward_algorithm[ Forward Backward is an algorithm of Machine Learning-I am not talking on that
>
> I am just trying to figure out a query on its Python coding.]
>
>
>
> I came across the following codes.
>
>
>
> >>> states = ('Healthy', 'Fever')
>
> >>> end_state = 'E'
>
> >>> observations = ('normal', 'cold', 'dizzy')
>
> >>> start_probability = {'Healthy': 0.6, 'Fever': 0.4}
>
> >>> transition_probability = {
>
> 'Healthy' : {'Healthy': 0.69, 'Fever': 0.3, 'E': 0.01},
>
> 'Fever' : {'Healthy': 0.4, 'Fever': 0.59, 'E': 0.01},
>
> }
>
> >>> emission_probability = {
>
> 'Healthy' : {'normal': 0.5, 'cold': 0.4, 'dizzy': 0.1},
>
> 'Fever' : {'normal': 0.1, 'cold': 0.3, 'dizzy': 0.6},
>
> }
>
>
>
> def fwd_bkw(x, states, a_0, a, e, end_st):
>
> L = len(x)
>
> fwd = []
>
> f_prev = {} #THE PROBLEM
>
> # forward part of the algorithm
>
> for i, x_i in enumerate(x):
>
> f_curr = {}
>
> for st in states:
>
> if i == 0:
>
> # base case for the forward part
>
> prev_f_sum = a_0[st]
>
> else:
>
> prev_f_sum = sum(f_prev[k]*a[k][st] for k in states) ##
>
>
>
> f_curr[st] = e[st][x_i] * prev_f_sum
>
>
>
> fwd.append(f_curr)
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> f_prev = f_curr
>
>
>
> p_fwd = sum(f_curr[k]*a[k][end_st] for k in states)
>
>
>
> As this value was being called in prev_f_sum = sum(f_prev[k]*a[k][st] for k in states marked ##
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> I wanted to know what values it is generating.
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> So, I had made the following experiment, after
>
> for i, x_i in enumerate(x):
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> I had put print f_prev
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> but I am not getting how f_prev is getting the values.
>
>
>
> Here,
>
> x=observations,
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> states= states,
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> a_0=start_probability,
>
> a= transition_probability,
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> e=emission_probability,
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> end_st= end_state
>
>
>
> Am I missing any minor aspect?
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> Code is running fine.
>
>
>
> If any one of the esteemed members may kindly guide me.
>
>
>
> Regards,
>
> Subhabrata Banerjee.
Dear Sir,
Thank you for your kind reply. I will check.
Regards,
Subhabrata Banerjee.
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