def do_search(keywords):
global user_top_20_database
# Fetch the current session
request_session = request.environ["beaker.session"]
# Fetch the users email for their session
user_email = request_session.get("user_email", "Anonymous")
if reduce(and_, map(lambda c: c in math_chars, keywords)):
result = None
try:
result = eval(keywords.replace("^", "**").replace("[", "(").replace("]", ")"))
return result_template(
user_email,
keywords,
template(
"""
<p> {{keywords}} = {{result}} </p>
""",
keywords=keywords,
result=result,
),
)
except Exception as e:
pass
# A list of all keywords from the search query.
keyword_list = map(str.lower, keywords.split())
keywords = keyword_list
# -----------------------------------------------------------------------
counted_keyword_list = [(keyword_list.count(x), x) for x in set(keyword_list)]
# Sort the list in descending order of frequency.
counted_keyword_list.sort(key=wordCount, reverse=1)
page = request.query.get("page")
if user_email <> "anonymous" and page == None:
# Fetch the top 20 list for that users email
user_top_20 = user_top_20_database.get(user_email)
if user_top_20 != None:
# Add to the top 20 list and update totals.
# Iterate through the counted keyword list.
for keywords1 in counted_keyword_list:
# If any keywords are already in the top 20 list, merge them into the top 20 list.
if any(keywords1[1] in element for element in user_top_20):
# Iterator to keep track of which keyword in the top 20 list we are at.
i = 0
# Iterate through the keyword pairs and add the values from the counted_keyword_list into the top20 list.
for keywords2 in user_top_20:
# If the keywords match.
if keywords2[1] == keywords1[1]:
# Save the count value of the user_top_20 version.
keyword_count = keywords2[0]
# Delete the old user_top_20 keyword and count.
del user_top_20[i]
# Add the keyword with updated count to the front of the top_20 list.
user_top_20.insert(0, ((keywords1[0] + keyword_count), keywords1[1]))
# Iterate
i = i + 1
# If the word isn't already in the top 20 list add it.
else:
user_top_20.append(keywords1)
# Organize the top 20 list in decending order by the frequency of a keyword.
user_top_20.sort(key=wordCount, reverse=1)
# Update the database of user search history
user_top_20_database["user_email"] = user_top_20
# If the user_top_20 list is longer than 20 keywords, trim it.
# while len(user_top_20) > 20:
# del user_top_20[-1]
# ------------------------------------------------------------------------
# Grab the first keyword that was inputted by the user
if keyword_list == []:
results_list = []
return generate_page_results(1, results_list, [], user_email)
if page == None:
page = 1
else:
page = int(page)
db = DBHandler()
# Get the word_ids through a getter in the database
word_ids = []
ignored_words = set(
[
"",
"the",
"of",
"at",
"on",
"in",
"is",
"it",
"a",
"b",
"c",
"d",
"e",
"f",
"g",
"h",
"i",
"j",
"k",
"l",
"m",
"n",
"o",
"p",
"q",
"r",
"s",
"t",
"u",
"v",
"w",
"x",
"y",
"z",
"and",
"or",
]
)
for keyword in keyword_list:
if keyword in ignored_words:
continue
word_ids.append(db.get_word_id(keyword))
# Get the doc_ids from the word_ids in the database
list_of_doc_id_lists = []
for word_id in word_ids:
if word_id == None:
list_of_doc_id_lists.append([])
else:
list_of_doc_id_lists.append(db.get_doc_ids(word_id))
# Find lists of doc_ids that intersect with each other, this will give us doc ids that contain both keywords
intersecting_doc_ids = find_intersections(list_of_doc_id_lists)
# Get the url_ranks from pagerank in the database
ranks = db.get_pageranks(intersecting_doc_ids)
# Zip the doc_ids with the corresponding url_ranks to make ranked_doc_ids
ranked_doc_ids = zip(ranks, intersecting_doc_ids)
# Sort the ranked_doc_ids to make sorted_doc_ids and get the sorted_urls from the database
ranked_sorted_doc_ids = sorted(ranked_doc_ids, key=itemgetter(0))
results_list = map(itemgetter(0), db.get_urls(map(itemgetter(1), ranked_sorted_doc_ids)))
return generate_page_results(page, results_list, keyword_list, user_email)
def do_search(keywords):
global user_top_20_database
# Fetch the current session
request_session = request.environ['beaker.session']
# Fetch the users email for their session
user_email = request_session.get('user_email', 'Anonymous')
if reduce(and_, map(lambda c: c in math_chars, keywords)):
result = None
try:
result = eval(
keywords.replace('^', '**').replace('[',
'(').replace(']', ')'))
return result_template(
user_email, keywords,
template('''
<p> {{keywords}} = {{result}} </p>
''',
keywords=keywords,
result=result))
except Exception as e:
pass
# A list of all keywords from the search query.
keyword_list = map(str.lower, keywords.split())
keywords = keyword_list
#-----------------------------------------------------------------------
counted_keyword_list = [(keyword_list.count(x), x)
for x in set(keyword_list)]
# Sort the list in descending order of frequency.
counted_keyword_list.sort(key=wordCount, reverse=1)
page = request.query.get('page')
if user_email <> 'anonymous' and page == None:
# Fetch the top 20 list for that users email
user_top_20 = user_top_20_database.get(user_email)
if user_top_20 != None:
# Add to the top 20 list and update totals.
# Iterate through the counted keyword list.
for keywords1 in counted_keyword_list:
# If any keywords are already in the top 20 list, merge them into the top 20 list.
if any(keywords1[1] in element for element in user_top_20):
# Iterator to keep track of which keyword in the top 20 list we are at.
i = 0
# Iterate through the keyword pairs and add the values from the counted_keyword_list into the top20 list.
for keywords2 in user_top_20:
# If the keywords match.
if keywords2[1] == keywords1[1]:
# Save the count value of the user_top_20 version.
keyword_count = keywords2[0]
# Delete the old user_top_20 keyword and count.
del user_top_20[i]
# Add the keyword with updated count to the front of the top_20 list.
user_top_20.insert(
0,
((keywords1[0] + keyword_count), keywords1[1]))
# Iterate
i = i + 1
# If the word isn't already in the top 20 list add it.
else:
user_top_20.append(keywords1)
# Organize the top 20 list in decending order by the frequency of a keyword.
user_top_20.sort(key=wordCount, reverse=1)
# Update the database of user search history
user_top_20_database["user_email"] = user_top_20
# If the user_top_20 list is longer than 20 keywords, trim it.
# while len(user_top_20) > 20:
# del user_top_20[-1]
#------------------------------------------------------------------------
# Grab the first keyword that was inputted by the user
if keyword_list == []:
results_list = []
return generate_page_results(1, results_list, [], user_email)
if page == None:
page = 1
else:
page = int(page)
db = DBHandler()
# Get the word_ids through a getter in the database
word_ids = []
ignored_words = set([
'',
'the',
'of',
'at',
'on',
'in',
'is',
'it',
'a',
'b',
'c',
'd',
'e',
'f',
'g',
'h',
'i',
'j',
'k',
'l',
'm',
'n',
'o',
'p',
'q',
'r',
's',
't',
'u',
'v',
'w',
'x',
'y',
'z',
'and',
'or',
])
for keyword in keyword_list:
if keyword in ignored_words:
continue
word_ids.append(db.get_word_id(keyword))
# Get the doc_ids from the word_ids in the database
list_of_doc_id_lists = []
for word_id in word_ids:
if word_id == None:
list_of_doc_id_lists.append([])
else:
list_of_doc_id_lists.append(db.get_doc_ids(word_id))
# Find lists of doc_ids that intersect with each other, this will give us doc ids that contain both keywords
intersecting_doc_ids = find_intersections(list_of_doc_id_lists)
# Get the url_ranks from pagerank in the database
ranks = db.get_pageranks(intersecting_doc_ids)
# Zip the doc_ids with the corresponding url_ranks to make ranked_doc_ids
ranked_doc_ids = zip(ranks, intersecting_doc_ids)
# Sort the ranked_doc_ids to make sorted_doc_ids and get the sorted_urls from the database
ranked_sorted_doc_ids = sorted(ranked_doc_ids, key=itemgetter(0))
results_list = map(itemgetter(0),
db.get_urls(map(itemgetter(1), ranked_sorted_doc_ids)))
return generate_page_results(page, results_list, keyword_list, user_email)
