So I wrote a program for my Kindle e-reader that searches my highlights and deletes repetitive text (it's usually information about the book title, author, page number, etc.). I thought it was functional but sometimes there would random be periods (.) on certain lines of the output. At first I thought the program was just buggy but then I realized that the regex I'm using to match the books title and author was also matching any sentence that ended in brackets.
This is the code for the regex that I'm using to detect the books title and author
titleRegex = re.compile('(.+)\((.+)\)')
Example
Desired book title and author match: Book title (Author name)
What would also get matched: *I like apples because they are green (they are sometimes red as well). *
In this case it would delete everything and leave just the period at the end of the sentence. This is obviously not ideal because it deletes the text I highlighted
Here is the unformatted text file that goes into my program
The program works by finding all of the matches for the regexes I wrote, looping through those matches and one by one replacing them with empty strings.
Would there be any ways to make my title regex more specific so that it only picks up author titles and not full sentences that end in brackets? If not, what steps would I have to take to restructure this program?
I've attached my code to the bottom of this post. I would greatly appreciate any help as I'm a total coding newbie. Thanks :)
import re
titleRegex = re.compile('(.+)\((.+)\)')
titleRegex2 = re.compile(r'\ufeff (.+)\((.+)\)')
infoRegex = re.compile(r'(.) ([a-zA-Z]+) (Highlight|Bookmark|Note) ([a-zA-Z]+) ([a-zA-Z]+) ([0-9]+) (\|)')
locationRegex = re.compile(r' Location (\d+)(-\d+)? (\|)')
dateRegex = re.compile(r'([a-zA-Z]+) ([a-zA-Z]+) ([a-zA-Z]+), ([a-zA-Z]+) ([0-9]+), ([0-9]+)')
timeRegex = re.compile(r'([0-9]+):([0-9]+):([0-9]+) (AM|PM)')
newlineRegex = re.compile(r'\n')
sepRegex = re.compile('==========')
regexList = [titleRegex, titleRegex2, infoRegex, locationRegex, dateRegex, timeRegex, sepRegex, newlineRegex]
string = open("/Users/devinnagami/myclippings.txt").read()
for x in range(len(regexList)):
newString = re.sub(regexList[x], ' ', string)
string = newString
finalText = newString.split(' ')
with open('booknotes.txt', 'w') as f:
for item in finalText:
f.write('%s\n' % item)
There isn't enough information to tell if "Book title (Book Author)" is different than something like "I like Books (Good Ones)" without context. Thankfully, the text you showed has plenty of context. Instead of creating several different regular expressions, you can combine them into one expression to encode that context.
For instance:
quoteInfoRegex = re.compile(
r"^=+\n(?P<title>.*?) \((?P<author>.*?)\)\n" +
r"- Your Highlight on page (?P<page>[\d]+) \| Location (?P<location>[\d-]+) \| Added on (?P<added>.*?)\n" +
r"\n" +
r"(?P<quote>.*?)\n", flags=re.MULTILINE)
for m in quoteInfoRegex.finditer(data):
print(m.groupdict())
This will pull out each line of the text, and parse it, knowing that the book title is the first line after the equals, and the quote itself is below that.
I have a text and I have got a task in python with reading module:
Find the names of people who are referred to as Mr. XXX. Save the result in a dictionary with the name as key and number of times it is used as value. For example:
If Mr. Churchill is in the novel, then include {'Churchill' : 2}
If Mr. Frank Churchill is in the novel, then include {'Frank Churchill' : 4}
The file is .txt and it contains around 10-15 paragraphs.
Do you have ideas about how can it be improved? (It gives me error after some words, I guess error happens due to the reason that one of the Mr. is at the end of the line.)
orig_text= open('emma.txt', encoding = 'UTF-8')
lines= orig_text.readlines()[32:16267]
counts = dict()
for line in lines:
wordsdirty = line.split()
try:
print (wordsdirty[wordsdirty.index('Mr.') + 1])
except ValueError:
continue
Try this:
text = "When did Mr. Churchill told Mr. James Brown about the fish"
m = [x[0] for x in re.findall('(Mr\.( [A-Z][a-z]*)+)', text)]
You get:
['Mr. Churchill', 'Mr. James Brown']
To solve the line issue simply read the entire file:
text = file.read()
Then, to count the occurrences, simply run:
Counter(m)
Finally, if you'd like to drop 'Mr. ' from all your dictionary entries, use x[0][4:] instead of x[0].
This can be easily done using regex and capturing group.
Take a look here for reference, in this scenario you might want to do something like
# retrieve a list of strings that match your regex
matches = re.findall("Mr\. ([a-zA-Z]+)", your_entire_file) # not sure about the regex
# then create a dictionary and count the occurrences of each match
# if you are allowed to use modules, this can be done using Counter
Counter(matches)
To access the entire file like that, you might want to map it to memory, take a look at this question
This question already has answers here:
How to replace multiple substrings of a string?
(28 answers)
Closed 6 years ago.
I'm writing a fairly simple python program to find and download videos from a particular site. I would like to have my script name the file by using the page title except the page title contains various strings i would like remove for e.g.,
The title is:
The Big Bang Theory S09E15 720p HDTV X264-DIMENSION
but the titles are not always consistent for e.g.,
The title is:
Triple 9 2016 READNFO HDRip AC3-EVO
How can I replace strings if they are present?
Maybe create a list or dictionary of possible strings and if they are present then remove them (or replace with empty string)? I have tried and tried to find an answer but cannot find anything that helps my situation.
Basically if "HDTV", "HDRip", "720p", "X264", etc are present then replace them otherwise carry on?
Simple example:
string = 'The Big Bang Theory S09E15 720p HDTV X264-DIMENSION'
dict = {'720p':'1080p'} # format 'substring':'replacement'
for key, value in dict.iteritems():
if key in string:
string.replace(key,value)
The only problem with this is that if you want to replace a word that could be part of another word. For example if you want to replace 'an' with a, then the string in this example would become 'The Big Bag Theory ... '. To fix this I would try breaking up the string into a set of words and compare the words to dictionary entries.
for undesired_word in ("HDTV", "HDRip", "720p", "X264"):
title = title.replace(undesired_word, "")
title = 'The Big Bang Theory S09E15 720p HDTV X264-DIMENSION'
if 'HDTV' in title:
title = title.replace('HDTV', '')
not very pythonic but it will do what you want
Kevins answer will work for you, but just in case you find yourself wanting to use a regex:
import re
string_to_replace = ["HDTV", "HDRip", "720p", "X264"]
regex_string = r"|".join(string_to_replace)
S = "The Big Bang Theory S09E15 720p HDTV X264-DIMENSION"
new_string = re.sub(regex_string, "", S, flags=re.I)
print(new_string)
prints:
The Big Bang Theory S09E15 -DIMENSION
Also, as you will notice the spaces that went after the strings you were replacing are still there, if you do not want that, you can change string_to_replace to include the spaces, like so: ["HDTV ", "HDRip ", "720p ", "X264 "] and this would result in the output being:
The Big Bang Theory S09E15 X264-DIMENSION
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)