Say I have a string
string = "{1/100}"
I want to use regular expressions in Python to convert it into
new_string = "\frac{1}{100}"
I think I would need to use something like this
new_string = re.sub(r'{.+/.+}', r'', string)
But I'm stuck on what I would put in order to preserve the characters in the fraction, in this example 1 and 100.
You can use () to capture the numbers. Then use \1 and \2 to refer to them:
new_string = re.sub(r'{(.+)/(.+)}', r'\\frac{\1}{\2}', string)
# \frac{1}{100}
Note: Don't forget to escape the backslash \\.
Capture the numbers using parens and then reference them in the replacement text using \1 and \2. For example:
>>> print re.sub(r'{(.+)/(.+)}', r'\\frac{\1}{\2}', "{1/100}")
\frac{1}{100}
Anything inside the braces would be a number/number. So in the regex place numbers([0-9]) instead of a .(dot).
>>> import re
>>> string = "{1/100}"
>>> new = re.sub(r'{([0-9]+)/([0-9]+)}', r'\\frac{\1}{\2}', string)
>>> print new
\frac{1}{100}
Use re.match. It's more flexible:
>>> m = re.match(r'{(.+)/(.+)}', string)
>>> m.groups()
('1', '100')
>>> new_string = "\\frac{%s}{%s}"%m.groups()
>>> print new_string
\frac{1}{100}
Related
I have a regex pattern with optional characters however at the output I want to remove those optional characters. Example:
string = 'a2017a12a'
pattern = re.compile("((20[0-9]{2})(.?)(0[1-9]|1[0-2]))")
result = pattern.search(string)
print(result)
I can have a match like this but what I want as an output is:
desired output = '201712'
Thank you.
You've already captured the intended data in groups and now you can use re.sub to replace the whole match with just contents of group1 and group2.
Try your modified Python code,
import re
string = 'a2017a12a'
pattern = re.compile(".*(20[0-9]{2}).?(0[1-9]|1[0-2]).*")
result = re.sub(pattern, r'\1\2', string)
print(result)
Notice, how I've added .* around the pattern, so any of the extra characters around your data is matched and gets removed. Also, removed extra parenthesis that were not needed. This will also work with strings where you may have other digits surrounding that text like this hello123 a2017a12a some other 99 numbers
Output,
201712
Regex Demo
You can just use re.sub with the pattern \D (=not a number):
>>> import re
>>> string = 'a2017a12a'
>>> re.sub(r'\D', '', string)
'201712'
Try this one:
import re
string = 'a2017a12a'
pattern = re.findall("(\d+)", string) # this regex will capture only digit
print("".join(p for p in pattern)) # combine all digits
Output:
201712
If you want to remove all character from string then you can do this
import re
string = 'a2017a12a'
re.sub('[A-Za-z]+','',string)
Output:
'201712'
You can use re module method to get required output, like:
import re
#method 1
string = 'a2017a12a'
print (re.sub(r'\D', '', string))
#method 2
pattern = re.findall("(\d+)", string)
print("".join(p for p in pattern))
You can also refer below doc for further knowledge.
https://docs.python.org/3/library/re.html
I want to write a line of regular expression that can match strings like "(2000)" with years in parentheses. then I can check if any string contains the substring "2000".
for example, I want the regex to match (2000) not 2000, or (20000),or (200).
That is to say: they have to have exactly four digits, the first digit between 1 and 2; they have to include the parentheses.
also 2000 is just an example I use but really I want to the regex to include all the possible years.
You have to escape the open and close paranthesis,
>>> import re
>>> str = """foo(2000)bar(1000)foobar2000"""
>>> regex = r'\(2000\)'
>>> result = re.findall(regex, str)
>>> print(result)
['(2000)']
OR
>>> import re
>>> str = """foo(2000)bar(1000)foobar(2014)barfoo(2020)"""
>>> regex = r'\([0-9]{4}\)'
>>> result = re.findall(regex, str)
>>> print(result)
['(2000)', '(1000)', '(2014)', '(2020)']
It matches all the four digit numbers(year's) present within the paranthesis.
Special characters need to be escaped with a backslash. A parenthesis ( becomes \(. Therefore (2000) becomes \(2000\).
Then you can do something like:
if re.search(r"\(2000\)", subject):
# Successful match
else:
# Match attempt failed
>>> import re
>>> x = re.match(r'\((\d*?)\)', "(2000)")
>>> x.group(1)
'2000'
This code below should be self explanatory. The regular expression is simple. Why doesn't it match?
>>> import re
>>> digit_regex = re.compile('\d')
>>> string = 'this is a string with a 4 digit in it'
>>> result = digit_regex.match(string)
>>> print result
None
Alternatively, this works:
>>> char_regex = re.compile('\w')
>>> result = char_regex.match(string)
>>> print result
<_sre.SRE_Match object at 0x10044e780>
Why does the second regex work, but not the first?
Here is what re.match() says If zero or more characters at the beginning of string match the regular expression pattern ...
In your case the string doesn't have any digit \d at the beginning. But for the \w it has t at the beginning at your string.
If you want to check for digit in your string using same mechanism, then add .* with your regex:
digit_regex = re.compile('.*\d')
The second finds a match because string starts with a word character. If you want to find matches within the string, use the search or findall methods (I see this was suggested in a comment too). Or change your regex (e.g. .*(\d).*) and use the .groups() method on the result.
Say I have three strings:
abc534loif
tvd645kgjf
tv96fjbd_gfgf
and three lists:
beginning captures just the first part of the string "the name"
middle captures just the number
end contains only the rest of the characters that are after the number portion
How do I accomplish this in the most efficent way?
Use regular expressions?
>>> import re
>>> strings = 'abc534loif tvd645kgjf tv96fjbd_gfgf'.split()
>>> for s in strings:
... for match in re.finditer(r'\b([a-z]+)(\d+)(.+?)\b', s):
... print match.groups()
...
('abc', '534', 'loif')
('tvd', '645', 'kgjf')
('tv', '96', 'fjbd_gfgf')
This is language agnostic approach that aims at higher efficiency:
find first digit in the string and save its position p0
find last digit in the string and save its position p1
extract substring from 0 to p0-1 into beginning
extract substring from p0 to p1 into middle
extract substring from p1+1 to length-1 into end
I guess you're looking for re.findall:
strs = """
abc534loif
tvd645kgjf
tv96fjbd_gfgf
"""
import re
print re.findall(r'\b(\w+?)(\d+)(\w+)', strs)
>> [('abc', '534', 'loif'), ('tvd', '645', 'kgjf'), ('tv', '96', 'fjbd_gfgf')]
>>> import itertools as it
>>> s="abc534loif"
>>> [''.join(j) for i,j in it.groupby(s, key=str.isdigit)]
['abc', '534', 'loif']
I'd something like this:
>>> import re
>>> l = ['abc534loif', 'tvd645kgjf', 'tv96fjbd_gfgf']
>>> regex = re.compile('([a-z_]+)(\d+)([a-z_]+)')
>>> beginning, middle, end = zip(*[regex.match(s).groups() for s in l])
>>> beginning
('abc', 'tvd', 'tv')
>>> middle
('534', '645', '96')
>>> end
('loif', 'kgjf', 'fjbd_gfgf')
I wouls use regualar expressions like:
(?P<beginning>[^0-9]*)(?P<middle>[^0-9]*)(?P<end>[^0-9]*)
and pull out the three matching sections.
import re
m = re.match(r"(?P<beginning>[^0-9]*)(?P<middle>[^0-9]*)(?P<end>[^0-9]*)", "abc534loif")
m.group('beginning')
m.group('middle')
m.group('end')
import re #You want to match a string against a pattern so you import the regular expressions module 're'
mystring = "abc1234def" #Just a string to test with
match = re.match(r"^(\D+)([0)9]+](\D+)$") #Our regular expression. Everything between brackets is 'captured', meaning that it is accessible as one of the 'groups' in the returned match object. The ^ sign matches at the beginning of a string, while the $ matches the end. the characters in between the square brackets [0-9] are character ranges, so [0-9] matches any digit character, \D is any non-digit character.
if match: # match will be None if the string didn't match the pattern, so we need to check for that, as None.group doesn't exist.
beginning = match.group(1)
middle = match.group(2)
end = match.group(3)
I am trying to remove chars from an unicode string. I have a whitelist of allowed unicode chars and I would like to remove everything that is not on the list.
allowed_list = ur'[\u0041-\u005A]|[\u0061-\u007A]|[\u00C0-\u00D6]|[\u00D8-\u00F6]|[\u00F8-\u012F]|\u0131|[\u0386]|[\u0388-\u038A]'
negated_list = ur'[^\u0041-\u005A]|[^\u0061-\u007A]|[^\u00C0-\u00D6]|[^\u00D8-\u00F6]|[^\u00F8-\u012F]|^\u0131|[^\u0386]|[^\u0388-\u038A]'
I am testing it with a subset of my list and I don't get why it is not working.
This removes all but lowercase latin chars:
>>> mystr = 'Arugg^]T'
>>> myre = re.compile(ur'[^\u0061-\u007A]', re.UNICODE)
>>> result = myre.sub('', mystr)
>>> result
'rugg'
This removes all but uppercase latin chars:
>>> mystr = 'Arugg^]T'
>>> myre = re.compile(ur'[^\u0041-\u005A]', re.UNICODE)
>>> result = myre.sub('', mystr)
>>> result
'AT'
But when I combine them, all chars get removed:
>>> mystr = 'Arugg^]T'
>>> myre = re.compile(ur'[^\u0041-\u005A]|[^\u0061-\u007A]', re.UNICODE)
>>> result = myre.sub('', mystr)
>>> result
''
When I tested the regex [^\u0041-\u005A]|[^\u0061-\u007A] on https://pythex.org/ it does what I am expecting, but when I atempt to use it in my code, it is not doing what I want it to. What am I missing?
Thank you in advance!
Your regex is not correct, you are using | which checks if either one is true.
You need to create one expression with multiple ranges,
[^\u0041-\u005A\u0061-\u007A] will match any characters except range \u0041-\u005A or \u0061-\u007A.
import re
regex = r"[^\u0041-\u005A\u0061-\u007A]"
test_str = "Arugg^]T"
myre = re.compile(regex, re.UNICODE)
result = myre.sub('', test_str)
print(result)
# output,
AruggT
Implicitly positive, regex class items are OR'd together.
Your regex is then the same as
[\u0041-\u005a\u0061-\u007a\u00c0-\u00d6\u00d8-\u00f6\u00f8-\u012f\u0131\u0386\u0388-\u038a]
But for the Negative regex class [^], items are individually negated then AND'ed together.
That regex is then
[^\u0041-\u005a\u0061-\u007a\u00c0-\u00d6\u00d8-\u00f6\u00f8-\u012f\u0131\u0386\u0388-\u038a]
which is logically the same as
[^\u0041-\u005A] and [^\u0061-\u007A] and [^\u00C0-\u00D6] and [^\u00D8-\u00F6] and [^\u00F8-\u012F] and [^\u0131] and [^\u0386] and [^\u0388-\u038A]
What you tried to do was negate each item, then OR them together
which is not the same.
You are replacing all characters that are
not in '[^\u0041-\u005A]' or not in [^\u0061-\u007A]' (due to the ^) .
If either one is true, all get replaced by '' - so its always true no matter what you have.
Use ur'[^\u0041-\u005A\u0061-\u007A]' instead (both ranges inside one [...].