Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Prob Set 3. I'm scratching my head about the starts1(2) part. Not sure how to proceed. I'm reading it as a fuzzy match type problem. Any thoughts?
 3 years ago
 3 years ago
Prob Set 3. I'm scratching my head about the starts1(2) part. Not sure how to proceed. I'm reading it as a fuzzy match type problem. Any thoughts?
 3 years ago
 3 years ago

This Question is Closed

AslanderBest ResponseYou've already chosen the best response.0
ps. My code so far on the string search part seems to work fine: def subStringMatchExact(w,l): """w = word, l = search string""" import string s = 0 index = () for i in w: z = string.find(w, l, s) s = z + 1 if z not in index and z != 1: index = index + (z,) print "Indexes of your search string are:", index Any help would be appreciated  Maybe I just need some sleep. ;)
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
I remember that my original solution to this problem was full of ugly kludges. If you want to work through it together later tonight I'd be in.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Cool  I won't have much time tonight, so I wrote up a quick bit of code that may work. I haven't assigned variables or anything, so it is very raw. Here is my basic idea: def fuzzymatching(): import string w = "abcaabaacabbaccacb" p = "abc" print "Word is:", w print "Search item is:", p s1 = p[0] s2 = p[1] s3 = p[2] s4 = p[:2] s5 = p[1:3] s6 = p[0]+p[2] lista = [s1,s2,s3,s4,s5,s6,] for d in lista: g = string.find(w,d,0) print d, "is at index", g
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Ok well, I quickly ran over my code and produced this instead, which works, albeit its ugly! lol def fuzzymatching(): import string w = "abcaabaacabbaccacb" p = "abc" print "Word is:", w print "Search item is:", p s1 = p[0] s2 = p[1] s3 = p[2] s4 = p[:2] s5 = p[1:3] s6 = p[0]+p[2] lista = [s1,s2,s3,s4,s5,s6,] print "list a is:", lista #print s1, s2, s3, s4, s5, s6 h = 0 for e in w: for d in lista: #while h < len(w): g = string.find(w,d,h) h = h + 1 if g != 1: print d, "is at index", g
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
Cool. I'm looking at a way to save the locations in lists or a big list
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Ya  right now my main problem is writing a function to break down the abc into "a,b,c,ab,ac" automatically, for any size string, and for any combination I specify...
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
yeah, that was the real nightmare last time I tried thistwo different segments meant two tuples of location tuples, then matching them up....This time I think i want to do it in sequenceso find [1:], record all locations, find [0] and [2:], record all locations, etc.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Well I can't be too hard on myself  my total programming experience is less than 3 weeks! I'm off to bed  have a good one...let me know how your attempts turns out. I'm still working on that ps.
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
cool, I'll put this away for the night after I finish this bit.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
I made some progress...I am working on this now: def fuzzier3(): """For W, add a word in quotes. For S, add a letter(s) to find in w.""" import string w = "abcabdabeabcmmmma" s = "a" i = 0 z = 0 print s while i <= len(w): j = string.find(w, s, i) i = i + 1 if j != 1: print "Index for each letter combination is:", j i = j + 1 z = z + 1 print "The amount of times the string is found in your word:", z You think I'm on the right track?
 3 years ago

zhentarBest ResponseYou've already chosen the best response.0
Getting the first one right is going to be crucial for the 3b, 3c,and 3d. First, I suggest using more descriptive names, and combining the while with the if j != 1. Also note that he wants you be using tuples and a specific function setup, Here is what I came up with for the iterative part for 3a: def subStringMatchExact(target,key): locations = () found = find(target,key) while found != 1: locations = locations[:] + (found,) found = find(target,key,found+1) return locations While it looks simple, there are a few tricks in there you should know, like how to dynamically create and append to tuples. Hopefully this will help you see the rest of the assignment a little more clearly
 3 years ago

zhentarBest ResponseYou've already chosen the best response.0
if you want to return the number of matches use this at the end instead: return len(locations)
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
I never thought of writing this: locations = locations[:] + (found,). Thanks! My other working trial has better descriptions of variables. I start by using very simple single letters, otherwise I find I get mixed up. Thanks for the advice...and I am still reading about tuples...haven't read everything yet though.
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
Aslander, I'm not entirely sure what part of the problem you're solving with the previous. I still think what you need is a way to generate all the different substrings you need to look for. I suggest using an integer to represent an index number in the original string, and search for two strings broken using that integer. Something like targetstring = the string I want to find stuff in keystring = the string I want to find indexnumber = 0 for indexnumber in range (length of keystring): firstsubstring = the part of the keystring from its beginning to the current indexnumber secondsubstring = the part of the keystring from the current indexnumber to the end then you can store those to look for them later, or you can look for them and figure out how to match them up to find strings with one element changed, or whatever.
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
Oh, and for either firstsubstring or secondsubstring you'll have to add/subtract from indexnumber to make sure you're skipping a letter, and you'll have to do something (if statement or whatever) to make sure you don't get an indexerror when that addition/subtraction puts the indexnumber out of the index of the keystring.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Somnamniac: What I was trying was to first solve the small problem of finding a search letter (s) in a word(w). If that code worked I would then tweak it to find a combination of letters in "w". If this worked, then I would try and enter the tuples into the code, and try finding combinations of the two letter combos in "w".
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
For some reason I can find the index of combos, but not make it recur...or iterate, for any combination of two letters in the key.
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
I'm not sure what you mean by "a combination of letters." is that one string, or two strings, or a series of strings of single letters? The way I see it, the first thing you posted in this thread will give you (or can be made to give you, I forget) a tuple of locations of a key string in a target string. If you just call that function a bunch of times with different key strings, you'll get a tuple of the locations of each keystring. so if your original keystring is '12345678', you can call that function with '' then '2345678', then '1' and '345678' etc, and the problem becomes one of taking the tuples of locations you get and matching them up to find where the first string is separated from the second by one index number.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Ok  I see now, I was reading the problem incorrectly, and got confused...my bad. :) I ended up writing code that would concatenate abc as abc a  ab ac  b  bc. But where is the ac? I'm working on that now...thanks for making me realize that somnamniac.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
ps. The "abc as abc a  ab ac  b  bc. ", should be without the ac....there should be a way to fix typos! :)
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
you don't want ac, you want 'a c', right?
 3 years ago

somnamniacBest ResponseYou've already chosen the best response.0
or rather, 'a' and 'c' with something else between them.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Exactly...a and c with something in between.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
Now if I use the key "abc". I would need to get "something"+bc or a+"something"+c, or ab+"something", where Something = a, b or c. Would that be right?
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
If anyone is interested  the code I got to work is this: def findmatch(word, search): """Finds a search item in a word.""" import string n = 0 searchtuple = () for i in word: index = string.find(word,search,n) n = index + 1 if index != 1 and index not in searchtuple: searchtuple = searchtuple + (index,) print "hi", checkkeys(search) return searchtuple, "Keys from search item:", checkkeys(search) ps. checkkeys(search) is another function I wrote to return all those substrings/keys. It works pretty good, but I could still write it better...or finetune it.
 3 years ago

KeenBest ResponseYou've already chosen the best response.0
Interesting, I made a different assumption than you did. I decided that overlapping results were not really separate results. So if you're searching for 'abab' in 'ababab', you'd just get 0. This is what I came up with: def subStringMatchExact(target, key) : lastfind = find(target, key), while find (target, key, lastfind[1] + len(key)) > 1 : lastfind += find(target, key, lastfind[1] + len(key)), if lastfind == (1,) : return () else: return lastfind
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
I understood the problem to demand returns of all instances of the key, regardless of overlapping. It also called for fuzzy matching, or if the key is abcd, then we would need to return any matches for a(n)cd, ab(n)d, abc, and bcd. In these cases, anything close might hit in the target string. My checkkeys(search) function tries to do that, returning everything in tuples.
 3 years ago

nodiggaBest ResponseYou've already chosen the best response.0
I'm slightly confused on part 4 of the problem. The sentence that says "If we keep only those elements of the second tuple that don't occur in the first tuple, we will have the matches with exactly one substitution" Using target1 and key12, if I input them into subStringExactMatch I get (5,15). If I input these into subStringMatchOneSub I get (5,15) for all subs except for when the key is broken up to "atg" and " ", where the possible matches are (0,5,15). From what I am understanding then the match with one substitution would be the tuple (0). Am I missing something here? Thanks.
 3 years ago

AslanderBest ResponseYou've already chosen the best response.0
What I understood was that the functions we wrote will return hits of keys with and without substitutions. What problem 4 says to me is: do not get exact matches, but only those with one sub, by using your answers(functions) to problem 2+3, and use only thoes elements from Tuple 2 that do not match (or occur) in Tuple1. Now if I misunderstood things as well, then I have a bigger problem than you do! :) Food for thought though, and I think I'll have to go over that problems set now!
 3 years ago

nodiggaBest ResponseYou've already chosen the best response.0
I'll give it another look a bit later. Thanks for the help :)
 3 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.