I have a list with address information
The placement of words in the list can be random.
address = [' South region', ' district KTS', ' 4', ' app. 106', ' ent. 1', ' st. 15']
I want to extract each item of a list in a new string.
r = re.compile(".region")
region = list(filter(r.match, address))
It works, but there are more than 1 pattern "region". For example, there can be "South reg." or "South r-n".
How can I combine a multiple patterns?
And digit 4 in list means building number. There can be onle didts, or smth like 4k1.
How can I extract building number?
Hopefully I understood the requirement correctly.
For extracting the region, I chose to get it by the first word, but if you can be sure of the regions which are accepted, it would be better to construct the regex based on the valid values, not first word.
Also, for the building extraction, I am not sure of which are the characters you want to keep, versus the ones which you may want to remove. In this case I chose to keep only alphanumeric, meaning that everything else would be stripped.
CODE
import re
list1 = [' South region', ' district KTS', ' -4k-1.', ' app. 106', ' ent. 1', ' st. 15']
def GetFirstWord(list2,column):
return re.search(r'\w+', list2[column].strip()).group()
def KeepAlpha(list2,column):
return re.sub(r'[^A-Za-z0-9 ]+', '', list2[column].strip())
print(GetFirstWord(list1,0))
print(KeepAlpha(list1,2))
OUTPUT
South
4k1
['[{"word":"meaning","phonetics":[{"text":"/ˈmiːnɪŋ/","audio":"https://lex-audio.useremarkable.com/mp3/meaning_gb_1.mp3"}],"meanings":[{"partOfSpeech":"noun","definitions":[{"definition":"What '
'is meant by a word, text, concept, or '
'action.","synonyms":["definition","sense","explanation","denotation","connotation","interpretation","elucidation","explication"],"example":"the '
'meaning of the Hindu word is ‘breakthrough, '
'release’"}]},{"partOfSpeech":"adjective","definitions":[{"definition":"Intended '
'to communicate something that is not directly '
'expressed.","synonyms":["meaningful","significant","pointed","eloquent","expressive","pregnant","speaking","telltale","revealing","suggestive"]}]}]}]']
This is the format.
I wanna extract:
"meanings":[{"partOfSpeech":"noun","definitions":[{"definition":"A single distinct meaningful element of speech or writing, used with others (or sometimes alone) to form a sentence and typically shown with a space on either side when written or printed.",
How may I do it, in Python.
I believe this is what you want
import json
data = ['[{"word":"meaning","phonetics":[{"text":"/ˈmiːnɪŋ/","audio":"https://lex-audio.useremarkable.com/mp3/meaning_gb_1.mp3"}],"meanings":[{"partOfSpeech":"noun","definitions":[{"definition":"What ' 'is meant by a word, text, concept, or ' 'action.","synonyms":["definition","sense","explanation","denotation","connotation","interpretation","elucidation","explication"],"example":"the ' 'meaning of the Hindu word is ‘breakthrough, ' 'release’"}]},{"partOfSpeech":"adjective","definitions":[{"definition":"Intended ' 'to communicate something that is not directly ' 'expressed.","synonyms":["meaningful","significant","pointed","eloquent","expressive","pregnant","speaking","telltale","revealing","suggestive"]}]}]}]']
json_data = json.loads(data[0])
meanings = json_data[0]['meanings']
print(meanings)
# [{'partOfSpeech': 'noun', 'definitions': [{'definition': 'What is meant by a word, text, concept, or action.', 'synonyms': ['definition', 'sense', 'explanation', 'denotation', 'connotation', 'interpretation', 'elucidation', 'explication'], 'example': 'the meaning of the Hindu word is ‘breakthrough, release’'}]}, {'partOfSpeech': 'adjective', 'definitions': [{'definition': 'Intended to communicate something that is not directly expressed.', 'synonyms': ['meaningful', 'significant', 'pointed', 'eloquent', 'expressive', 'pregnant', 'speaking', 'telltale', 'revealing', 'suggestive']}]}]
I have a dataframe of text where I want to replace the text of some substrings. For example:
"[' Foods are adequately protected from\\n contamination during handling and storage.', ' Food handler hygiene and hand washing is\\n properly followed.', ' Foods are cooked, cooled and stored at\\n proper temperatures.', ' Garbage and/or waste is properly stored\\n and removed.', ' Pest control practices are properly maintained.', ' Equipment and utensils are properly cleaned,\\n sanitized and maintained.', ' Food premise is properly maintained in a clean\\n and sanitary condition.']"
I want to replace '\n' with ''.
[sub.replace('\\n', '') for sub in abc_test]
where abc_test is just the first row of the dataframe content. When I apply this function the result turns out to be different than what I was hoping for.
['[',
"'",
' ',
'F',
'o',
'o',
'd',
's',
' ',
'a',
'r',
'e',
'
Any help would be appreciated.
The point here is that your strings contain combinations of a backslash and n char, not newline chars. Thus, neither "\n" (an LF, line feed, char) nor "\\n" (a \n regex escape that matches a newline, LF, char) work.
You can use
df['res'] = df['text'].str.replace(r"\\n", "")
Pandas test:
>>> import pandas as pd
>>> df = pd.DataFrame({'text': [' Foods are adequately protected from\\n contamination during handling and storage.', ' Food handler hygiene and hand washing is\\n properly followed.', ' Foods are cooked, cooled and stored at\\n proper temperatures.', ' Garbage and/or waste is properly stored\\n and removed.', ' Pest control practices are properly maintained.', ' Equipment and utensils are properly cleaned,\\n sanitized and maintained.', ' Food premise is properly maintained in a clean\\n and sanitary condition.']})
>>> df['res'] = df['text'].str.replace(r"\\n", "")
>>> df
text res
0 Foods are adequately protected from\n contami... Foods are adequately protected from contamina...
1 Food handler hygiene and hand washing is\n pr... Food handler hygiene and hand washing is prop...
2 Foods are cooked, cooled and stored at\n prop... Foods are cooked, cooled and stored at proper...
3 Garbage and/or waste is properly stored\n and... Garbage and/or waste is properly stored and r...
4 Pest control practices are properly maintained. Pest control practices are properly maintained.
5 Equipment and utensils are properly cleaned,\... Equipment and utensils are properly cleaned, ...
6 Food premise is properly maintained in a clea... Food premise is properly maintained in a clea...
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)