Basically, I print a long message but I want to group all of those words into 5 character long strings.
For example "iPhone 6 isn’t simply bigger — it’s better in every way. Larger, yet dramatically thinner." I want to make that
"iPhon 6isn' tsimp lybig ger-i t'sbe terri never yway. Large r,yet drama tical lythi nner. "
As suggested by #vaultah, this is achieved by splitting the string by a space and joining them back without spaces; then using a for loop to append the result of a slice operation to an array. An elegant solution is to use a comprehension.
text = "iPhone 6 isn’t simply bigger — it’s better in every way. Larger, yet dramatically thinner."
joined_text = ''.join(text.split())
splitted_to_six = [joined_text[char:char+6] for char in range(0,len(joined_text),6)]
' '.join(splitted_to_six)
I'm sure you can use the re module to get back dashes and apostrophes as they're meant to be
Simply do the following.
import re
sentence="iPhone 6 isn't simply bigger - it's better in every way. Larger, yet dramatically thinner."
sentence = re.sub(' ', '', sentence)
count=0
new_sentence=''
for i in sentence:
if(count%5==0 and count!=0):
new_sentence=new_sentence+' '
new_sentence=new_sentence+i
count=count+1
print new_sentence
Output:
iPhon e6isn 'tsim plybi gger- it'sb etter ineve ryway .Larg er,ye tdram atica llyth inner .
Actually i am coming from C++ and i am new here as well, I am having iteration problem.I am using python 2.7.8 and unable to solve which is what i am wanting. I have a file name called "foo.txt". Through code i am trying to find using how many "a e i o u" are in the file. I have created array: vowel[] = {'a','e','i','o',u} and my code should shd give me the combine count of all vowels. But i am facing
error:
TypeError: list indices must be integers, not str
file foo.txt
Chronobiology might sound a little futuristic – like something from a science fiction novel, perhaps – but it’s actually a field of study that concerns one of the oldest processes life on this planet has ever known: short-term rhythms of time and their effect on flora and fauna.
This can take many forms. Marine life, for example, is influenced by tidal patterns. Animals tend to be active or inactive depending on the position of the sun or moon. Numerous creatures, humans included, are largely diurnal – that is, they like to come out during the hours of sunlight. Nocturnal animals, such as bats and possums, prefer to forage by night. A third group are known as crepuscular: they thrive in the low-light of dawn and dusk and remain inactive at other hours.
When it comes to humans, chronobiologists are interested in what is known as the circadian rhythm. This is the complete cycle our bodies are naturally geared to undergo within the passage of a twenty-four hour day. Aside from sleeping at night and waking during the day, each cycle involves many other factors such as changes in blood pressure and body temperature. Not everyone has an identical circadian rhythm. ‘Night people’, for example, often describe how they find it very hard to operate during the morning, but become alert and focused by evening. This is a benign variation within circadian rhythms known as a chronotype.
my code:
fo = open("foo.txt", "r")
count = 0
for i in fo:
word = i
vowels = ['a','e','i','o','u','y']
word = word.lower().strip(".:;?!")
#print word
for j in word: # wanting that loop shd iterate till the end of file
for k in vowels: # wanting to index string array until **vowels.length()**
if (vowels[k] == word[j]):
count +=1
#print word[0]
print count
Python has a wonderful module called collections with a function Counter. You can use it like this:
import collections
with open('foo.txt') as f:
letters = collections.Counter(f.read())
vowels = ['a','e','i','o','u','y']
## you just want the sum
print(sum(letters[vowel] for vowel in vowels))
You can also do it without collections.Counter():
import itertools
vowels = {'a','e','i','o','u','y'}
with open("foo.txt") as f:
print(sum(1 for char in itertools.chain.from_iterable(f) if char in vowels))
Please note that the time complexity of a set {} lookup is O(1), whereas the time complexity for a list [] lookup is O(n) according to this page on wiki.python.org.
I tested both methods with the module timeit and as expected the first method using collections.Counter() is slightly faster:
0.13573385099880397
0.16710168996360153
Do in range(len()) instead, because if you use for k in vowels , k will be 'a' then 'b' then 'c'... etc. However, the syntax for getting objects via indexes is vowels[index_number], not vowels[content]. So, you have to iterate over the length of the array, and use vowels[0] to get 'a', then vowels[1]' to get 'b' etc.
fo = open("foo.txt", "r")
count = 0
for i in fo:
word = i
vowels = ['a','e','i','o','u','y']
word = word.lower().strip(".:;?!")
#print word
for j in range(len(word)): # wanting that loop shd iterate till the end of file
if (word[j] in vowels):
count +=1
#print word[0]
print count
Python prides itself on its abstraction and standard library data structures. Check out collections.Counter. It takes an iterable and returns a dict of value -> frequency.
with open('foo.txt') as f:
string = f.read()
counter = collections.Counter(string) # a string is an iterable of characters
vowel_counts = {vowel: counter[vowel] for vowel in "aeiou"}
This question already has answers here:
How can I split a text into sentences?
(20 answers)
Closed 3 years ago.
I want to make a list of sentences from a string and then print them out. I don't want to use NLTK to do this. So it needs to split on a period at the end of the sentence and not at decimals or abbreviations or title of a name or if the sentence has a .com This is attempt at regex that doesn't work.
import re
text = """\
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.
"""
sentences = re.split(r' *[\.\?!][\'"\)\]]* *', text)
for stuff in sentences:
print(stuff)
Example output of what it should look like
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it.
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with a probability of .9 it isn't.
(?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)\s
Try this. split your string this.You can also check demo.
http://regex101.com/r/nG1gU7/27
Ok so sentence-tokenizers are something I looked at in a little detail, using regexes, nltk, CoreNLP, spaCy. You end up writing your own and it depends on the application. This stuff is tricky and valuable and people don't just give their tokenizer code away. (Ultimately, tokenization is not a deterministic procedure, it's probabilistic, and also depends very heavily on your corpus or domain, e.g. legal/financial documents vs social-media posts vs Yelp reviews vs biomedical papers...)
In general you can't rely on one single Great White infallible regex, you have to write a function which uses several regexes (both positive and negative); also a dictionary of abbreviations, and some basic language parsing which knows that e.g. 'I', 'USA', 'FCC', 'TARP' are capitalized in English.
To illustrate how easily this can get seriously complicated, let's try to write you that functional spec for a deterministic tokenizer just to decide whether single or multiple period ('.'/'...') indicates end-of-sentence, or something else:
function isEndOfSentence(leftContext, rightContext)
Return False for decimals inside numbers or currency e.g. 1.23 , $1.23, "That's just my $.02" Consider also section references like 1.2.A.3.a, European date formats like 09.07.2014, IP addresses like 192.168.1.1, MAC addresses...
Return False (and don't tokenize into individual letters) for known abbreviations e.g. "U.S. stocks are falling" ; this requires a dictionary of known abbreviations. Anything outside that dictionary you will get wrong, unless you add code to detect unknown abbreviations like A.B.C. and add them to a list.
Ellipses '...' at ends of sentences are terminal, but in the middle of sentences are not. This is not as easy as you might think: you need to look at the left context and the right context, specifically is the RHS capitalized and again consider capitalized words like 'I' and abbreviations. Here's an example proving ambiguity which : She asked me to stay... I left an hour later. (Was that one sentence or two? Impossible to determine)
You may also want to write a few patterns to detect and reject miscellaneous non-sentence-ending uses of punctuation: emoticons :-), ASCII art, spaced ellipses . . . and other stuff esp. Twitter. (Making that adaptive is even harder). How do we tell if #midnight is a Twitter user, the show on Comedy Central, text shorthand, or simply unwanted/junk/typo punctuation? Seriously non-trivial.
After you handle all those negative cases, you could arbitrarily say that any isolated period followed by whitespace is likely to be an end of sentence. (Ultimately, if you really want to buy extra accuracy, you end up writing your own probabilistic sentence-tokenizer which uses weights, and training it on a specific corpus(e.g. legal texts, broadcast media, StackOverflow, Twitter, forums comments etc.)) Then you have to manually review exemplars and training errors. See Manning and Jurafsky book or Coursera course [a].
Ultimately you get as much correctness as you are prepared to pay for.
All of the above is clearly specific to the English-language/ abbreviations, US number/time/date formats. If you want to make it country- and language-independent, that's a bigger proposition, you'll need corpora, native-speaking people to label and QA it all, etc.
All of the above is still only ASCII, which is practically speaking only 96 characters. Allow the input to be Unicode, and things get harder still (and the training-set necessarily must be either much bigger or much sparser)
In the simple (deterministic) case, function isEndOfSentence(leftContext, rightContext) would return boolean, but in the more general sense, it's probabilistic: it returns a float 0.0-1.0 (confidence level that that particular '.' is a sentence end).
References: [a] Coursera video: "Basic Text Processing 2-5 - Sentence Segmentation - Stanford NLP - Professor Dan Jurafsky & Chris Manning" [UPDATE: an unofficial version used to be on YouTube, was taken down]
Try to split the input according to the spaces rather than a dot or ?, if you do like this then the dot or ? won't be printed in the final result.
>>> import re
>>> s = """Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't."""
>>> m = re.split(r'(?<=[^A-Z].[.?]) +(?=[A-Z])', s)
>>> for i in m:
... print i
...
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it.
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with a probability of .9 it isn't.
sent = re.split('(?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)(\s|[A-Z].*)',text)
for s in sent:
print s
Here the regex used is : (?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)(\s|[A-Z].*)
First block: (?<!\w\.\w.) : this pattern searches in a negative feedback loop (?<!) for all words (\w) followed by fullstop (\.) , followed by other words (\.)
Second block: (?<![A-Z][a-z]\.): this pattern searches in a negative feedback loop for anything starting with uppercase alphabets ([A-Z]), followed by lower case alphabets ([a-z]) till a dot (\.) is found.
Third block: (?<=\.|\?): this pattern searches in a feedback loop of dot (\.) OR question mark (\?)
Fourth block: (\s|[A-Z].*): this pattern searches after the dot OR question mark from the third block. It searches for blank space (\s) OR any sequence of characters starting with a upper case alphabet ([A-Z].*).
This block is important to split if the input is as
Hello world.Hi I am here today.
i.e. if there is space or no space after the dot.
Naive approach for proper english sentences not starting with non-alphas and not containing quoted parts of speech:
import re
text = """\
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.
"""
EndPunctuation = re.compile(r'([\.\?\!]\s+)')
NonEndings = re.compile(r'(?:Mrs?|Jr|i\.e)\.\s*$')
parts = EndPunctuation.split(text)
sentence = []
for part in parts:
if len(part) and len(sentence) and EndPunctuation.match(sentence[-1]) and not NonEndings.search(''.join(sentence)):
print(''.join(sentence))
sentence = []
if len(part):
sentence.append(part)
if len(sentence):
print(''.join(sentence))
False positive splitting may be reduced by extending NonEndings a bit. Other cases will require additional code. Handling typos in a sensible way will prove difficult with this approach.
You will never reach perfection with this approach. But depending on the task it might just work "enough"...
I'm not great at regular expressions, but a simpler version, "brute force" actually, of above is
sentence = re.compile("([\'\"][A-Z]|([A-Z][a-z]*\. )|[A-Z])(([a-z]*\.[a-z]*\.)|([A-Za-z0-9]*\.[A-Za-z0-9])|([A-Z][a-z]*\. [A-Za-z]*)|[^\.?]|[A-Za-z])*[\.?]")
which means
start acceptable units are '[A-Z] or "[A-Z]
please note, most regular expressions are greedy so the order is very important when we do |(or). That's, why I have written i.e. regular expression first, then is come forms like Inc.
Try this:
(?<!\b(?:[A-Z][a-z]|\d|[i.e]))\.(?!\b(?:com|\d+)\b)
I wrote this taking into consideration smci's comments above. It is a middle-of-the-road approach that doesn't require external libraries and doesn't use regex. It allows you to provide a list of abbreviations and accounts for sentences ended by terminators in wrappers, such as a period and quote: [.", ?', .)].
abbreviations = {'dr.': 'doctor', 'mr.': 'mister', 'bro.': 'brother', 'bro': 'brother', 'mrs.': 'mistress', 'ms.': 'miss', 'jr.': 'junior', 'sr.': 'senior', 'i.e.': 'for example', 'e.g.': 'for example', 'vs.': 'versus'}
terminators = ['.', '!', '?']
wrappers = ['"', "'", ')', ']', '}']
def find_sentences(paragraph):
end = True
sentences = []
while end > -1:
end = find_sentence_end(paragraph)
if end > -1:
sentences.append(paragraph[end:].strip())
paragraph = paragraph[:end]
sentences.append(paragraph)
sentences.reverse()
return sentences
def find_sentence_end(paragraph):
[possible_endings, contraction_locations] = [[], []]
contractions = abbreviations.keys()
sentence_terminators = terminators + [terminator + wrapper for wrapper in wrappers for terminator in terminators]
for sentence_terminator in sentence_terminators:
t_indices = list(find_all(paragraph, sentence_terminator))
possible_endings.extend(([] if not len(t_indices) else [[i, len(sentence_terminator)] for i in t_indices]))
for contraction in contractions:
c_indices = list(find_all(paragraph, contraction))
contraction_locations.extend(([] if not len(c_indices) else [i + len(contraction) for i in c_indices]))
possible_endings = [pe for pe in possible_endings if pe[0] + pe[1] not in contraction_locations]
if len(paragraph) in [pe[0] + pe[1] for pe in possible_endings]:
max_end_start = max([pe[0] for pe in possible_endings])
possible_endings = [pe for pe in possible_endings if pe[0] != max_end_start]
possible_endings = [pe[0] + pe[1] for pe in possible_endings if sum(pe) > len(paragraph) or (sum(pe) < len(paragraph) and paragraph[sum(pe)] == ' ')]
end = (-1 if not len(possible_endings) else max(possible_endings))
return end
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1:
return
yield start
start += len(sub)
I used Karl's find_all function from this entry: Find all occurrences of a substring in Python
My example is based on the example of Ali, adapted to Brazilian Portuguese. Thanks Ali.
ABREVIACOES = ['sra?s?', 'exm[ao]s?', 'ns?', 'nos?', 'doc', 'ac', 'publ', 'ex', 'lv', 'vlr?', 'vls?',
'exmo(a)', 'ilmo(a)', 'av', 'of', 'min', 'livr?', 'co?ls?', 'univ', 'resp', 'cli', 'lb',
'dra?s?', '[a-z]+r\(as?\)', 'ed', 'pa?g', 'cod', 'prof', 'op', 'plan', 'edf?', 'func', 'ch',
'arts?', 'artigs?', 'artg', 'pars?', 'rel', 'tel', 'res', '[a-z]', 'vls?', 'gab', 'bel',
'ilm[oa]', 'parc', 'proc', 'adv', 'vols?', 'cels?', 'pp', 'ex[ao]', 'eg', 'pl', 'ref',
'[0-9]+', 'reg', 'f[ilí]s?', 'inc', 'par', 'alin', 'fts', 'publ?', 'ex', 'v. em', 'v.rev']
ABREVIACOES_RGX = re.compile(r'(?:{})\.\s*$'.format('|\s'.join(ABREVIACOES)), re.IGNORECASE)
def sentencas(texto, min_len=5):
# baseado em https://stackoverflow.com/questions/25735644/python-regex-for-splitting-text-into-sentences-sentence-tokenizing
texto = re.sub(r'\s\s+', ' ', texto)
EndPunctuation = re.compile(r'([\.\?\!]\s+)')
# print(NonEndings)
parts = EndPunctuation.split(texto)
sentencas = []
sentence = []
for part in parts:
txt_sent = ''.join(sentence)
q_len = len(txt_sent)
if len(part) and len(sentence) and q_len >= min_len and \
EndPunctuation.match(sentence[-1]) and \
not ABREVIACOES_RGX.search(txt_sent):
sentencas.append(txt_sent)
sentence = []
if len(part):
sentence.append(part)
if sentence:
sentencas.append(''.join(sentence))
return sentencas
Full code in: https://github.com/luizanisio/comparador_elastic
If you want to break up sentences at 3 periods (not sure if this is what you want) you can use this regular expresion:
import re
text = """\
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.
"""
sentences = re.split(r'\.{3}', text)
for stuff in sentences:
print(stuff)
Say I have a list of movie names with misspellings and small variations like this -
"Pirates of the Caribbean: The Curse of the Black Pearl"
"Pirates of the carribean"
"Pirates of the Caribbean: Dead Man's Chest"
"Pirates of the Caribbean trilogy"
"Pirates of the Caribbean"
"Pirates Of The Carribean"
How do I group or find such sets of words, preferably using python and/or redis?
Have a look at "fuzzy matching". Some great tools in the thread below that calculates similarities between strings.
I'm especially fond of the difflib module
>>> get_close_matches('appel', ['ape', 'apple', 'peach', 'puppy'])
['apple', 'ape']
>>> import keyword
>>> get_close_matches('wheel', keyword.kwlist)
['while']
>>> get_close_matches('apple', keyword.kwlist)
[]
>>> get_close_matches('accept', keyword.kwlist)
['except']
https://stackoverflow.com/questions/682367/good-python-modules-for-fuzzy-string-comparison
You might notice that similar strings have large common substring, for example:
"Bla bla bLa" and "Bla bla bRa" => common substring is "Bla bla ba" (notice the third word)
To find common substring you may use dynamic programming algorithm. One of algorithms variations is Levenshtein distance (distance between most similar strings is very small, and between more different strings distance is bigger) - http://en.wikipedia.org/wiki/Levenshtein_distance.
Also for quick performance you may try to adapt Soundex algorithm - http://en.wikipedia.org/wiki/Soundex.
So after calculating distance between all your strings, you have to clusterize them. The most simple way is k-means (but it needs you to define number of clusters). If you actually don't know number of clusters, you have to use hierarchical clustering. Note that number of clusters in your situation is number of different movies titles + 1(for totally bad spelled strings).
I believe there is in fact two distinct problems.
The first is spell correction. You can have one in Python here
http://norvig.com/spell-correct.html
The second is more functional. Here is what I'd do after the spell correction. I would make a relation function.
related( sentence1, sentence2 ) if and only if sentence1 and sentence2 have rare common words. By rare, I mean words different than (The, what, is, etc...). You can take a look at the TF/IDF system to determine if two document are related using their words. Just googling a bit I found this:
https://code.google.com/p/tfidf/
To add another tip to Fredrik's answer, you could also get inspired from search engines like code, such as this one :
def dosearch(terms, searchtype, case, adddir, files = []):
found = []
if files != None:
titlesrch = re.compile('>title<.*>/title<')
for file in files:
title = ""
if not (file.lower().endswith("html") or file.lower().endswith("htm")):
continue
filecontents = open(BASE_DIR + adddir + file, 'r').read()
titletmp = titlesrch.search(filecontents)
if titletmp != None:
title = filecontents.strip()[titletmp.start() + 7:titletmp.end() - 8]
filecontents = remove_tags(filecontents)
filecontents = filecontents.lstrip()
filecontents = filecontents.rstrip()
if dofind(filecontents, case, searchtype, terms) > 0:
found.append(title)
found.append(file)
return found
Source and more information: http://www.zackgrossbart.com/hackito/search-engine-python/
Regards,
Max
One approach would be to pre-process all the strings before you compare them: convert all to lowercase, standardize whitespace (eg, replace any whitespace with single spaces). If punctuation is not important to your end goal, you can remove all punctuation characters as well.
Levenshtein distance is commonly-used to determine similarity of a string, this should help you group strings which differ by small spelling errors.
in Python, I have created a text generator that acts on certain parameters but my code is -at most of the time- slow and performs below my expectations. I expect one sentence per every 3-4 minutes but it fails to comply if the database it works on is large -I use the project Gutenberg's 18-book corpus and I will create my custom corpus and add further books so performance is vital.- The algorithm and the implementation is below:
ALGORITHM
1- Enter the trigger sentence -only once, at the beginning of the program-
2- Get the longest word in the trigger sentence
3- Find all the sentences of the corpus that contain the word at step2
4- Randomly select one of those sentences
5- Get the sentence (named sentA to resolve the ambiguity in description) that follows the sentence picked at step4 -so long as sentA is longer than 40 characters-
6- Go to step 2, now the trigger sentence is the sentA of step5
IMPLEMENTATION
from nltk.corpus import gutenberg
from random import choice
triggerSentence = raw_input("Please enter the trigger sentence:")#get input sentence from user
previousLongestWord = ""
listOfSents = gutenberg.sents()
listOfWords = gutenberg.words()
corpusSentences = [] #all sentences in the related corpus
sentenceAppender = ""
longestWord = ""
#this function is not mine, code courtesy of Dave Kirby, found on the internet about sorting list without duplication speed tricks
def arraySorter(seq):
seen = set()
return [x for x in seq if x not in seen and not seen.add(x)]
def findLongestWord(longestWord):
if(listOfWords.count(longestWord) == 1 or longestWord.upper() == previousLongestWord.upper()):
longestWord = sortedSetOfValidWords[-2]
if(listOfWords.count(longestWord) == 1):
longestWord = sortedSetOfValidWords[-3]
doappend = corpusSentences.append
def appending():
for mysentence in listOfSents: #sentences are organized into array so they can actually be read word by word.
sentenceAppender = " ".join(mysentence)
doappend(sentenceAppender)
appending()
sentencesContainingLongestWord = []
def getSentence(longestWord, sentencesContainingLongestWord):
for sentence in corpusSentences:
if sentence.count(longestWord):#if the sentence contains the longest target string, push it into the sentencesContainingLongestWord list
sentencesContainingLongestWord.append(sentence)
def lengthCheck(sentenceIndex, triggerSentence, sentencesContainingLongestWord):
while(len(corpusSentences[sentenceIndex + 1]) < 40):#in case the next sentence is shorter than 40 characters, pick another trigger sentence
sentencesContainingLongestWord.remove(triggerSentence)
triggerSentence = choice(sentencesContainingLongestWord)
sentenceIndex = corpusSentences.index(triggerSentence)
while len(triggerSentence) > 0: #run the loop as long as you get a trigger sentence
sentencesContainingLongestWord = []#all the sentences that include the longest word are to be inserted into this set
setOfValidWords = [] #set for words in a sentence that exists in a corpus
split_str = triggerSentence.split()#split the sentence into words
setOfValidWords = [word for word in split_str if listOfWords.count(word)]
sortedSetOfValidWords = arraySorter(sorted(setOfValidWords, key = len))
longestWord = sortedSetOfValidWords[-1]
findLongestWord(longestWord)
previousLongestWord = longestWord
getSentence(longestWord, sentencesContainingLongestWord)
triggerSentence = choice(sentencesContainingLongestWord)
sentenceIndex = corpusSentences.index(triggerSentence)
lengthCheck(sentenceIndex, triggerSentence, sentencesContainingLongestWord)
triggerSentence = corpusSentences[sentenceIndex + 1]#get the sentence that is next to the previous trigger sentence
print triggerSentence
print "\n"
corpusSentences.remove(triggerSentence)#in order to view the sentence index numbers, you can remove this one so index numbers are concurrent with actual gutenberg numbers
print "End of session, please rerun the program"
#initiated once the while loop exits, so that the program ends without errors
The computer I run the code on is a bit old, dual-core CPU was bought in Feb. 2006 and 2x512 RAM was bought in Sept. 2004 so I'm not sure if my implementation is bad or the hardware is a reason for the slow runtime. Any ideas on how I can rescue this from its hazardous form ? Thanks in advance.
I think my first advice must be: Think carefully about what your routines do, and make sure the name describes that. Currently you have things like:
arraySorter which neither deals with arrays nor sorts (it's an implementation of nub)
findLongestWord which counts things or selects words by criteria not present in the algorithm description, yet ends up doing nothing at all because longestWord is a local variable (argument, as it were)
getSentence which appends an arbitrary number of sentences onto a list
appending which sounds like it might be a state checker, but operates only through side effects
considerable confusion between local and global variables, for instance the global variable sentenceAppender is never used, nor is it an actor (for instance, a function) like the name suggests
For the task itself, what you really need are indices. It might be overkill to index every word - technically you should only need index entries for words that occur as the longest word of a sentence. Dictionaries are your primary tool here, and the second tool is lists. Once you have those indices, looking up a random sentence containing any given word takes only a dictionary lookup, a random.choice, and a list lookup. Perhaps a few list lookups, given the sentence length restriction.
This example should prove a good object lesson that modern hardware or optimizers like Psyco do not solve algorithmic problems.
Maybe Psyco speeds up the execution?