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Given a set of phrases, i would like to filter the set of all phrases that contain any of the other phrases. Contained here means that if a phrase contains all the words of another phrase it should be filtered out. Order of the words within the phrase...
Started by yoav.aviram on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
It could be pretty well iterating over words[....
Like:
let matched = set of all phrases for each word in the searched phrase let wordMatch = all this, matched would contain all phrases matching all words in the target phrase.
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First, a disclaimer. I know a little about regex's but I'm no expert. They seem to be something that I really need twice a year so they just don't stay "on top" of my brain.
The situation: I'd like to write a regex to match a certain word, let's call ...
Started by jcollum on
, 6 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
(where the second "Ostrich" does match)
Note that the lookahead assertion: (?![^{]*}) is... .
I believe this will work, using lookahead and lookbehind assertions:
(?<!{[^}]*)Ostrich(?![^{]*})
I also tested the case My {Ostrich} went to the Ostrich Race .
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Input: phrase 1, phrase 2
output: semantic similarity value (between 0 and 1), or the probability these two phrases are talking about the same thing
Started by btw0 on
, 9 posts
by 9 people.
Answer Snippets (Read the full thread at stackoverflow):
To this would be to first run each phrase through a stop-word list (to remove "common" words such as "a", "to", "the", etc.) Then for each of the remaining words in each phrase, you could compute the semantic "similarity" between each of....
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Hi
I have a database table with around 1000 keywords/phrases (one to four words long) - This table changes rarely, so I could extract the data into something more useful (like a regular expression?) - So this is not finding / guessing at keywords based...
Answer Snippets (Read the full thread at stackoverflow):
StringIO("""inputting some text extract the data a phrase not here""") keywords = set(line.strip() for line in keyword_file) results = defaultdict(int) for phrase in keywords: if userinput.find(phrase) != -1: results[phrase....
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Given an arbitrary string, what is an efficient method of finding duplicate phrases? We can say that phrases must be longer than a certain length to be included.
Ideally, you would end up with the number of occurrences for each phrase.
Started by Larsenal on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
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I'm trying to write a filter for Lucene, similar to StopWordsFilter (thus implementing TokenFilter), but I need to remove phrases (sequence of tokens) instead of words.
The "stop phrases" are represented themselves as a sequence of tokens: punctuation...
Started by Enrico Detoma on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
Of words is a "phrase" is dependent on cues, such as punctuation, that are not available after.
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How can I scan a bunch of Microsoft Word (2003) documents? I am searching for a certain phrase in the documents and want to return the file names of those which contain the phrase.
A code sample would be especially helpful.
Started by Josh Stodola on
, 9 posts
by 9 people.
Answer Snippets (Read the full thread at stackoverflow):
Has_phrase(filename, phrase): found = False app = win32com.client.Dispatch('Word.Application') doc = app.Documents.Open(filename, False, False, False) if phrase in doc.Content.Text.lower(): found = True app.Quit() return found ....
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When I created my SSH key while setting up git, I chose to use a pass-phrase for the key. Now, whenever I push to my remote repository (unfuddle), I get prompted for this pass-phrase. Is there some way to automate this?
Started by ProfK on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
Use ssh-agent: http://www.sourcemage.org/Git_Guide#I.27m_tired_of_typing_my_SSH_key_passphrase..
Use SSH Agent.
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Is there any method in .net that wraps phrases given a maximum length for each line?
Example:
Phrase: The quick red fox jumps over the lazy cat Length: 20
Result:
The quick red fox jumps over the lazy cat
thanks!
Started by Schwertz on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
You can use a regular expression:
string text = "The quick brown fox jumps over the lazy dog."; int minLength = 1; int maxLength = 20; MatchCollection lines = Regex.Matches(text, "(.{"+minLength.ToString()+","+maxLength... .
There is no built in method for that.
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I have a database full of phrases (80-100 characters), and some longish documents (50-100Kb), and I would like a ranked list of phrases for a given document; rather than the usual output of a search engine, list of documents for a given phrase.
I've used...
Started by Tourch on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
Go of the phrases and count occurrences....
"XX") if the word occurs at the end of the phrase.
Would it be too slow to turn each phrase into a regex and run each one on the document, counting, to reduce matches, or some default character (e.g.
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