def get_words(letters):
# Index-pairs of letter with similar median stroke widths and similar heights
# We use log2 for linear distance comparison in KDTree
# (i.e. if log2(x) - log2(y) > 1, we know that x > 2*y)
s_ix_letr_pairs = KDTree(np.log2(letters[:, 0:1])).query_pairs(1)
h_ix_letr_pairs = KDTree(np.log2(letters[:, 1:2])).query_pairs(1)
# Calc the angle (direction of text) for all letter-pairs which are
# similar and close to each other
pairs = []
for ix_letr1, ix_letr2 in h_ix_letr_pairs.intersection(s_ix_letr_pairs):
diff = letters[ix_letr1, 3:5] - letters[ix_letr2, 3:5]
# Distance between letters smaller than
# 3 times the width of the wider letter
dist = np.linalg.norm(diff)
if dist < max(letters[ix_letr1, 2], letters[ix_letr2, 2]) * 3:
angle = math.atan2(diff[0], diff[1])
angle += math.pi if angle < 0 else 0
pairs.append([ix_letr1, ix_letr2, angle])
pairs = np.asarray(pairs)
# Pairs of letter-pairs with a similar angle (direction of text)
a_ix_pair_pairs = KDTree(pairs[:, 2:3]).query_pairs(math.pi / 12)
chains = []
for ix_pair_a, ix_pair_b in a_ix_pair_pairs:
pair_a_letr1, pair_a_letr2 = int(pairs[ix_pair_a,
0]), int(pairs[ix_pair_a, 1])
pair_b_letr1, pair_b_letr2 = int(pairs[ix_pair_b,
0]), int(pairs[ix_pair_b, 1])
added = False
for c in chains:
if pair_a_letr1 in c:
c.add(pair_a_letr2)
added = True
elif pair_a_letr2 in c:
c.add(pair_a_letr1)
added = True
if not added:
chains.append({pair_a_letr1, pair_a_letr2})
added = False
for c in chains:
if pair_b_letr1 in c:
c.add(pair_b_letr2)
added = True
elif pair_b_letr2 in c:
c.add(pair_b_letr1)
added = True
if not added:
chains.append({pair_b_letr1, pair_b_letr2})
chains = np.asarray(chains)
vecfunc = np.vectorize(len)
chains = chains[vecfunc(chains) > 3]
_, uniq_ix = np.unique(chains.astype(str), return_index=True)
return chains[uniq_ix]
def _find_words(letr_inf): # swts, heights, widths, topleft_pts, images):
# Index-pairs of letter with similar median stroke widths and similar heights
# We use log2 for linear distance comparison in KDTree
# (i.e. if log2(x) - log2(y) > 1, we know that x > 2*y)
s_ix_letr_pairs = KDTree(np.log2(letr_inf[:, 0:1])).query_pairs(1)
h_ix_letr_pairs = KDTree(np.log2(letr_inf[:, 1:2])).query_pairs(1)
# Calc the angle (direction of text) for all letter-pairs which are
# similar and close to each other
pairs = []
for ix_letr1, ix_letr2 in h_ix_letr_pairs.intersection(s_ix_letr_pairs):
diff = letr_inf[ix_letr1, 3:5] - letr_inf[ix_letr2, 3:5]
# Distance between letters smaller than
# 3 times the width of the wider letter
dist = np.linalg.norm(diff)
if dist < max(letr_inf[ix_letr1, 2], letr_inf[ix_letr2, 2]) * 3:
angle = math.atan2(diff[0], diff[1])
angle += math.pi if angle < 0 else 0
pairs.append([ix_letr1, ix_letr2, angle])
pairs = np.asarray(pairs)
# Pairs of letter-pairs with a similar angle (direction of text)
a_ix_pair_pairs = KDTree(pairs[:, 2:3]).query_pairs(math.pi / 12)
chains = []
for ix_pair_a, ix_pair_b in a_ix_pair_pairs:
# Letter pairs [a] & [b] have a similar angle and each pair consists of
# letter [1] & [2] which meet the similarity-requirements.
pair_a_letr1, pair_a_letr2 = int(pairs[ix_pair_a, 0]), int(pairs[ix_pair_a, 1])
pair_b_letr1, pair_b_letr2 = int(pairs[ix_pair_b, 0]), int(pairs[ix_pair_b, 1])
# TODO: not correct?
added = False
for c in chains:
if pair_a_letr1 in c:
c.add(pair_a_letr2)
added = True
elif pair_a_letr2 in c:
c.add(pair_a_letr1)
added = True
if not added:
chains.append({pair_a_letr1, pair_a_letr2})
added = False
for c in chains:
if pair_b_letr1 in c:
c.add(pair_b_letr2)
added = True
elif pair_b_letr2 in c:
c.add(pair_b_letr1)
added = True
if not added:
chains.append({pair_b_letr1, pair_b_letr2})
chains = np.asarray(chains)
# List of sets of letters with possibly many duplicates
# return chains
# Single list of unique letters
# return np.unique([int(ix) for chain in chains if len(chain) >= 3 for ix in chain])
vecfunc = np.vectorize(len)
chains = chains[vecfunc(chains) > 3]
_, uniq_ix = np.unique(chains.astype(str), return_index=True)
return chains[uniq_ix]
