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I need to find all English words which can be formed from the letters in a string
sentence="Ziegler's Giant Bar"
I can make an array of letters by
sentence.split(//)
How can I make more than 4500 English words from the sentence in Ruby?
[edit]
It may ...
Started by Masi on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
Count}}[^#{letter}]*$)/i end # combine the regexes, to form a regex that will only # match words/words") do |f| f.map { |word| word.chomp }.find_all { |word| regex =~ word } end words.length #=>=[] splitArray....
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I need to have all 6 letter Latin words in a list.
I would also like to have words which follow the pattern Xyzzyx in a list.
I have used little Python.
Started by Masi on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
Python would actually be a pretty good to what....
Declensions and verbs' conjugations, your program won't produce anything like all the six-letter words and conjugates Latin words given enough metainformation about each word.
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I'm not sure how to use regular expressions in a function so that I could grab all the words in a sentence starting with a particular letter. I know that I can do:
word =~ /^#{letter}/
to check if the word starts with the letter, but how do I go from ...
Started by TenJack on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
The \b indicates a word boundary - we want to only match starting....
Then we have as many as possible letter characters, followed by another word/\ba[a-z]*\b/i
will match any word starting with 'a'.
word to start with.
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Ask your Facebook Friends
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Using Python, I'm trying to convert a sentence of words into a flat list of all distinct letters in that sentence.
Here's my current code:
words = 'She sells seashells by the seashore' ltr = [] # Convert the string that is "words" to a list of its component...
Started by Art Metzer on
, 8 posts
by 8 people.
Answer Snippets (Read the full thread at stackoverflow):
Set([letter.lower() for letter in words if letter != ' '])
Edit : I just tried it and found in words if letter != ' ')
And if you need to have a list rather than a set, you can
list(set(letter.lower() for letter....
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How do you replace the first letter of a word into Capital letter, e.g.
Trouble me Gold rush brides
into
Trouble Me Gold Rush Brides
Started by neversaint on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
This line should do it:
sed -e "s/\b\(.\)/\u\1/g"
# awk '{for(i=1;i<=NF;i++){ $i=toupper(substr($i,1,1)) substr($i,2) }}1' file Trouble Me Gold Rush Brides .
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Lately I have been playing a game on my iPhone called Scramble. Some of you may know this game as Boggle. Essentially, when the game starts you get a matrix of letters like so:
F X I E A M L O E W B X A S T U
The goal of the game is to find as many words...
Started by Paolo Bergantino on
, 17 posts
by 17 people.
Answer Snippets (Read the full thread at stackoverflow):
As you add more letter permutations it would further decrease the available word sets decreasing = open(dictfile) root = TrieNode(None, '') for word in dict: curNode = root for letter in word.lower doesn't output 1-letter....
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I tried this Lucene code example, which worked:
http://snippets.dzone.com/posts/show/8965
However changing:
Query query = parser.parse("st.");
to
Query query = parser.parse("t");
returned zero hits.
How to write a Lucene query that returns all words containing...
Answer Snippets (Read the full thread at stackoverflow):
You're only option at this point appears to be iterating through all terms, doing you're punctuation and breaks words according....
You can ask Lucene to return terms that begin with a certain letter.
The letter 't' at an arbitrary location.
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Hi, so I need to write an efficient algorithm for looking up words with missing letters in a dictionary and I want the set of possible words.
For example, if I have th??e, I might get back these, those, theme, there.etc.
I was wondering if anyone can ...
Started by SuperString on
, 21 posts
by 20 people.
Answer Snippets (Read the full thread at stackoverflow):
For length 5 words-words").read().split....
Improve on this for large numbers of queries by building an additional list of words for each letter, so length containing an index of each index=character matching word sets.
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How can I delete the \n and the following letters? Thanks a lot.
wordlist = ['Schreiben\nEs', 'Schreiben', 'Schreiben\nEventuell', 'Schreiben\nHaruki'] for x in wordlist: ...?
Started by kame on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
NEventuell', 'Schreiben\nHaruki'] >>> [ re.sub("\n.*", "", word) for word in wordlist ] ['Schreiben wordlist = [re.sub("\n.*", "", word) for word in wordlist]
The regular expression \n.* matches the first the comment....
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I'm not sure exactly how to explain this, so I'll just start with an example.
Given the following data:
Apple Apricot Blackberry Blueberry Cherry Crabapple Cranberry Elderberry Grapefruit Grapes Kiwi Mulberry Nectarine Pawpaw Peach Pear Plum Raspberry...
Started by BrianH on
, 6 posts
by 6 people.
Answer Snippets (Read the full thread at stackoverflow):
Frequency['Z'+1]=threshold_high+1 for letter in range('A' to 'Z'): count += frequency[letter]; if (count>=threshold_low or count+frequency[letter+1]>threshold_high): if (inrange): print rangeStart+'-' print letter+' ....
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