Hello and thanks for helping.
String examples:
"Hello 43543" ---> "43543"
"John Doe 434-234" ---> "434-234"
I need a regex to extract the examples on the right.
I would do it following way:
import re
pattern = r'\d[0-9\-]*'
number1 = re.findall(pattern,'Hello 43543')
number2 = re.findall(pattern,'John Doe 434-234')
print(number1[0]) #43543
print(number2[0]) #434-234
My solution assumes that you are looking for any string starting with digit and with all other characters being digit or -, this mean it will also grab for example 4--- or 9-2-4--- and so on, however this might be not issue in your use case.
I want to note that before writing pattern, you should answer question: what it should match exactly? My pattern works as intended for examples you given, but keep in mind that this do NOT automatically mean it would give desired output with all data you might want to process using it.
If all your strings are like this, you can achieve the same without re:
s = "John Doe 434-234"
n = s.split()[-1]
print(n)
>>> "434-234"
It will split your string on spaces and give you the last field.
Related
I'm really sorry for asking because there are some questions like this around. But can't get the answer fixed to make problem.
This are the input lines (e.g. from a config file)
profile2.name=share2
profile8.name=share8
profile4.name=shareSSH
profile9.name=share9
I just want to extract the values behind the = sign with Python 3.9. regex.
I tried this on regex101.
^profile[0-9]\.name=(.*?)
But this gives me the variable name including the = sign as result; e.g. profile2.name=. But I want exactly the inverted opposite.
The expected results (what Pythons re.find_all() return) are
['share2', 'share8', 'shareSSH', 'share9']
Try pattern profile\d+\.name=(.*), look at Regex 101 example
import re
re.findall('profile\d+\.name=(.*)', txt)
# output
['share2', 'share8', 'shareSSH', 'share9']
But this problem doesn't necessarily need regex, split should work absolutely fine:
Try removing the ? quantifier. It will make your capture group match an empty st
regex101
I have a string:
s3://tester/test.pdf
I want to exclude s3://tester/ so even if i have s3://tester/folder/anotherone/test.pdf I am getting the entire path after s3://tester/
I have attempted to use the split & partition method but I can't seem to get it.
Currently am trying:
string.partition('/')[3]
But i get an error saying that it out of index.
EDIT: I should have specified that the name of the bucket will not always be the same so I want to make sure that it is only grabbing anything after the 3rd '/'.
You can use str.split():
path = 's3://tester/test.pdf'
print(path.split('/', 3)[-1])
Output:
test.pdf
UPDATE: With regex:
import re
path = 's3://tester/test.pdf'
print(re.split('/',path,3)[-1])
Output:
test.pdf
Have you tried .replace?
You could do:
string = "s3://tester/test.pdf"
string = string.replace("s3://tester/", "")
print(string)
This will replace "s3://tester/" with the empty string ""
Alternatively, you could use .split rather than .partition
You could also try:
string = "s3://tester/test.pdf"
string = "/".join(string.split("/")[3:])
print(string)
To answer "How to get everything after x amount of characters in python"
string[x:]
PLEASE SEE UPDATE
ORIGINAL
Using the builtin re module.
p = re.search(r'(?<=s3:\/\/tester\/).+', s).group()
The pattern uses a lookbehind to skip over the part you wish to ignore and matches any and all characters following it until the entire string is consumed, returning the matched group to the p variable for further processing.
This code will work for any length path following the explicit s3://tester/ schema you provided in your question.
UPDATE
Just saw updates duh.
Got the wrong end of the stick on this one, my bad.
Below re method should work no matter S3 variable, returning all after third / in string.
p = ''.join(re.findall(r'\/[^\/]+', s)[1:])[1:]
I'm trying to replace {x;y} patterns in a text corpus with "x or y", except that the number of elements is variable, so sometimes there will be 3 or more elements i.e. {x;y;z} (max is 9).
I'm trying to do this with regex, but I'm not sure how to do this such that I can replace according to the number of elements present. So I mean like, if I use a regex with a variable component like the following
part = '(;[\w\s]+)'
regex = '\(([\w\s]+);([\w\s]+){}?\)'.format(part)
re.sub(regex,/1 or /2 or /3, text)
I will sometimes get an additional 'or' (and more if I increase the number of variable elements) when there are only 2 elements present in the braces, which I don't want. The alternative is to do this many times with different number of variable parts but the code would be very clunky. I'm wondering if there are any ways I could achieve this with regex methods? Would appreciate any ideas.
I'm using python3.5 with spyder.
The scenario is just a bit too much for a regular search-and-replace action, so I would recommend passing in a function to dynamically generate the replacement string.
import re
text = 'There goes my {cat;dog} playing in the {street;garden}.'
def replacer(m):
return m.group(1).replace(';', ' or ')
output = re.sub(r'\{((\w;?)*\w)\}', replacer, text)
print(output)
Output:
There goes my cat or dog playing in the street or garden.
Programming in Python3.
I am having difficulty in controlling whether a string meets a specific format.
So, I know that Python does not have a .contain() method like Java but that we can use regex.
My code hence will probably look something like this, where lowpan_headers is a dictionary with a field that is a string that should meet a specific format.
So the code will probably be like this:
import re
lowpan_headers = self.converter.lowpan_string_to_headers(lowpan_string)
pattern = re.compile("^([A-Z][0-9]+)+$")
pattern.match(lowpan_headers[dest_addrS])
However, my issue is in the format and I have not been able to get it right.
The format should be like bbbb00000000000000170d0000306fb6, where the first 4 characters should be bbbb and all the rest, with that exact length, should be hexadecimal values (so from 0-9 and a-f).
So two questions:
(1) any easier way of doing this except through importing re
(2) If not, can you help me out with the regex?
As for the regex you're looking for I believe that
^bbbb[0-9a-f]{28}$
should validate correctly for your requirements.
As for if there is an easier way than using the re module, I would say that there isn't really to achieve the result you're looking for. While using the in keyword in python works in the way you would expect a contains method to work for a string, you are actually wanting to know if a string is in a correct format. As such the best solution, as it is relatively simple, is to use a regular expression, and thus use the re module.
Here is a solution that does not use regex:
lowpan_headers = 'bbbb00000000000000170d0000306fb6'
if lowpan_headers[:4] == 'bbbb' and len(lowpan_headers) == 32:
try:
int(lowpan_headers[4:], 16) # tries interpreting the last 28 characters as hexadecimal
print('Input is valid!')
except ValueError:
print('Invalid Input') # hex test failed!
else:
print('Invalid Input') # either length test or 'bbbb' prefix test failed!
In fact, Python does have an equivalent to the .contains() method. You can use the in operator:
if 'substring' in long_string:
return True
A similar question has already been answered here.
For your case, however, I'd still stick with regex as you're indeed trying to evaluate a certain String format. To ensure that your string only has hexadecimal values, i.e. 0-9 and a-f, the following regex should do it: ^[a-fA-F0-9]+$. The additional "complication" are the four 'b' at the start of your string. I think an easy fix would be to include them as follows: ^(bbbb)?[a-fA-F0-9]+$.
>>> import re
>>> pattern = re.compile('^(bbbb)?[a-fA-F0-9]+$')
>>> test_1 = 'bbbb00000000000000170d0000306fb6'
>>> test_2 = 'bbbb00000000000000170d0000306fx6'
>>> pattern.match(test_1)
<_sre.SRE_Match object; span=(0, 32), match='bbbb00000000000000170d0000306fb6'>
>>> pattern.match(test_2)
>>>
The part that is currently missing is checking for the exact length of the string for which you could either use the string length method or extend the regex -- but I'm sure you can take it from here :-)
As I mentioned in the comment Python does have contains() equivalent.
if "blah" not in somestring:
continue
(source) (PythonDocs)
If you would prefer to use a regex instead to validate your input, you can use this:
^b{4}[0-9a-f]{28}$ - Regex101 Demo with explanation
I'm working with some poorly formatted HTML and I need to find every instance of a certain type of pattern. The issue is as follows:
a space, followed by a 1 to 3 digit number, followed by letters (a word, usually). Here are some examples of what I mean.
hello 7Out
how 99In
are 123May
So I would be looking for the expression to get the "7Out", "99In", "123May", etc. The initial space does not need to be included. I hope this is descriptive enough, as I am literally just starting to expose myself to regular expressions and am still struggling a bit. In the end, I will want to count the total number of these instances and add the total count to a df that already exists, so if you have any suggestions on how to do that I would be open to that as well. Thanks for your help in advance!
Your regular expression will be: r'\w\s(\d{1,3}[a-zA-Z]+)'
So in order to get count you can use len() upon list returned by findall. The code will be
import re
string='hello 70qwqeqwfwe123 12wfgtr123 34wfegr123 dqwfrgb'
result=re.findall(r'\w\s(\d{1,3}[a-zA-Z]+)',string)
print "result = ",result #this will give you all the found occurances as list
print "len(result) = ",len(result) #this will give you total no of occurances.
The result will be:
result = ['70qwqeqwfwe', '12wfgtr', '34wfegr']
len(result) = 3
Hint: findall will evaluate regular expression and returns results based on grouping. I'm using that to solve this problem.
Try these:
re.findall(r'(\w\s((\d{1,3})[a-zA-Z]+))',string)
re.findall(r'\w\s((\d{1,3})[a-zA-Z]+)',string)
To get an idea about regular expressions refer python re, tutorials point and to play with the matching characters use this.