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I have these strings:
Phone: 3396222
Phone: +33333388
I want to extract the numbers.
I tried this regular expression:
Phone:\s*(\d+\.\d+)
But I got an empty result
I am using scrapy so my code is like this: sel.xpath(..).re(..)
please don't suggest using other feature in python than regular expression
Your regular expression requires a . dot in the text, but your sample input has none.
Demo:
>>> import re
>>> re.search(r'Phone:\s*(\d+\.\d+)', 'Phone: 3396222') is None
True
>>> re.search(r'Phone:\s*(\d+\.\d+)', 'Phone: 339.6222').group(1)
'339.6222'
If you wanted to make either of your sample phone numbers match, remove the \. (instead adding it to a character set) and add an optional + to the expression:
r'Phone:\s*(\+?[\d.]+)'
Demo:
>>> re.search(r'Phone:\s*(\+?[\d.]+)', 'Phone: 3396222').group(1)
'3396222'
>>> re.search(r'Phone:\s*(\+?[\d.]+)', 'Phone: +33333388').group(1)
'+33333388'
This pattern also allows for any number of dots in the number:
>>> re.search(r'Phone:\s*(\+?[\d.]+)', 'Phone: +333.333.88').group(1)
'+333.333.88'
You are asking for mandatory dot(.) inside your regex. Mate it optional:
Phone:\s*\+?(\d+\.?\d+)
^^^ ^
I have updated by adding optional \+ as you added + in your input.
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What is wrong with the below used regular expression? Why does it not match the password?
import re
pattern = re.compile ('^\w##$%{8,}')
password = '12345abcd##$%'
x = pattern.search(password)
print (x)
print (len(password))
You didn't escape the $ which has a special meaning in a regular expression and didn't put the allowed characters in square brackets to allow any of them.
This: ^[\w##\$%]{8,} is the modified version of the regex which matches the password.
Escaping the $ character isn't really necessary within square brackets so ^[\w##$%]{8,} will work as well.
I suggest you check your regular expressions here: https://regex101.com/r/ldvJLf/1 . This site explains in detail the meaning of all single elements of the regular expression, so you can directly see what is wrong if things doesn't work as you expected.
Tip:
check your regexes online https://regexr.com/
I think you want:
pattern = re.compile ('^[\w##$%]{8,}')
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I am learning Python, and following along sentdex's videos. I just got to regular expressions, and copied the code he used. While the ages print out fine, when I try to print out the names I just get '[]' as an output.
import re
examplestring = ''' Jessica is 15 years old, and Daniel is 27 years old.
Edward is 97, and his grandfather, Oscar, is 102
'''
ages = re.findall(r'\d{1,3}',examplestring)
name = re.findall(r'[A-Z], [a-z]*',examplestring)
print(ages)
print(name)
There are multiple scenarios can be possible to match name. In you case, if name is Oscar then your regex should look like this.
Regex: [A-Z][a-z]+ there should be no comma and then space as it will try to find as CoryKramer mentioned.
[A-Z] means first letter is word and it is Capital.
[a-z] means from second letter onwards all letters are lowercase.
I have mentioned + instead of *. Difference between + and * is,
+ denotes at least one time so if you have word just O it will not match, your data should be at least two character like Os.
* denotes zero or more time so if you have word just O it will match, so if your name is any letter from Alphabet it will match. So if you think that your name can be only one letter use * else use +.
Example for *: https://regex101.com/r/n9HSIu/1
Example for +: https://regex101.com/r/hL4Pd8/1
The problem here is with you using comma(,) while writing the expression.
According to it, it will be looking for a Capital Letter(A-Z) followed by comma(,) and then space followed by n number of alphabets which your string doesn't satisfy.
For your desired result you need to eliminate comma(,) and use this instead:
name = re.findall(r'[A-Z][a-z]*',examplestring)
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I have multiple strings like this 90'4 I want to extract the digits from the string and sum them up to get 94.
I tried compiling the pattern.
pattern="\d'\d"
re.compile(pattern)
I tried the methods findall and match, but did not get what I wanted.
I need to use regex I cannot use .split()
Use \d+ with findall to extract numbers and then find their sum:
import re
s = "this is 90'4"
numbers = re.findall(r'\d+', s)
print(sum(map(int, numbers)))
# 94
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Objective
I'm looking to use the regular expression \d+ to extract just the digits from the string, answer_40194.
Problem
I'm targeting a form element with Selenium and I'm printing the formID to the Terminal, but after the line re.findall('\d+', formID) I expect formID to be just the numbers 40194, but instead I'm getting the entire string answer_40194.
script.py
import selenium
import re
form = browser.find_element_by_tag_name('form')
formID = form.get_attribute('id')
re.findall('\d+', formID)
print formIDNumber
You need to assign the result to a variable, e.g.
var1 = re.findall('\d+', formID)
print(var1)
This will generate a list, if you only want one result, use
var1 = re.search('\d+', formID)
print(var1.group(0))
The latter is called a regular expression object, hence the .group(0), see the documentation on python.org for more information.
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I am trying to use regular expressions in python to say a 4 characters string with 1st character being a digit and 3 other characters being either a digit or a capital letter.
Here's examples of patterns that should match 1CTT, 2IR8, 35TR, 4T1R
I tried many ways, here's the last code I tried :
exp=re.compile("[0-9]{1}([A-Z0-9]{3})")
Thank you for your help !
The expression you've tried last, looks correct and should match the provided test strings. Though you don't have to specify {1} and there is no need for a capturing group (the parenthesis):
>>> import re
>>> text = "text, 1CTT, 2IR8, 35TR, 4T1R, smth else"
>>> pattern = re.compile(r"[0-9][A-Z0-9]{3}")
>>> pattern.findall(text)
['1CTT', '2IR8', '35TR', '4T1R']
You might need to additionally add the word boundary constraint (thanks to #Jon Clements):
>>> text = "text, 1CTT, 2IR8, 35TR, 4T1R, smth else, 35TT35XYZ"
>>> pattern = re.compile(r"\b[0-9][A-Z0-9]{3}\b")
>>> pattern.findall(text)
['1CTT', '2IR8', '35TR', '4T1R']