Here is an example string:
text = "hello, i like to eat beef 'sandwiches' and beef 'jerky' and chicken 'patties' and chicken 'burgers' and also chicken 'fingers' and other chicken 'meat' too."
I am trying to separate the words "patties", "burgers",
fingers", and "meat" from this text. I want to separate the words after chicken but before the closing quotation.
I have gotten stumped on how to even separate a single one. I can split after "chicken ' but then how can i select the text up until the next ' ?
I would like to iterate through a list to save the variables to an array. Thanks for any help you can provide.
You can use regular expressions:
import re
text = "hello, i like to eat beef 'sandwiches' and beef 'jerky' and chicken 'patties' and chicken 'burgers' and also chicken 'fingers' and other chicken 'meat' too."
match = re.findall(r'chicken \'(\S+)\'', text)
print (match)
Outputs:
['patties', 'burgers', 'fingers', 'meat']
This is a good use-case for regex.
import re
print(re.findall(r"chicken '(.*?)'", text))
Here's an explanation of the regex: https://regex101.com/r/8IdseD/1
Here's the python code running: https://repl.it/repls/SquareQuerulousModes
The regex, part by part:
chicken ' - matches that literal text
( - starts a capture group - the part that re.findall will spit out.
. - matches any character...
*? - ...any number of times, but as few possible (this is to ensure we don't capture the final ')
) - end the capture group
' - match a literal '.
So re.findall will give you a list of all the substrings that are captured in the group.
You can use zero-width lookarounds to match the surroundings:
(?<=chicken\s')[^']+(?=')
(?<=chicken\s') is zero-width positive lookbehind that matches chicken '
[^']+ matches the portion upto next single quote i.e. the desired substring
(?=') is zero-width positive lookahead that matches ' after the desired substring
Example:
In [713]: text = "hello, i like to eat beef 'sandwiches' and beef 'jerky' and chicken 'patties' and chicken 'burgers' and also chicken 'fingers' and other chicken 'meat' too."
In [714]: re.findall(r"(?<=chicken\s')[^']+(?=')", text)
Out[714]: ['patties', 'burgers', 'fingers', 'meat']
Select just the portion of the sentence from the first occurrence of "chicken":
chicken_text = text[text.find("chicken"):]
Split that text on spaces:
chicken_words = chicken_text.split(" ")
Scan the list for words that begin and end with a single quote:
for word in chicken_words:
if word[0] == "'" and word[-1] == "'":
print word[1:-1]
This won't work if the single-quoted words themselves contain spaces, but that isn't the case in the sample text you gave.
Related
I am trying to extract the comma delimited numbers inside () brackets from a string. I can get the numbers if that are alone in a line. But i cant seem to find a solution to get the numbers when other surrounding text is involved. Any help will be appreciated. Below is the code that I current use in python.
line = """
Abuta has a history of use in the preparation of curares, an arrow poison to cause asphyxiation in hunting
It has also been used in traditional South American and Indian Ayurvedic medicines (101065,101066,101067)
The genus name Cissampelos is derived from the Greek words for ivy and vine (101065)
"""
line = each.strip()
regex_criteria = r'"^([1-9][0-9]*|\([1-9][0-9]*\}|\(([1-9][0-9]*,?)+[1-9][0-9]*\))$"gm'
if (line.__contains__('(') and line.__contains__(')') and not re.search('[a-zA-Z]', refline)):
refline = line[line.find('(')+1:line.find(')')]
if not re.search('[a-zA-Z]', refline):
Remove the ^, $ is whats preventing you from getting all the numbers. And gm flags wont work in python re.
You can change your regex to :([1-9][0-9]*|\([1-9][0-9]*\}|\(?:([1-9][0-9]*,?)+[1-9][0-9]*\)) if you want to get each number separately.
Or you can simplify your pattern to (?<=[(,])[1-9][0-9]+(?=[,)])
Test regex here: https://regex101.com/r/RlGwve/1
Python code:
import re
line = """
Abuta has a history of use in the preparation of curares, an arrow poison to cause asphyxiation in hunting
It has also been used in traditional South American and Indian Ayurvedic medicines (101065,101066,101067)
The genus name Cissampelos is derived from the Greek words for ivy and vine (101065)
"""
print(re.findall(r'(?<=[(,])[1-9][0-9]+(?=[,)])', line))
# ['101065', '101066', '101067', '101065']
(?<=[(,])[1-9][0-9]+(?=[,)])
The above pattern tells to match numbers which begin with 1-9 followed by one or more digits, only if the numbers begin with or end with either comma or brackets.
Here's another option:
pattern = re.compile(r"(?<=\()[1-9]+\d*(?:,[1-9]\d*)*(?=\))")
results = [match[0].split(",") for match in pattern.finditer(line)]
(?<=\(): Lookbehind for (
[1-9]+\d*: At least one number (would \d+ work too?)
(?:,[1-9]\d*)*: Zero or multiple numbers after a ,
(?=\)): Lookahead for )
Result for your line:
[['101065', '101066', '101067'], ['101065']]
If you only want the comma separated numbers:
pattern = re.compile(r"(?<=\()[1-9]+\d*(?:,[1-9]\d*)+(?=\))")
results = [match[0].split(",") for match in pattern.finditer(line)]
(?:,[1-9]\d*)+: One or more numbers after a ,
Result:
[['101065', '101066', '101067']]
Now, if your line could also look like
line = """
Abuta has a history of use in the preparation of curares, an arrow poison to cause asphyxiation in hunting
It has also been used in traditional South American and Indian Ayurvedic medicines ( 101065,101066, 101067 )
The genus name Cissampelos is derived from the Greek words for ivy and vine (101065)
"""
then you have to sprinkle the pattern with \s* and remove the whitespace afterwards (here with str.translate and str.maketrans):
pattern = re.compile(r"(?<=\()\s*[1-9]+\d*(?:\s*,\s*[1-9]\d*\s*)*(?=\))")
table = str.maketrans("", "", " ")
results = [match[0].translate(table).split(",") for match in pattern.finditer(line)]
Result:
[['101065', '101066', '101067'], ['101065']]
Using the pypi regex module you could also use capture groups:
\((?P<num>\d+)(?:,(?P<num>\d+))*\)
The pattern matches:
\( Match (
(?P<num>\d+) Capture group, match 1+ digits
(?:,(?P<num>\d+))* Optionally repeat matching , and 1+ digits in a capture group
\) Match )
Regex demo | Python demo
Example code
import regex
pattern = r"\((?P<num>\d+)(?:,(?P<num>\d+))*\)"
line = """
Abuta has a history of use in the preparation of curares, an arrow poison to cause asphyxiation in hunting
It has also been used in traditional South American and Indian Ayurvedic medicines (101065,101066,101067)
The genus name Cissampelos is derived from the Greek words for ivy and vine (101065)
"""
matches = regex.finditer(pattern, line)
for _, m in enumerate(matches, start=1):
print(m.capturesdict())
Output
{'num': ['101065', '101066', '101067']}
{'num': ['101065']}
I have a text like this Cat In A Tea Cup by New Yorker cover artist Gurbuz Dogan Eksioglu,Handsome cello wrapped hard magnet, Ideal for home or office.
I removed punctuations from this text by the following code.
import string
string.punctuation
def remove_punctuation(text):
punctuationfree="".join([i for i in text if i not in string.punctuation])
return punctuationfree
#storing the puntuation free text
df_Train['BULLET_POINTS']= df_Train['BULLET_POINTS'].apply(lambda x:remove_punctuation(x))
df_Train.head()
here in the above code df_Train is a pandas dataframe in which "BULLET_POINTS" column contains the kind of text data mentioned above.
The result I got is Cat In A Tea Cup by New Yorker cover artist Gurbuz Dogan EksiogluHandsome cello wrapped hard magnet Ideal for home or office
Notice how two words Eksioglu and Handsome are combing due to no space after , . I need a way to overcome this issue.
In these case, it makes sense to replace all the special chars with a space, and then strip the result and shrink multiple spaces to a single space:
df['BULLET_POINTS'] = df['BULLET_POINTS'].str.replace(r'(?:[^\w\s]|_)+', ' ', regex=True).str.strip()
Or, if you have chunks of punctuation + whitespace to handle:
df['BULLET_POINTS'].str.replace(r'[\W_]+', ' ', regex=True).str.strip()
Output:
>>> df['BULLET_POINTS'].str.replace(r'(?:[^\w\s]|_)+', ' ', regex=True).str.strip()
0 Cat In A Tea Cup by New Yorker cover artist Gurbuz Dogan Eksioglu Handsome cello wrapped hard magnet Ideal for home or office
Name: BULLET_POINTS, dtype: object
The (?:[^\w\s]|_)+ regex matches one or more occurrences of any char other than word and whitespace chars or underscores (i.e. one or more non-alphanumeric chars), and replaces them with a space.
The [\W_]+ pattern is similar but includes whitespace.
The .str.strip() part is necessary as the replacement might result in leading/trailing spaces.
How to removing non English words from text in df.columns words contain letters and numbers
Ex
df['text']
'the interiors nrd studio | happy mothers day ”there is no influence so powerful as that of the mother.” —sara josepha hale... happy mother’s day mom & to all the mothers around the world! lots of light natasha
0wet3bxtfl'
'but still missing you every day happy mothers day francis mcclafferty (mccool) 9wlhju7cxf'
from the above 2 rows I need to remove the word '0wet3bxtfl' & '9wlhju7cxf'
The example includes to retain some strings that would not be found in a list of English words ("nrd", "mcclafferty", "mccool") while removing "0wet3bxtfl" and "9wlhju7cxf", so the expected result is probably best achieved by removing any non-whitespace sequences that contain either a letter followed by digit or a digit followed by letter (together with any spaces that follow), without regard to whether words are "English" or not.
The following would do this:
import re
...
filtered = re.sub('[^\s]*(\d[a-zA-Z]|[a-zA-Z]\d)[^\s]* *', '', df['text'])
I've searched and searched, but can't find an any relief for my regex woes.
I wrote the following dummy sentence:
Watch Joe Smith Jr. and Saul "Canelo" Alvarez fight Oscar de la Hoya and Genaddy Triple-G Golovkin for the WBO belt GGG. Canelo Alvarez and Floyd 'Money' Mayweather fight in Atlantic City, New Jersey. Conor MacGregor will be there along with Adonis Superman Stevenson and Mr. Sugar Ray Robinson. "Here Goes a String". 'Money Mayweather'. "this is not a-string", "this is not A string", "This IS a" "Three Word String".
I'm looking for a regular expression that will return the following when used in Python 3.6:
Canelo, Money, Money Mayweather, Three Word String
The regex that has gotten me the closest is:
(["'])[A-Z](\\?.)*?\1
I want it to only match strings of 3 capitalized words or less immediately surrounded by single or double quotes. Unfortunately, so far it seem to match any string in quotes, no matter what the length, no matter what the content, as long is it begins with a capital letter.
I've put a lot of time into trying to hack through it myself, but I've hit a wall. Can anyone with stronger regex kung-fu give me an idea of where I'm going wrong here?
Try to use this one: (["'])((?:[A-Z][a-z]+ ?){1,3})\1
(["']) - opening quote
([A-Z][a-z]+ ?){1,3} - Capitalized word repeating 1 to 3 times separated by space
[A-Z] - capital char (word begining char)
[a-z]+ - non-capital chars (end of word)
_? - space separator of capitalized words (_ is a space), ? for single word w/o ending space
{1,3} - 1 to 3 times
\1 - closing quote, same as opening
Group 2 is what you want.
Match 1
Full match 29-37 `"Canelo"`
Group 1. 29-30 `"`
Group 2. 30-36 `Canelo`
Match 2
Full match 146-153 `'Money'`
Group 1. 146-147 `'`
Group 2. 147-152 `Money`
Match 3
Full match 318-336 `'Money Mayweather'`
Group 1. 318-319 `'`
Group 2. 319-335 `Money Mayweather`
Match 4
Full match 398-417 `"Three Word String"`
Group 1. 398-399 `"`
Group 2. 399-416 `Three Word String`
RegEx101 Demo: https://regex101.com/r/VMuVae/4
Working with the text you've provided, I would try to use regular expression lookaround to get the words surrounded by quotes and then apply some conditions on those matches to determine which ones meet your criterion. The following is what I would do:
[p for p in re.findall('(?<=[\'"])[\w ]{2,}(?=[\'"])', txt) if all(x.istitle() for x in p.split(' ')) and len(p.split(' ')) <= 3]
txt is the text you've provided here. The output is the following:
# ['Canelo', 'Money', 'Money Mayweather', 'Three Word String']
Cleaner:
matches = []
for m in re.findall('(?<=[\'"])[\w ]{2,}(?=[\'"])', txt):
if all(x.istitle() for x in m.split(' ')) and len(m.split(' ')) <= 3:
matches.append(m)
print(matches)
# ['Canelo', 'Money', 'Money Mayweather', 'Three Word String']
Here's my go at it: ([\"'])(([A-Z][^ ]*? ?){1,3})\1
Using https://regex101.com/
MY current regex Expression: ^.*'(\d\s*.*)'*$
which doesnt seem to be working. What is the right combination formula that i should use?
I want to able to parse out 4 variable namely
items, quantity, cost and Total
MY CODE:
import re
str = "xxxxxxxxxxxxxxxxxx"
match = re.match(r"^.*'(\d\s*.*)'*$",str)
print match.group(1)
The following regex matches each ingredient string and stores wanted informations into groups: r'^(\d+)\s+([A-Za-z ]+)\s+(\d+(?:\.\d*))$'
It defines 3 groups each separated from other by spaces:
^ marks the string start
(\d+) is the first group and looks for at least one digit
\s+ is the first separation between groups and looks for at least one white character
([A-Za-z ]+) is the second group and looks for a least one alphabetical character or space
\s+ is the second separation beween groups and looks for at least one white character
(\d+(?:\.\d*) is the third group and looks for at least one digit with eventually a decimal point and some other digits
$ marks the string end
A regex to obtain the total does not need to be explained I think.
Here is a test code using your test data. Is should be a good starting point:
import re
TEST_DATA = ['Table: Waiter: kenny',
'======================================',
'1 SAUSAGE WRAPPED WITH B 10.00',
'1 ESCARGOT WITH GARLIC H 12.00',
'1 PAN SEARED FOIE GRAS 15.00',
'1 SAUTE FIELD MUSHROOM W 9.00',
'1 CRISPY CHICKEN WINGS 7.00',
'1 ONION RINGS 6.00',
'----------------------------------',
'TOTAL 59.00',
'CASH 59.00',
'CHANGE 0.00',
'Signature:__________________________',
'Thank you & see you again soon!']
INGREDIENT_RE = re.compile(r'^(\d+)\s+([A-Za-z ]+)\s+(\d+(?:\.\d*))$')
TOTAL_RE = re.compile(r'^TOTAL (.+)$')
ingredients = []
total = None
for string in TEST_DATA:
match = INGREDIENT_RE.match(string)
if match:
ingredients.append(match.groups())
continue
match = TOTAL_RE.match(string)
if match:
total = match.groups()[0]
break
print(ingredients)
print(total)
this prints:
[('1', 'SAUSAGE WRAPPED WITH B', '10.00'), ('1', 'ESCARGOT WITH GARLIC H', '12.00'), ('1', 'PAN SEARED FOIE GRAS', '15.00'), ('1', 'SAUTE FIELD MUSHROOM W', '9.00'), ('1', 'CRISPY CHICKEN WINGS', '7.00'), ('1', 'ONION RINGS', '6.00')]
59.00
Edit on Python raw strings:
The r character before a Python string indicates that it is a raw string, which means that spécial characters (like \t, \n, etc...) are not interpreted.
To be clear, and for example, in a standard string \t is one tabulation character. It a raw string it is two characters: \ and t.
r'\t' is equivalent to '\\t'.
more details in the doc