<filename>decoders/Loss.py<gh_stars>0
from __future__ import division
from __future__ import print_function
import math
import numpy as np
import decoders.Common as Common

# reference: https://github.com/githubharald/CTCDecoder, advancement raraz15,
# everything is converted to log probabilities.


def recLabelingProb(t, s, mat, labelingWithBlanks, blank, cache):
	"recursively compute probability of labeling, save results of sub-problems in cache to avoid recalculating them"

	# check index of labeling
	if s < 0:
		return 0.0

	# sub-problem already computed
	if cache[t][s] != None:
		return cache[t][s]

	# initial values
	if t == 0:
		if s == 0:
			res = mat[0, blank]
		elif s == 1:
			res = mat[0, labelingWithBlanks[1]]
		else:
			res = 0.0

		cache[t][s] = res
		return res

	A = sum_probs(recLabelingProb(t-1, s, mat, labelingWithBlanks, blank, cache), recLabelingProb(t-1, s-1, mat, labelingWithBlanks, blank, cache))

	res =  multiply_probs(A, mat[t, labelingWithBlanks[s]])

	# in case of a blank or a repeated label, we only consider s and s-1 at t-1, so we're done
	if labelingWithBlanks[s] == blank or (s >= 2 and labelingWithBlanks[s-2] == labelingWithBlanks[s]):
		cache[t][s] = res
		return res

	# otherwise, in case of a non-blank and non-repeated label, we additionally add s-2 at t-1
	res = sum_probs(res,multiply_probs(recLabelingProb(t-1, s-2, mat, labelingWithBlanks, blank, cache),mat[t, labelingWithBlanks[s]]))
	cache[t][s] = res
	return res


def emptyCache(maxT, labelingWithBlanks):
	"create empty cache"
	return [[None for _ in range(len(labelingWithBlanks))] for _ in range(maxT)]


def ctcLabelingProb(mat, gt, classes, blank):
	"calculate probability p(gt|mat) of a given labeling gt and a matrix mat according to section 'The CTC Forward-Backward Algorithm' in Graves paper"

	maxT, _ = mat.shape # size of input matrix

	labelingWithBlanks = Common.extendByBlanks(Common.wordToLabelSeq(gt, classes), blank) # ground truth text as label string extended by blanks

	cache = emptyCache(maxT, labelingWithBlanks) # cache subresults to avoid recalculating  subproblems over and over again
	A = recLabelingProb(maxT-1, len(labelingWithBlanks)-1, mat, labelingWithBlanks, blank, cache)
	B = recLabelingProb(maxT-1, len(labelingWithBlanks)-2, mat, labelingWithBlanks, blank, cache)
	prob =  sum_probs(A,B)
	return prob

def multiply_probs(ln_pa, ln_pb):
	"""
	Returns the log probability of multiplied probabilities.
	Pa * Pb = Pc => ln(Pc) = ln(Pa)+ln(Pb)
	"""
	return ln_pa+ln_pb

def sum_probs(ln_pa,ln_pb):
	"""
	Returns the log prob. of two added probs.
	Pa + Pb = Pc => ln(Pc) = ln(Pa) + ln(1+exp(ln(Pb)-ln(Pa))), by Graves dissertation eq 7.18
	"""
	return ln_pa + math.log(1+math.exp(ln_pb-ln_pa))


def ctcLoss(mat, gt, classes): 
	"calculate CTC loss"
	try:
		return -math.log(ctcLabelingProb(mat, gt, classes))
	except:
		return float('inf')
		

<filename>decoders/Loss.py<gh_stars>0
from __future__ import division
from __future__ import print_function
import math
import numpy as np
import decoders.Common as Common

# reference: https://github.com/githubharald/CTCDecoder, advancement raraz15,
# everything is converted to log probabilities.


def recLabelingProb(t, s, mat, labelingWithBlanks, blank, cache):
	"recursively compute probability of labeling, save results of sub-problems in cache to avoid recalculating them"

	# check index of labeling
	if s < 0:
		return 0.0

	# sub-problem already computed
	if cache[t][s] != None:
		return cache[t][s]

	# initial values
	if t == 0:
		if s == 0:
			res = mat[0, blank]
		elif s == 1:
			res = mat[0, labelingWithBlanks[1]]
		else:
			res = 0.0

		cache[t][s] = res
		return res

	A = sum_probs(recLabelingProb(t-1, s, mat, labelingWithBlanks, blank, cache), recLabelingProb(t-1, s-1, mat, labelingWithBlanks, blank, cache))

	res =  multiply_probs(A, mat[t, labelingWithBlanks[s]])

	# in case of a blank or a repeated label, we only consider s and s-1 at t-1, so we're done
	if labelingWithBlanks[s] == blank or (s >= 2 and labelingWithBlanks[s-2] == labelingWithBlanks[s]):
		cache[t][s] = res
		return res

	# otherwise, in case of a non-blank and non-repeated label, we additionally add s-2 at t-1
	res = sum_probs(res,multiply_probs(recLabelingProb(t-1, s-2, mat, labelingWithBlanks, blank, cache),mat[t, labelingWithBlanks[s]]))
	cache[t][s] = res
	return res


def emptyCache(maxT, labelingWithBlanks):
	"create empty cache"
	return [[None for _ in range(len(labelingWithBlanks))] for _ in range(maxT)]


def ctcLabelingProb(mat, gt, classes, blank):
	"calculate probability p(gt|mat) of a given labeling gt and a matrix mat according to section 'The CTC Forward-Backward Algorithm' in Graves paper"

	maxT, _ = mat.shape # size of input matrix

	labelingWithBlanks = Common.extendByBlanks(Common.wordToLabelSeq(gt, classes), blank) # ground truth text as label string extended by blanks

	cache = emptyCache(maxT, labelingWithBlanks) # cache subresults to avoid recalculating  subproblems over and over again
	A = recLabelingProb(maxT-1, len(labelingWithBlanks)-1, mat, labelingWithBlanks, blank, cache)
	B = recLabelingProb(maxT-1, len(labelingWithBlanks)-2, mat, labelingWithBlanks, blank, cache)
	prob =  sum_probs(A,B)
	return prob

def multiply_probs(ln_pa, ln_pb):
	"""
	Returns the log probability of multiplied probabilities.
	Pa * Pb = Pc => ln(Pc) = ln(Pa)+ln(Pb)
	"""
	return ln_pa+ln_pb

def sum_probs(ln_pa,ln_pb):
	"""
	Returns the log prob. of two added probs.
	Pa + Pb = Pc => ln(Pc) = ln(Pa) + ln(1+exp(ln(Pb)-ln(Pa))), by Graves dissertation eq 7.18
	"""
	return ln_pa + math.log(1+math.exp(ln_pb-ln_pa))


def ctcLoss(mat, gt, classes): 
	"calculate CTC loss"
	try:
		return -math.log(ctcLabelingProb(mat, gt, classes))
	except:
		return float('inf')
		

# reference: https://github.com/githubharald/CTCDecoder, advancement raraz15, # everything is converted to log probabilities. # check index of labeling # sub-problem already computed # initial values # in case of a blank or a repeated label, we only consider s and s-1 at t-1, so we're done # otherwise, in case of a non-blank and non-repeated label, we additionally add s-2 at t-1 # size of input matrix # ground truth text as label string extended by blanks # cache subresults to avoid recalculating  subproblems over and over again Returns the log probability of multiplied probabilities. 	Pa * Pb = Pc => ln(Pc) = ln(Pa)+ln(Pb) Returns the log prob. of two added probs. 	Pa + Pb = Pc => ln(Pc) = ln(Pa) + ln(1+exp(ln(Pb)-ln(Pa))), by Graves dissertation eq 7.18