I'm attempting to string match 5-digit coupon codes spread throughout a HTML web page. For example, 53232, 21032, 40021 etc... I can handle the simpler case of any string of 5 digits with [0-9]{5}, though this also matches 6, 7, 8... n digit numbers. Can someone please suggest how I would modify this regular expression to match only 5 digit numbers?
>>> import re
>>> s="four digits 1234 five digits 56789 six digits 012345"
>>> re.findall(r"\D(\d{5})\D", s)
['56789']
if they can occur at the very beginning or the very end, it's easier to pad the string than mess with special cases
>>> re.findall(r"\D(\d{5})\D", " "+s+" ")
Without padding the string for special case start and end of string, as in John La Rooy answer one can use the negatives lookahead and lookbehind to handle both cases with a single regular expression
>>> import re
>>> s = "88888 999999 3333 aaa 12345 hfsjkq 98765"
>>> re.findall(r"(?<!\d)\d{5}(?!\d)", s)
['88888', '12345', '98765']
full string: ^[0-9]{5}$
within a string: [^0-9][0-9]{5}[^0-9]
Note: There is problem in using \D since \D matches any character that is not a digit , instead use \b.
\b is important here because it matches the word boundary but only at end or beginning of a word .
import re
input = "four digits 1234 five digits 56789 six digits 01234,56789,01234"
re.findall(r"\b\d{5}\b", input)
result : ['56789', '01234', '56789', '01234']
but if one uses
re.findall(r"\D(\d{5})\D", s)
output : ['56789', '01234']
\D is unable to handle comma or any continuously entered numerals.
\b is important part here it matches the empty string but only at end or beginning of a word .
More documentation: https://docs.python.org/2/library/re.html
More Clarification on usage of \D vs \b:
This example uses \D but it doesn't capture all the five digits number.
This example uses \b while capturing all five digits number.
Cheers
A very simple way would be to match all groups of digits, like with r'\d+', and then skip every match that isn't five characters long when you process the results.
You probably want to match a non-digit before and after your string of 5 digits, like [^0-9]([0-9]{5})[^0-9]. Then you can capture the inner group (the actual string you want).
You could try
\D\d{5}\D
or maybe
\b\d{5}\b
I'm not sure how python treats line-endings and whitespace there though.
I believe ^\d{5}$ would not work for you, as you likely want to get numbers that are somewhere within other text.
I use Regex with easier expression :
re.findall(r"\d{5}", mystring)
It will research 5 numerical digits. But you have to be sure not to have another 5 numerical digits in the string
Related
I want to match strings in which every second character is same.
for example 'abababababab'
I have tried this : '''(([a-z])[^/2])*'''
The output should return the complete string as it is like 'abababababab'
This is actually impossible to do in a real regular expression with an amount of states polynomial to the alphabet size, because the expression is not a Chomsky level-0 grammar.
However, Python's regexes are not actually regular expressions, and can handle much more complex grammars than that. In particular, you could put your grammar as the following.
(..)\1*
(..) is a sequence of 2 characters. \1* matches the exact pair of characters an arbitrary (possibly null) number of times.
I interpreted your question as wanting every other character to be equal (ababab works, but abcbdb fails). If you needed only the 2nd, 4th, ... characters to be equal you can use a similar one.
.(.)(.\1)*
You could match the first [a-z] followed by capturing ([a-z]) in a group. Then repeat 0+ times matching again a-z and a backreference to group 1 to keep every second character the same.
^[a-z]([a-z])(?:[a-z]\1)*$
Explanation
^ Start of the string
[a-z]([a-z]) Match a-z and capture in group 1 matching a-z
)(?:[a-z]\1)* Repeat 0+ times matching a-z followed by a backreference to group 1
$ End of string
Regex demo
Though not a regex answer, you could do something like this:
def all_same(string):
return all(c == string[1] for c in string[1::2])
string = 'abababababab'
print('All the same {}'.format(all_same(string)))
string = 'ababacababab'
print('All the same {}'.format(all_same(string)))
the string[1::2] says start at the 2nd character (1) and then pull out every second character (the 2 part).
This returns:
All the same True
All the same False
This is a bit complicated expression, maybe we would start with:
^(?=^[a-z]([a-z]))([a-z]\1)+$
if I understand the problem right.
Demo
Example strings:
I am a numeric string 75698
I am a alphanumeric string A14-B32-C7D
So far my regex works: (\S+)$
I want to add a way (probably look ahead) to check if the result generated by above regex contains any digit (0-9) one or more times?
This is not working: (\S+(?=\S*\d\S*))$
How should I do it?
Look ahead is not necessary for this, this is simply :
(\S*\d+\S*)
Here is a test case :
http://regexr.com?34s7v
permute it and use the \D class instead of \S:
((?=\D*\d)\S+)$
explanation: \D = [^\d] in other words it is all that is not a digit.
You can be more explicit (better performances for your examples) with:
((?=[a-zA-Z-]*\d)\[a-zA-Z\d-]+)$
and if you have only uppercase letters, you know what to do. (smaller is the class, better is the regex)
text = '''
I am a numeric string 75698 \t
I am a alphanumeric string A14-B32-C7D
I am a alphanumeric string A14-B32-C74578
I am an alphabetic number: three
'''
import re
regx = re.compile('\s(?=.*\d)([\da-zA-Z-]+)\s*$',re.MULTILINE)
print regx.findall(text)
# result ['75698', 'A14-B32-C7D', 'A14-B32-C74578']
Note the presence of \s* in front of $ in order to catch alphanumeric portions that are separated with whitespazces from the end of the lines.
I want to make sure using regex that a string is of the format- "999.999-A9-Won" and without any white spaces or tabs or newline characters.
There may be 2 or 3 numbers in the range 0 - 9.
Followed by a period '.'
Again followed by 2 or 3 numbers in the range 0 - 9
Followed by a hyphen, character 'A' and a number between 0 - 9 .
This can be followed by anything.
Example: 87.98-A8-abcdef
The code I have come up until now is:
testString = "87.98-A1-help"
regCompiled = re.compile('^[0-9][0-9][.][0-9][0-9][-A][0-9][-]*');
checkMatch = re.match(regCompiled, testString);
if checkMatch:
print ("FOUND")
else:
print("Not Found")
This doesn't seem to work. I'm not sure what I'm missing and also the problem here is I'm not checking for white spaces, tabs and new line characters and also hard-coded the number for integers before and after decimal.
With {m,n} you can specify the number of times a pattern can repeat, and the \d character class matches all digits. The \S character class matches anything that is not whitespace. Using these your regular expression can be simplified to:
re.compile(r'\d{2,3}\.\d{2,3}-A\d-\S*\Z')
Note also the \Z anchor, making the \S* expression match all the way to the end of the string. No whitespace (newlines, tabs, etc.) are allowed here. If you combine this with the .match() method you assure that all characters in your tested string conform to the pattern, nothing more, nothing less. See search() vs. match() for more information on .match().
A small demonstration:
>>> import re
>>> pattern = re.compile(r'\d{2,3}\.\d{2,3}-A\d-\S*\Z')
>>> pattern.match('87.98-A1-help')
<_sre.SRE_Match object at 0x1026905e0>
>>> pattern.match('123.45-A6-no whitespace allowed')
>>> pattern.match('123.45-A6-everything_else_is_allowed')
<_sre.SRE_Match object at 0x1026905e0>
Let's look at your regular expression. If you want:
"2 or 3 numbers in the range 0 - 9"
then you can't start your regular expression with '^[0-9][0-9][.] because that will only match strings with exactly two integers at the beginning. A second issue with your regex is at the end: [0-9][-]* - if you wish to match anything at the end of the string then you need to finish your regular expression with .* instead. Edit: see Martijn Pieters's answer regarding the whitespace in the regular expressions.
Here is an updated regular expression:
testString = "87.98-A1-help"
regCompiled = re.compile('^[0-9]{2,3}\.[0-9]{2,3}-A[0-9]-.*');
checkMatch = re.match(regCompiled, testString);
if checkMatch:
print ("FOUND")
else:
print("Not Found")
Not everything needs to be enclosed inside [ and ], in particular when you know the character(s) that you wish to match (such as the part -A). Furthermore:
the notation {m,n} means: match at least m times and at most n times, and
to explicitly match a dot, you need to escape it: that's why there is \. in the regular expression above.
I have the string.
st = "12345 hai how r u #3456? Awer12345 7890"
re.findall('([0-9]+)',st)
It should not come like :
['12345', '3456', '12345', '7890']
I should get
['12345','7890']
I should only take the numeric values
and
It should not contain any other chars like alphabets,special chars
No need to use a regular expression:
[i for i in st.split(" ") if i.isdigit()]
Which I think is much more readable than using a regex
Corey's solution is really the right way to go here, but since the question did ask for regex, here is a regex solution that I think is simpler than the others:
re.findall(r'(?<!\S)\d+(?!\S)', st)
And an explanation:
(?<!\S) # Fail if the previous character (if one exists) isn't whitespace
\d+ # Match one or more digits
(?!\S) # Fail if the next character (if one exists) isn't whitespace
Some examples:
>>> re.findall(r'(?<!\S)\d+(?!\S)', '12345 hai how r u #3456? Awer12345 7890')
['12345', '7890']
>>> re.findall(r'(?<!\S)\d+(?!\S)', '12345 hai how r u #3456? Awer12345 7890123ER%345 234 456 789')
['12345', '234', '456', '789']
In [21]: re.findall(r'(?:^|\s)(\d+)(?=$|\s)', st)
Out[21]: ['12345', '7890']
Here,
(?:^|\s) is a non-capture group that matches the start of the string, or a space.
(\d+) is a capture group that matches one or more digits.
(?=$|\s) is lookahead assertion that matches the end of the string, or a space, without consuming it.
use this: (^|\s)[0-9]+(\s|$) pattern. (^|\s) means that your number must be at the start of the string or there must be a whitespace character before the number. And (\s|$) means that there must be a whitespace after number or the number is at the end of the string.
As said Jan Pöschko, 456 won't be found in 123 456. If your "bad" parts (#, Awer) are always prefixes, you can use this (^|\s)[0-9]+ pattern and everything will be OK. It will match all numbers, which have only whitespaces before or are at the start of the string. Hope this helped...
Your expression finds all sequences of digits, regardless of what surrounds them. You need to include a specification of what comes before and after the sequence to get the behavior you want:
re.findall(r"[\D\b](\d+)[\D\b]", st)
will do what you want. In English, it says "match all sequences of one or more digits that are surrounded by a non-digit character.or a word boundary"
I have such string:
something: 20 kg/ something: 120 kg
I have this regex ("[0-9]{1,2} kg", string), but it returns 20kg both times. I need to return 20kg only in first case.
Try this:
(?<!\d)\d{1,2}\s+kg
The (?<!...) is a negative look behind. So it matches one or two digits not preceded by a digit. I also changed the literal space with one or more space chars.
Seeing you've asked Python questions, here's a demo in Python:
#!/usr/bin/env python
import re
string = 'something: 20 kg/ something: 120 kg'
print re.findall(r'(?<!\d)\d{1,2}\s+kg', string)
which will print ['20 kg']
edit
As #Tim mentioned, a word boundary \b is enough: r'\b\d{1,2}\s+kg'