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build: simplify compute_huffman_coding()

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No functional change.
This commit is contained in:
Artyom Skrobov 2021-04-09 08:36:26 -04:00
parent 68920682b6
commit dcee89ade7
1 changed files with 7 additions and 14 deletions
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21
py/makeqstrdata.py
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@ -166,7 +166,7 @@ def compute_huffman_coding(translations, compression_filename):
sum_len = 0 sum_len = 0
while True: while True:
# Until the dictionary is filled to capacity, use a heuristic to find # Until the dictionary is filled to capacity, use a heuristic to find
# the best "word" (2- to 9-gram) to add to it. # the best "word" (3- to 9-gram) to add to it.
# #
# The TextSplitter allows us to avoid considering parts of the text # The TextSplitter allows us to avoid considering parts of the text
# that are already covered by a previously chosen word, for example # that are already covered by a previously chosen word, for example
@ -178,32 +178,25 @@ def compute_huffman_coding(translations, compression_filename):
for t in texts: for t in texts:
for (found, word) in extractor.iter_words(t): for (found, word) in extractor.iter_words(t):
if not found: if not found:
for substr in iter_substrings(word, minlen=2, maxlen=9): for substr in iter_substrings(word, minlen=3, maxlen=9):
counter[substr] += 1 counter[substr] += 1
# Score the candidates we found. This is an empirical formula only, # Score the candidates we found. This is an empirical formula only,
# chosen for its effectiveness. # chosen for its effectiveness.
scores = sorted( scores = sorted(
((s, (len(s) - 1) ** (occ + 4), occ) for (s, occ) in counter.items()), ((s, (len(s) - 1) ** (occ + 4)) for (s, occ) in counter.items() if occ > 4),
key=lambda x: x[1], key=lambda x: x[1],
reverse=True, reverse=True,
) )
# Do we have a "word" that occurred 5 times and got a score of at least # Pick the one with the highest score.
# 5? Horray. Pick the one with the highest score. if not scores:
word = None
for (s, score, occ) in scores:
if occ < 5:
continue
if score < 5:
break
word = s
break break
word = scores[0][0]
# If we can successfully add it to the dictionary, do so. Otherwise, # If we can successfully add it to the dictionary, do so. Otherwise,
# we've filled the dictionary to capacity and are done. # we've filled the dictionary to capacity and are done.
if not word:
break
if sum_len + len(word) - 2 > max_words_len: if sum_len + len(word) - 2 > max_words_len:
break break
if len(words) == max_words: if len(words) == max_words: