On Jan 7, 3:50 pm, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote: > On Jan 7, 5:10 pm, dgoldsmith_89 <[EMAIL PROTECTED]> wrote: > > > > > On Jan 7, 2:54 pm, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote: > > > > On Jan 7, 4:37 pm, dgoldsmith_89 <[EMAIL PROTECTED]> wrote: > > > > > Can anyone point me to a downloadable open source English dictionary > > > > suitable for programmatic use with python: I'm programming a puzzle > > > > generator, and I need to be able to generate more or less complete > > > > lists of English words, alphabetized. Thanks! DG > > > >www.puzzlers.orghasnumerousword lists & dictionarys in text > > > format that can be downloaded. I recommend you insert them into > > > some form of database. I have most of them in an Access db and > > > it's 95 MB. That's a worse case as I also have some value-added > > > stuff, the OSPD alone would be a lot smaller. > > > > <http://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:start> > > > Sorry for my ignorance: I can query an Access DB w/ standard SQL > > queries (and this is how I would access it w/ Python)? > > Yes, if you have the appropriate way to link to the DB. > I use Windows and ODBC from Win32. I don't know what you > would use on a Mac. > > As Paul McGuire said, you could easily do this with SqlLite3. > Personnaly, I always use Access since my job requires it > and I find it much more convenient. I often use Crosstab > tables which I think SqlLite3 doesn't support. Typically, > I'll write complex queries in Access and simple select SQL > statements in Python to grab them. > > Here's my anagram locator. (the [signature] is an example > of the value-added I mentioned). > > ## I took a somewhat different approach. Instead of in a file, > ## I've got my word list (562456 words) in an MS-Access database. > ## And instead of calculating the signature on the fly, I did it > ## once and added the signature as a second field: > ## > ## TABLE CONS_alpha_only_signature_unique > ## -------------------------------------- > ## CONS text 75 > ## signature text 26 > ## > ## The signature is a 26 character string where each character is > ## the count of occurences of the matching letter. Luckily, in > ## only a single case was there more than 9 occurences of any > ## given letter, which turned not to be a word but a series of > ## words concatenated so I just deleted it from the database > ## (lots of crap in the original word list I used). > ## > ## Example: > ## > ## CONS signature > ## aah 20000001000000000000000000 # 'a' occurs twice & 'h' once > ## aahed 20011001000000000000000000 > ## aahing 20000011100001000000000000 > ## aahs 20000001000000000010000000 > ## aaii 20000000200000000000000000 > ## aaker 20001000001000000100000000 > ## aal 20000000000100000000000000 > ## aalborg 21000010000100100100000000 > ## aalesund > 20011000000101000010100000 > ## > ## Any words with identical signatures must be anagrams. > ## > ## Once this was been set up, I wrote a whole bunch of queries > ## to use this table. I use the normal Access drag and drop > ## design, but the SQL can be extracted from each, so I can > ## simply open the query from Python or I can grab the SQL > ## and build it inside the program. The example > ## > ## signatures_anagrams_select_signature > ## > ## is hard coded for criteria 9 & 10 and should be cast inside > ## Python so the criteria can be changed dynamically. > ## > ## > ## QUERY signatures_anagrams_longest > ## --------------------------------- > ## SELECT Len([CONS]) AS Expr1, > ## Count(Cons_alpha_only_signature_unique.CONS) AS > CountOfCONS, > ## Cons_alpha_only_signature_unique.signature > ## FROM Cons_alpha_only_signature_unique > ## GROUP BY Len([CONS]), > ## Cons_alpha_only_signature_unique.signature > ## HAVING (((Count(Cons_alpha_only_signature_unique.CONS))>1)) > ## ORDER BY Len([CONS]) DESC , > ## Count(Cons_alpha_only_signature_unique.CONS) DESC; > ## > ## This is why I don't use SQLite3, must have crosstab queries. > ## > ## QUERY signatures_anagram_summary > ## -------------------------------- > ## TRANSFORM Count(signatures_anagrams_longest.signature) AS > CountOfsignature > ## SELECT signatures_anagrams_longest.Expr1 AS [length of word] > ## FROM signatures_anagrams_longest > ## GROUP BY signatures_anagrams_longest.Expr1 > ## PIVOT signatures_anagrams_longest.CountOfCONS; > ## > ## > ## QUERY signatures_anagrams_select_signature > ## ------------------------------------------ > ## SELECT Len([CONS]) AS Expr1, > ## Count(Cons_alpha_only_signature_unique.CONS) AS > CountOfCONS, > ## Cons_alpha_only_signature_unique.signature > ## FROM Cons_alpha_only_signature_unique > ## GROUP BY Len([CONS]), > ## Cons_alpha_only_signature_unique.signature > ## HAVING (((Len([CONS]))=9) AND > ## ((Count(Cons_alpha_only_signature_unique.CONS))=10)) > ## ORDER BY Len([CONS]) DESC , > ## Count(Cons_alpha_only_signature_unique.CONS) DESC; > ## > ## QUERY signatures_lookup_by_anagram_select_signature > ## --------------------------------------------------- > ## SELECT signatures_anagrams_select_signature.Expr1, > ## signatures_anagrams_select_signature.CountOfCONS, > ## Cons_alpha_only_signature_unique.CONS, > ## Cons_alpha_only_signature_unique.signature > ## FROM signatures_anagrams_select_signature > ## INNER JOIN Cons_alpha_only_signature_unique > ## ON signatures_anagrams_select_signature.signature > ## = Cons_alpha_only_signature_unique.signature; > ## > ## > ## Now it's a simple matter to use the ODBC from Win32 to extract > ## the query output into Python. > > import dbi > import odbc > > con = odbc.odbc("words") > cursor = con.cursor() > > ## This first section grabs the anagram summary. Note that > ## queries act just like tables (as long as they don't have > ## internal dependencies. I read somewhere you can get the > ## field names, but here I put them in by hand. > > ##cursor.execute("SELECT * FROM signature_anagram_summary") > ## > ##results = cursor.fetchall() > ## > ##for i in results: > ## for j in i: > ## print '%4s' % (str(j)), > ## print > > ## (if this wraps, each line is 116 characters) > ## 2 3 4 5 6 7 8 9 10 11 12 13 > 14 15 16 17 18 23 > ## 2 259 None None None None None None None None None None None > None None None None None None > ## 3 487 348 218 150 102 None None None None None None None > None None None None None None > ## 4 1343 718 398 236 142 101 51 26 25 9 8 3 > 2 None None None None None > ## 5 3182 1424 777 419 274 163 106 83 53 23 20 10 > 6 4 5 1 3 1 > ## 6 5887 2314 1051 545 302 170 114 54 43 21 15 6 > 5 4 4 2 None None > ## 7 7321 2251 886 390 151 76 49 37 14 7 5 1 > 1 1 None None None None > ## 8 6993 1505 452 166 47 23 8 6 4 2 2 None > None None None None None None > ## 9 5127 830 197 47 17 6 None None 1 None None None > None None None None None None > ## 10 2975 328 66 8 2 None None None None None None None > None None None None None None > ## 11 1579 100 5 4 2 None None None None None None None > None None None None None None > ## 12 781 39 2 1 None None None None None None None None > None None None None None None > ## 13 326 11 2 None None None None None None None None None > None None None None None None > ## 14 166 2 None None None None None None None None None None > None None None None None None > ## 15 91 None 1 None None None None None None None None None > None None None None None None > ## 16 60 None None None None None None None None None None None > None None None None None None > ## 17 35 None None None None None None None None None None None > None None None None None None > ## 18 24 None None None None None None None None None None None > None None None None None None > ## 19 11 None None None None None None None None None None None > None None None None None None > ## 20 6 None None None None None None None None None None None > None None None None None None > ## 21 6 None None None None None None None None None None None > None None None None None None > ## 22 4 None None None None None None None None None None None > None None None None None None > > ## From the query we have the word size as row header and size of > ## anagram set as column header. The data value is the count of > ## how many different anagram sets match the row/column header. > ## > ## For example, there are 7321 different 7-letter signatures that > ## have 2 anagram sets. There is 1 5-letter signature having a > ## 23 member anagram set. > ## > ## We can then pick any of these, say the single 10 member anagram > ## set of 9-letter words, and query out out the anagrams: > > cursor.execute("SELECT * FROM > signatures_lookup_by_anagram_select_signature") > results = cursor.fetchall() > for i in results: > for j in i: > print j, > print > > ## 9 10 anoretics 10101000100001100111000000 > ## 9 10 atroscine 10101000100001100111000000 > ## 9 10 certosina 10101000100001100111000000 > ## 9 10 creations 10101000100001100111000000 > ## 9 10 narcotise 10101000100001100111000000 > ## 9 10 ostracine 10101000100001100111000000 > ## 9 10 reactions 10101000100001100111000000 > ## 9 10 secration 10101000100001100111000000 > ## 9 10 tinoceras 10101000100001100111000000 > ## 9 10 tricosane 10101000100001100111000000 > > ## Nifty, eh? > > > > > DG
Yes, nifty. Thanks for all the help, all! DG -- http://mail.python.org/mailman/listinfo/python-list
