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how to add options to inflect number-to-word?

I have a program which calculates the number of minutes of a person's age. It works correctly. However, I want to ask if I can print the first letter capitalized.
from datetime import datetime, date
import sys
import inflect
inflector = inflect.engine()
def main():
# heute = date.today()
user = input('Date of birth: ')
min_preter(user)
def min_preter(data):
try:
data != datetime.strptime(data, '%Y-%m-%d')
# Get the y-m-d in current time
today = date.today()
# die y-m-d teilen
year, month , day = data.split('-')
# Convert to datetime
data = date(year=int(year), month=int(month), day=int(day))
# And valla
end = (today - data).total_seconds() / 60
# Convert to words
words = inflector.number_to_words(end).replace('point zero','minutes').upper()
return words
except:
sys.exit('Invalid date')
# convert from string format to datetime format
if __name__ == "__main__":
main()
Here is the output when I enter e.g 1999-01-01:
twelve million, four hundred and fifty-seven thousand, four hundred and forty point zero
where I expected
Twelve million, four hundred and fifty-seven thousand, four hundred and forty minutes
first word 'Twelve'(first letter capitalize)
I don't know what this point zero is. I just want the minutes at the end.
Thank you

You can use string.capitalize(). So you can do that:
return words.capitalize()
... and as for the "point zero", try converting the result to int before running your function, like
end = int((today - data).total_seconds() / 60)

Just replace .upper() by capitalize() in your code
An alternative to your replace would be to obtain the total number of minutes as an integer (point zero is because end is a float number) :
end = int((today - data).total_seconds() / 60)
In that case, your words variable would be :
words = inflector.number_to_words(end).capitalize() + " minutes"

You can use .capitalize() to capitalize the first word of the string.
EXAMPLE: words.capitalize()
"twelve million, four hundred and fifty-seven thousand, four hundred and forty point zero".capitalize()
OUTPUT
'Twelve million, four hundred and fifty-seven thousand, four hundred and forty point zero'
Regarding point zero
This particular code end = (today - data).total_seconds() / 60 is giving output as float which is leading to point zero so instead of division use floor division i.e. // instead of / which will return integer and hence point zero will be gone or else convert end to int.
Lastly add minutes string i.e. end + ' minutes' to final result.

Using regex to split a column

The regex I am using is \d+-\d+, but I'm not quite sure about how to separate the Roman numbers and how to create a new column with them.
I have this dataset:
Date_Title Date Copies
05-21 I. Don Quixote 1605 252
21-20 IV. Macbeth 1629 987
10-12 ML. To Kill a Mockingbird 1960 478
12 V. Invisible Man 1897 136
Basically, I would like to split the "Date Title", so, when I print a row, I would get this:
('05-21 I', 'I', 'Don Quixote', 1605, 252)
Or
('10-12 ML', 'ML', 'To Kill a Mockingbird',1960, 478)
In the first place, the numbers and the roman numeral, in the second; only the Roman numeral, in the third the name, and the fourth and fifth would be the same as the dataset.

You can use
df = pd.DataFrame({'Date_Title':['05-21 I. Don Quixote','21-20 IV. Macbeth','10-12 ML. To Kill a Mockingbird','12 V. Invisible Man'], 'Date':[1605,1629,1960,1897], 'Copies':[252,987,478,136]})
rx = r'^(\d+(?:-\d+)?\s*(M{0,4}(?:C[MD]|D?C{0,3})(?:X[CL]|L?X{0,3})(?:I[XV]|V?I{0,3})))\.\s*(.*)'
df[['NumRoman','Roman','Name']] = df.pop('Date_Title').str.extract(rx)
df = df[['NumRoman','Roman','Name', 'Date', 'Copies']]
>>> df
NumRoman Roman Name Date Copies
0 05-21 I I Don Quixote 1605 252
1 21-20 IV IV Macbeth 1629 987
2 10-12 ML ML To Kill a Mockingbird 1960 478
3 12 V V Invisible Man 1897 136
See the regex demo. Details:
^ - start of string
(\d+(?:-\d+)?\s*(M{0,4}(?:C[MD]|D?C{0,3})(?:X[CL]|L?X{0,3})(?:I[XV]|V?I{0,3}))) - Group 1 ("NumRoman"):
\d+(?:-\d+)? - one or more digits followed with an optional sequence of a - and one or more digits
\s* - zero or more whitespaces
(M{0,4}(?:C[MD]|D?C{0,3})(?:X[CL]|L?X{0,3})(?:I[XV]|V?I{0,3})) - Group 2 ("Roman"): see How do you match only valid roman numerals with a regular expression? for explanation
\. - a dot
\s* - zero or more whitespaces
(.*) - Group 3 ("Name"): any zero or more chars other than line break chars, as many as possible
Note df.pop('Date_Title') removes the Date_Title column and yields it as input for the extract method. df = df[['NumRoman','Roman','Name', 'Date', 'Copies']] is necessary if you need to keep the original column order.

I am pretty sure there might be a more optimal solution, but this is would be a fast way of solving it:
df['Date_Title'] = df['Date_Title'].apply(lambda x: (x.split()[0],x.split()[1],' '.join(x.split()[2:])
Or:
df['Date_Title'] = (df['Date_Title'].str.split().str[0],
df['Date_Title'].str.split().str[1],
' '.join(df['Date_Title'].str.split().str[2:])

Focusing on the string split:
string = "21-20 IV. Macbeth"
i = string.index(".") # Finds the first point
date, roman = string[:i].split() # 21-20, IV
title = string[i+2:] # Macbeth

df=df.assign(x=df['Date_Title'].str.split('\.').str[0],y=df['Date_Title'].str.extract('(\w+(?=\.))'),z=df['Date_Title'].str.split('\.').str[1:].str.join(','))

Python Regular expression of group to match text before amount

I am trying to write a python regular expression which captures multiple values from a few columns in dataframe. Below regular expression attempts to do the same. There are 4 parts of the string.
group 1: Date - month and day
group 2: Date - month and day
group 3: description text before amount i.e. group 4
group 4: amount - this group is optional
Some peculiar conditions for group 3 - text that
(1)the text itself might contain characters like "-" , "$". So we cannot use - & $ as the boundary of text.
(2) The text (group 3) sometimes may not be followed by amount.
(3) Empty space between group 3 and 4 is optional
Below is python function code which takes in a dataframe having 4 columns c1,c2,c3,c4 adds the columns dt, txt and amt after processing to dataframe.
def parse_values(args):
re_1='(([JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC]{3}\s{0,}[\d]{1,2})\s{0,}){2}(.*[\s]|.*[^\$]|.*[^-]){1}([-+]?\$[\d|,]+(?:\.\d+)?)?'
srch=re.search(re_1, args[0])
if srch is None:
return args
m = re.match(re_1, args[0])
args['dt']=m.group(1)
args['txt']=m.group(3)
args['amt']=m.group(4)
if m.group(4) is None:
if pd.isnull(args['c3']):
args['amt']=args.c2
else:
args['amt']=args.c3
return args
And in order to test the results I have below 6 rows which needs to return a properly formatted amt column in return.
tt=[{'c1':'OCT 7 OCT 8 HURRY CURRY THORNHILL ','c2':'$16.84'},
{'c1':'OCT 7 OCT 8 HURRY CURRY THORNHILL','c2':'$16.84'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK -$80,00,7770.70'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK-$2070.70'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK$2070.70'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK $80,00,7770.70'}
]
t=pd.DataFrame(tt,columns=['c1','c2','c3','c4'])
t=t.apply(parse_values,1)
t
However due to the error in my regular expression in re_1 I am not getting the amt column and txt column parsed properly as they return NaN or miss some words (as dipicted in some rows of the output image below).

How about this:
(((?:JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)\s*[\d]{1,2})\s*){2}(.*?)\s*(?=[\-$])([-+]?\$[\d|,]+(?:\.\d+)?)
As seen at regex101.com
Explanation:
First off, I've shortened the regex by changing a few minor details like using \s* instead of \s{0,}, which mean the exact same thing.
The whole [Jan|...|DEC] code was using a character class i.e. [], whcih only takes a single character from the entire set. Using non capturing groups is the correct way of selecting from different groups of multiple letters, which in your case are 'months'.
The meat of the regex: LOOKAHEADS
(?=[\-$]) tells the regex that the text before it in (.*) should match as much as it can until it finds a position followed by a dash or a dollar sign. Lookaheads don't actually match whatever they're looking for, they just tell the regex that the lookahead's arguments should be following that position.

Python RegEx for Australian Phone Numbers - False negative - 2 matches in the same substring

I am trying to extract phone numbers from a web page using Python & RegEx
Australian number format
+61 (international code - shown below as 'i')
02, 03, 07 or 08 (state codes - shown below as 's')
1234-5678 (8 digit local number - shown below as 'x')
Common variations of format (in order of commonality):
Format 1: ss xxxx xxxx (e.g. 02 1234 5678)
Format 2: +ii s xxxx xxxx (e.g. +61 2 1234 5678) (note the first 's' digit is removed here)
Format 3: (seen rarely) +ii (s)s xxxx-xxxx (e.g. +61 (0)2 1234 5678
My RegEx:
re.findall(r'[0][2]\d{8}|[0][3]\d{8}|[0][7]\d{8}|[0][8]\d{8}|[6][1][2]\d{8}|[6][1][3]\d{8}|[6][1][7]\d{8}|[6][1][8]\d{8}|[0][4]\d{8}|[6][1][4]\d{8}|[1][3][0][0]\d{6}|[1][8][0][0]\d{6}', re.sub(r'\W+', '', sample_text))
works well on a simple sample_text:
sample_text =
"610212345678ABC##610312345678ABC##610712345678ABC##610812345678ABC##0212345678ABC##0312345678ABC##0712345678ABC##0812345678ABC##61212345678ABC##61312345678ABC##61712345678ABC##61812345678ABC##0412345678ABC##61412345678ABC##130012345678ABC##180012345678ABC##"
Result:
['0212345678', '0312345678', '0712345678', '0812345678',
'0212345678', '0312345678', '0712345678', '0812345678',
'61212345678', '61312345678', '61712345678', '61812345678',
'0412345678', '61412345678', '1300123456', '1800123456']
The Goal
Using http://www.outware.com.au/contact as an example ...
The 2 actual numbers on the page are:
+61 (0)3 8684 9912 and +61 (0)2 8064 7043 (both numbers appear twice - once in the main section of the page and once in the footer)
The Problem
#take HTML markup from body tags
b = driver.find_element_by_css_selector('body').text
#remove all non-alpha + white space.
b = re.sub(r'\W+', '', b)
Result:
"PORTFOLIOINNOVATIONSERVICESCAREERSINSIGHTSNEWSABOUTCONTACTCONTACTOUTWAREMelbourneLe......AFRFast100Nov92017EXPLOREOUTWAREPortfolioInnovationWorkingatOutwareAboutSitemapCONNECTMELBOURNELevel3469LaTrobeStMelbourneVIC3000610386849912SYDNEYLevel41SmailStUltimoNSW2007610280647043"
Now if I apply my regex to this string
re.findall(r'[0][2]\d{8}|[0][3]\d{8}|[0][7]\d{8}|[0][8]\d{8}|[6][1][2]\d{8}|[6][1][3]\d{8}|[6][1][7]\d{8}|[6][1][8]\d{8}|[0][4]\d{8}|[6][1][4]\d{8}|[1][3][0][0]\d{6}|[1][8][0][0]\d{6}', re.sub(r'\W+', '', b))
Result:
[u'0386849912', u'0761028064', u'0386849912', u'0761028064']
I am getting a false positive because I have concatenated a postcode "NSW2007" onto the start of the phone number.
I presume because the regex has parsed the first part of "NSW2007610280647043" matching "0761028064" it doesn't then match "0280647043" which is also part of the same substring
I actually don't mind the false positive (i.e. getting "0761028064") but I do need to solve the false negative (i.e. not getting "0280647043")
I know there's some RegEx gurus here who can help on this. :-)
Please help!!

Don't search/replace any text prior to using the regex. That will make your input unusable. Try this:
(?:(?:\+?61 )?(?:0|\(0\))?)?[2378] \d{4}[ -]?\d{4}
https://regex101.com/r/1Q4HuD/3

It might help if you use a negative look ahead to check to see make sure the following character is not a number. For example: (?!\d).
This could create a problem though if some data following a phone number starts with a number.
The look behind looks like this when implemented in your regex:
(02\d{8}|03\d{8}|07\d{8}|08\d{8}|612\d{8}|613\d{8}|617\d{8}|618\d{8}|04\d{8}|614\d{8}|1300\d{6}|1800\d{6})(?!\d)
(I removed the square brackets as you do not need them when trying to match a single character)

This answer should be a comment, it isn't because of my low reputation!
I've seen you're updating the regex and I think this variation can help you. It should match very uncommon formats!
(\+61 )?(?:0|\(0\))?[2378] (?:[\s-]*\d){8}

How to use regular expression extract data not followed by something with pandas

I just want to extract the years, but not the number. How can I define not followed by XXX?
I made the following example, but the result is always a literal more than I expected.
text = ["hi2017", "322017"]
text = pd.Series(text)
myPat = "([^\d]\d{4})"
res = text.str.extract(myPat)
res
Then I get the result:
0 i2017
1 NaN
dtype: object
Actually, I just want to get "2017", but not "i2017", how can I do it?
PS. The "322017" should not be extracted, because it is not a year, but a number

Give this a try:
(?<!\d)(\d{4})(?!\d)
which returns 2017 and is based almost entirely on the comment by #PauloAlmeida

As I understand, you need only year, defined as 4 digits followed by non-number.
"(?:[a-z]+)(\d{4})$" works for me. (which means 4 digits followed by more than one character & the 4 digits are the last characters of the string)
text = ["hi2017", "322017"]
text = pd.Series(text)
myPat = "(?:[a-z]+)(\d{4})$"
res = text.str.extract(myPat)
Output:
print(res)
'''
0 2017
1 NaN
'''

You want 4-digit numbers where the first digit is either a 1 or a 2. This translates to all the numbers between 1000 to 2999, inclusive.
The regex for this is: (1[0-9]{3})|(2[0-9]{3})
This will get all the numbers between 1000 and 2999, inclusive within a string.
In your case, hi2017 will result in 2017. Additionally, 322017 will result in 2201. This is also a valid year as per your definition.

Regexr is a great online tool http://regexr.com/3ghcq
myPat = "(\d{4})"