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Fix parsing #18

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Fix parsing #18

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f4a4763
Use regex and json to parse synonyms
absterr08 Jul 11, 2018
b1ee506
Use regex and json to parse antonyms
absterr08 Jul 11, 2018
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26 changes: 14 additions & 12 deletions PyDictionary/core.py
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Original file line number Diff line number Diff line change
@@ -1,5 +1,5 @@
from __future__ import print_function
import sys, re, goslate
import sys, re, goslate, json
try:
from .utils import _get_soup_object
except:
Expand All @@ -19,7 +19,7 @@ def __init__(self, *args):
except:
self.args = args


def printMeanings(self):
dic = self.getMeanings()
for key in dic.keys():
Expand Down Expand Up @@ -69,12 +69,13 @@ def synonym(term, formatted=False):
else:
try:
data = _get_soup_object("http://www.thesaurus.com/browse/{0}".format(term))
terms = data.select("div#filters-0")[0].findAll("li")
if len(terms) > 5:
terms = terms[:5:]
re_synonyms = re.search('"synonyms":(\[.*?\])', data.get_text()).group(1)
synonyms = json.loads(re_synonyms)
if len(synonyms) > 5:
synonyms = synonyms[:5:]
li = []
for t in terms:
li.append(t.select("span.text")[0].getText())
for s in synonyms:
li.append(s["term"])
if formatted:
return {term: li}
return li
Expand All @@ -100,12 +101,13 @@ def antonym(word, formatted=False):
else:
try:
data = _get_soup_object("http://www.thesaurus.com/browse/{0}".format(word))
terms = data.select("section.antonyms")[0].findAll("li")
if len(terms) > 5:
terms = terms[:5:]
re_antonyms = re.search('"antonyms":(\[.*?\])', data.get_text()).group(1)
antonyms = json.loads(re_antonyms)
if len(antonyms) > 5:
antonyms = antonyms[:5:]
li = []
for t in terms:
li.append(t.select("span.text")[0].getText())
for a in antonyms:
li.append(a["term"])
if formatted:
return {word: li}
return li
Expand Down