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Python: Finding out if certain words in a list are actual English words or close to English words

I am working on a problem where I get a lot of words with their frequency of occurrence listed. Here is a sample of what I get:
drqsQAzaQ:1
OnKxnXecCINJ:1
QoGzQpg:1
cordially:1
Sponsorship:1
zQnpzQou:1
Thriving:1
febrero:1
rzaye:1
VseKEX:1
contributed:1
SNfXQoWV:1
hRwzmPR:1
Happening:1
TzJYAMWAQUIJTkWYBX:1
DYeUIqf:1
formats:1
eiizh:1
wIThY:1
infonewsletter:8
BusinessManager:10
MailScanner:12
As you can see, words like 'cordially' are actual English words, while words like 'infonewsletter' are not actual English words by themselves, but we can see that they are actually in English and mean something. However, words like 'OnKxnXecCINJ' do not mean anything (actually they are words from another charset, but I am ignoring them in my exercise and sticking to English) - I can discard them as junk
What would be the best method in Python to detect and eliminate such junk words from a given dictionary such as the one above?
I tried examining each word using nltk.corpus.word.words(), but it is killing my performance as my data set is very huge. Moreover, I am not certain whether this will give me a True for words like 'infonewsletter'
Please help.
Thanks,
Mahesh.

If the words are from completely different script within Unicode like CJK characters or Greek, Cyrillic, Thai, you could use unicodedata.category to see if they're letters to begin with (category starts with L):
>>> import unicodedata
>>> unicodedata.category('a')
'Ll'
>>> unicodedata.category('E')
'Lu'
>>> unicodedata.category('中')
'Lo'
>>> [unicodedata.category(i).startswith('L') for i in 'aE中,']
[True, True, True, False]
Then you can use the unicodedata.name to see that they're Latin letters:
>>> 'LATIN' in unicodedata.name('a')
True
>>> 'LATIN' in unicodedata.false('中')
False
Presumably it is not an English-language word if it has non-Latin letters in it.
Otherwise, you could use a letter bigram/trigram classifier to find out if there is a high probability these are English words. For example OnKxnXecCINJ contains Kxn which is a trigram that neither very probably exist in any single English language word, nor any concatenation of 2 words.
You can build one yourself from the corpus by splitting words into character trigrams, or you can use any of the existing libraries like langdetect or langid or so.
Also, see that the corpus is a set for fast in operations; only after the algorithm tells that there is a high probability it is in English, and the word fails to be found in the set, consider that it is alike to infonewsletter - a concatenation of several words; split it recursively into smaller chunks and see that each part thereof is found in the corpus.

Thank you. I am trying out this approach. However, I have a question. I have a word 'vdgutumvjaxbpz'. I know this is junk. I wrote some code to get all grams of this word, 4-gram and higher. This was the result:
['vdgu', 'dgut', 'gutu', 'utum', 'tumv', 'umvj', 'mvja', 'vjax', 'jaxb', 'axbp', 'xbpz', 'vdgut', 'dgutu', 'gutum', 'utumv', 'tumvj', 'umvja', 'mvjax', 'vjaxb', 'jaxbp', 'axbpz', 'vdgutu', 'dgutum', 'gutumv', 'utumvj', 'tumvja', 'umvjax', 'mvjaxb', 'vjaxbp', 'jaxbpz', 'vdgutum', 'dgutumv', 'gutumvj', 'utumvja', 'tumvjax', 'umvjaxb', 'mvjaxbp', 'vjaxbpz', 'vdgutumv', 'dgutumvj', 'gutumvja', 'utumvjax', 'tumvjaxb', 'umvjaxbp', 'mvjaxbpz', 'vdgutumvj', 'dgutumvja', 'gutumvjax', 'utumvjaxb', 'tumvjaxbp', 'umvjaxbpz', 'vdgutumvja', 'dgutumvjax', 'gutumvjaxb', 'utumvjaxbp', 'tumvjaxbpz', 'vdgutumvjax', 'dgutumvjaxb', 'gutumvjaxbp', 'utumvjaxbpz', 'vdgutumvjaxb', 'dgutumvjaxbp', 'gutumvjaxbpz', 'vdgutumvjaxbp', 'dgutumvjaxbpz', 'vdgutumvjaxbpz']
Now, I compared each gram result with nltk.corpus.words.words() and found the intersection of the 2 sets.
vocab = nltk.corpus.words.words()
vocab = set(w.lower().strip() for w in vocab)
def GetGramsInVocab(listOfGrams, vocab):
text_vocab = set(w.lower() for w in listOfGrams if w.isalpha())
common = text_vocab & vocab
return list(common)
However, the intersection contains 'utum', whereas I expected it to be NULL.
Also,
print("utum" in vocab)
returned true.
This does not make sense to me. I peeked into the vocabulary and found 'utum' in a few words like autumnian and metascutum
However, 'utum' is not a word by itself and I expected nltk to return false. Is there a more accurate corpus I can check against that would do whole word comparisons?
Also, I did a simple set operations test:
set1 = {"cutums" "acutum"}
print("utum" in set1)
This returned False as expected.
I guess I am confused as to why the code says 'utum' is present in the nltk words corpus.
Thanks,
Mahesh.

Python string grouping?

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 .

Python - RegEx for splitting text into sentences (sentence-tokenizing) [duplicate]

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)

What is a good strategy to group similar words?

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.

Python String Cleanup + Manipulation (Accented Characters)

I have a database full of names like:
John Smith
Scott J. Holmes
Dr. Kaplan
Ray's Dog
Levi's
Adrian O'Brien
Perry Sean Smyre
Carie Burchfield-Thompson
Björn Árnason
There are a few foreign names with accents in them that need to be converted to strings with non-accented characters.
I'd like to convert the full names (after stripping characters like " ' " , "-") to user logins like:
john.smith
scott.j.holmes
dr.kaplan
rays.dog
levis
adrian.obrien
perry.sean.smyre
carie.burchfieldthompson
bjorn.arnason
So far I have:
Fullname.strip() # get rid of leading/trailing white space
Fullname.lower() # make everything lower case
... # after bad chars converted/removed
Fullname.replace(' ', '.') # replace spaces with periods

Take a look at this link [redacted]
Here is the code from the page
def latin1_to_ascii (unicrap):
"""This replaces UNICODE Latin-1 characters with
something equivalent in 7-bit ASCII. All characters in the standard
7-bit ASCII range are preserved. In the 8th bit range all the Latin-1
accented letters are stripped of their accents. Most symbol characters
are converted to something meaningful. Anything not converted is deleted.
"""
xlate = {
0xc0:'A', 0xc1:'A', 0xc2:'A', 0xc3:'A', 0xc4:'A', 0xc5:'A',
0xc6:'Ae', 0xc7:'C',
0xc8:'E', 0xc9:'E', 0xca:'E', 0xcb:'E',
0xcc:'I', 0xcd:'I', 0xce:'I', 0xcf:'I',
0xd0:'Th', 0xd1:'N',
0xd2:'O', 0xd3:'O', 0xd4:'O', 0xd5:'O', 0xd6:'O', 0xd8:'O',
0xd9:'U', 0xda:'U', 0xdb:'U', 0xdc:'U',
0xdd:'Y', 0xde:'th', 0xdf:'ss',
0xe0:'a', 0xe1:'a', 0xe2:'a', 0xe3:'a', 0xe4:'a', 0xe5:'a',
0xe6:'ae', 0xe7:'c',
0xe8:'e', 0xe9:'e', 0xea:'e', 0xeb:'e',
0xec:'i', 0xed:'i', 0xee:'i', 0xef:'i',
0xf0:'th', 0xf1:'n',
0xf2:'o', 0xf3:'o', 0xf4:'o', 0xf5:'o', 0xf6:'o', 0xf8:'o',
0xf9:'u', 0xfa:'u', 0xfb:'u', 0xfc:'u',
0xfd:'y', 0xfe:'th', 0xff:'y',
0xa1:'!', 0xa2:'{cent}', 0xa3:'{pound}', 0xa4:'{currency}',
0xa5:'{yen}', 0xa6:'|', 0xa7:'{section}', 0xa8:'{umlaut}',
0xa9:'{C}', 0xaa:'{^a}', 0xab:'<<', 0xac:'{not}',
0xad:'-', 0xae:'{R}', 0xaf:'_', 0xb0:'{degrees}',
0xb1:'{+/-}', 0xb2:'{^2}', 0xb3:'{^3}', 0xb4:"'",
0xb5:'{micro}', 0xb6:'{paragraph}', 0xb7:'*', 0xb8:'{cedilla}',
0xb9:'{^1}', 0xba:'{^o}', 0xbb:'>>',
0xbc:'{1/4}', 0xbd:'{1/2}', 0xbe:'{3/4}', 0xbf:'?',
0xd7:'*', 0xf7:'/'
}
r = ''
for i in unicrap:
if xlate.has_key(ord(i)):
r += xlate[ord(i)]
elif ord(i) >= 0x80:
pass
else:
r += i
return r
# This gives an example of how to use latin1_to_ascii().
# This creates a string will all the characters in the latin-1 character set
# then it converts the string to plain 7-bit ASCII.
if __name__ == '__main__':
s = unicode('','latin-1')
for c in range(32,256):
if c != 0x7f:
s = s + unicode(chr(c),'latin-1')
print 'INPUT:'
print s.encode('latin-1')
print
print 'OUTPUT:'
print latin1_to_ascii(s)

If you are not afraid to install third-party modules, then have a look at the python port of the Perl module Text::Unidecode (it's also on pypi).
The module does nothing more than use a lookup table to transliterate the characters. I glanced over the code and it looks very simple. So I suppose it's working on pretty much any OS and on any Python version (crossingfingers). It's also easy to bundle with your application.
With this module you don't have to create your lookup table manually ( = reduced risk it being incomplete).
The advantage of this module compared to the unicode normalization technique is this: Unicode normalization does not replace all characters. A good example is a character like "æ". Unicode normalisation will see it as "Letter, lowercase" (Ll). This means using the normalize method will give you neither a replacement character nor a useful hint. Unfortunately, that character is not representable in ASCII. So you'll get errors.
The mentioned module does a better job at this. This will actually replace the "æ" with "ae". Which is actually useful and makes sense.
The most impressive thing I've seen is that it goes much further. It even replaces Japanese Kana characters mostly properly. For example, it replaces "は" with "ha". Wich is perfectly fine. It's not fool-proof though as the current version replaces "ち" with "ti" instead of "chi". So you'll have to handle it with care for the more exotic characters.
Usage of the module is straightforward::
from unidecode import unidecode
var_utf8 = "æは".decode("utf8")
unidecode( var_utf8 ).encode("ascii")
>>> "aeha"
Note that I have nothing to do with this module directly. It just happens that I find it very useful.
Edit: The patch I submitted fixed the bug concerning the Japanese kana. I've only fixed the one's I could spot right away. I may have missed some.

The following function is generic:
import unicodedata
def not_combining(char):
return unicodedata.category(char) != 'Mn'
def strip_accents(text, encoding):
unicode_text= unicodedata.normalize('NFD', text.decode(encoding))
return filter(not_combining, unicode_text).encode(encoding)
# in a cp1252 environment
>>> print strip_accents("déjà", "cp1252")
deja
# in a cp1253 environment
>>> print strip_accents("καλημέρα", "cp1253")
καλημερα
Obviously, you should know the encoding of your strings.

I would do something like this
# coding=utf-8
def alnum_dot(name, replace={}):
import re
for k, v in replace.items():
name = name.replace(k, v)
return re.sub("[^a-z.]", "", name.strip().lower())
print alnum_dot(u"Frédrik Holmström", {
u"ö":"o",
" ":"."
})
Second argument is a dict of the characters you want replaced, all non a-z and . chars that are not replaced will be stripped

The translate method allows you to delete characters. You can use that to delete arbitrary characters.
Fullname.translate(None,"'-\"")
If you want to delete whole classes of characters, you might want to use the re module.
re.sub('[^a-z0-9 ]', '', Fullname.strip().lower(),)