I am trying to write a generic pattern using regex so that it fetches only particular things from the string. Let's say we have strings like GigabitEthernet0/0/0/0 or FastEthernet0/4 or Ethernet0/0.222. The regex should fetch the first 2 characters and all the numerals. Therefore, the fetched result should be something like Gi0000 or Fa04 or Et00222 depending on the above cases.
x = 'GigabitEthernet0/0/0/2
m = re.search('([\w+]{2}?)[\\\.(\d+)]{0,}',x)
I am not able to understand how shall I write the regular expression. The values can be fetched in the form of a list also. I write few more patterns but it isn't helping.
In regex, you may use re.findall function.
>>> import re
>>> s = 'GigabitEthernet0/0/0/0 '
>>> s[:2]+''.join(re.findall(r'\d', s))
'Gi0000'
OR
>>> ''.join(re.findall(r'^..|\d', s))
'Gi0000'
>>> ''.join(re.findall(r'^..|\d', 'Ethernet0/0.222'))
'Et00222'
OR
>>> s = 'GigabitEthernet0/0/0/0 '
>>> s[:2]+''.join([i for i in s if i.isdigit()])
'Gi0000'
z="Ethernet0/0.222."
print z[:2]+"".join(re.findall(r"(\d+)(?=[\d\W]*$)",z))
You can try this.This will make sure only digits from end come into play .
Here is another option:
s = 'Ethernet0/0.222'
"".join(re.findall('^\w{2}|[\d]+', s))
Related
I have the following file names that exhibit this pattern:
000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...
I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:
time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
But this gives me two extra empty strings in the returned list:
time_info=['', '20111007T084734', '20111008T023142', '']
How do I get only the two time stamp information? i.e. I want:
time_info=['20111007T084734', '20111008T023142']
I'm no Python expert but maybe you could just remove the empty strings from your list?
str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)
Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.
>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')
You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')
If the timestamps are always after the second _ then you can use str.split and str.strip:
>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']
Since this came up on google and for completeness, try using re.findall as an alternative!
This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.
Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.
>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']
or, somewhat more general:
>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']
I have this string /1B5DB40?full and I want to convert it to 1B5DB40.
I need to remove the ?full and the front /
My site won't always have ?full at the end so I need something that will still work even if the ?full is not there.
Thanks and hopefully this isn't too confusing to get some help :)
EDIT:
I know I could slice at 0 and 8 or whatever, but the 1B5DB40 could be longer or shorter. For example it could be /1B5DB4000?full or /1B5
Using str.lstrip (to remove leading /) and str.split (to remove optinal part after ?):
>>> '/1B5DB40?full'.lstrip('/').split('?')[0]
'1B5DB40'
>>> '/1B5DB40'.lstrip('/').split('?')[0]
'1B5DB40'
or using urllib.parse.urlparse:
>>> import urllib.parse
>>> urllib.parse.urlparse('/1B5DB40?full').path.lstrip('/')
'1B5DB40'
>>> urllib.parse.urlparse('/1B5DB40').path.lstrip('/')
'1B5DB40'
You can use lstrip and rstrip:
>>> data.lstrip('/').rstrip('?full')
'1B5DB40'
This only works as long as you don't have the characters f, u, l, ?, / in the part that you want to extract.
You can use regular expressions:
>>> import re
>>> extract = re.compile('/?(.*?)\?full')
>>> print extract.search('/1B5DB40?full').group(1)
1B5DB40
>>> print extract.search('/1Buuuuu?full').group(1)
1Buuuuu
What about regular expressions?
import re
re.search(r'/(?P<your_site>[^\?]+)', '/1B5DB40?full').group('your_site')
In this case it matches everything that is between '/' and '?', but you can change it to your specific requirements
>>> '/1B5DB40?full'split('/')[1].split('?')[0]
'1B5DB40'
>>> '/1B5'split('/')[1].split('?')[0]
'1B5'
>>> '/1B5DB40000?full'split('/')[1].split('?')[0]
'1B5DB40000'
Split will simply return a single element list containing the original string if the separator is not found.
I need to extract the IP address from the following string.
>>> mydns='ec2-54-196-170-182.compute-1.amazonaws.com'
The text to the left of the dot needs to be returned. The following works as expected.
>>> mydns[:18]
'ec2-54-196-170-182'
But it does not work in all cases. For e.g.
mydns='ec2-666-777-888-999.compute-1.amazonaws.com'
>>> mydns[:18]
'ec2-666-777-888-99'
How to I use regular expressions in python?
No need for regex... Just use str.split
mydns.split('.', 1)[0]
Demo:
>>> mydns='ec2-666-777-888-999.compute-1.amazonaws.com'
>>> mydns.split('.', 1)[0]
'ec2-666-777-888-999'
If you wanted to use regex for this:
Regex String
ec2-([0-9]{1,3})-([0-9]{1,3})-([0-9]{1,3})-([0-9]{1,3}).*
Alternative (EC2 Agnostic):
.*\b([0-9]{1,3})-([0-9]{1,3})-([0-9]{1,3})-([0-9]{1,3}).*
Replacement String
Regular: \1.\2.\3.\4
Reverse: \4.\3.\2.\1
Python code
import re
subject = 'ec2-54-196-170-182.compute-1.amazonaws.com'
result = re.sub("ec2-([0-9]{1,3})-([0-9]{1,3})-([0-9]{1,3})-([0-9]{1,3}).*", r"\1.\2.\3.\4", subject)
print result
This regex will match (^[^.]+:
So Try this:
import re
string = "ec2-54-196-170-182.compute-1.amazonaws.com"
ip = re.findall('^[^.]+',string)[0]
print ip
Output:
ec2-54-196-170-182
Best thing is this will match even if the instance was ec2,ec3 so this regex is actually very much similar to the code of #mgilson
I am rather new to Python Regex (regex in general) and I have been encountering a problem. So, I have a few strings like so:
str1 = r'''hfo/gfbi/mytag=a_17014b_82c'''
str2 = r'''/bkyhi/oiukj/game/?mytag=a_17014b_82c&'''
str3 = r'''lkjsd/image/game/mytag=a_17014b_82c$'''
the & and the $ could be any symbol.
I would like to have a single regex (and replace) which replaces:
mytag=a_17014b_82c
to:
mytag=myvalue
from any of the above 3 strings. Would appreciate any guidance on how I can achieve this.
UPDATE: the string to be replaced is always not the same. So, a_17014b_82c could be anything in reality.
If the string to be replaced is constant you don't need a regex. Simply use replace:
>>> str1 = r'''hfo/gfbi/mytag=a_17014b_82c'''
>>> str1.replace('a_17014b_82c','myvalue')
'hfo/gfbi/mytag=myvalue'
Use re.sub:
>>> import re
>>> r = re.compile(r'(mytag=)(\w+)')
>>> r.sub(r'\1myvalue', str1)
'hfo/gfbi/mytag=myvalue'
>>> r.sub(r'\1myvalue', str2)
'/bkyhi/oiukj/game/?mytag=myvalue&'
>>> r.sub(r'\1myvalue', str3)
'lkjsd/image/game/mytag=myvalue$'
import re
r = re.compile(r'(mytag=)\w+$')
r.sub(r'\1myvalue', str1)
This is based on #Ashwini's answer, two small changes are we are saying the mytag=a_17014b part should be at the end of input, so that even inputs such as
str1 = r'''/bkyhi/mytag=blah/game/?mytag=a_17014b_82c&'''
will work fine, substituting the last mytag instead of the the first.
Another small change is we are not unnecessarily capturing the \w+, since we aren't using it anyway. This is just for a bit of code clarity.
I have string: './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
I need string: '27-10-2011 17:07:02'
How can i do this in python?
There are many ways to do this, one way is to use str.partition:
text='./money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
before,_,after = text.partition('[')
print(after[:-1])
# 27-10-2011 17:07:02
Another is to use str.split:
before,after = text.split('[',1)
print(after[:-1])
# 27-10-2011 17:07:02
or str.find and str.rfind:
ind1 = text.find('[')+1
ind2 = text.rfind(']')
print(text[ind1:ind2])
All these methods rely on the desired substring immediately following the first left-bracket [.
The first two methods also rely on the desired substring ending at the next-to-last character in text. The last method (using rfind) searches from the right for the index of the right-bracket, so it is a little more general, and does not depend on quite so many (potential off-by-one) constants.
If your string has always the same structure this is probably the simplest solution:
s = r'./money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
s[s.find("[")+1:s.find("]")]
Update:
After seeing some of the other answers this is a slight improvement:
s[s.find("[")+1:-1]
Exploiting the fact that the closing square bracket is the last character in your string.
If the format is "fixed", you can also use this
>>> s = './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
>>> s[-20:-1:]
'27-10-2011 17:07:02'
>>>
You can also use regular expression:
import re
s = './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
print re.search(r'\[(.*?)\]', s).group(1)
Try with a regex :
import re
re.findall(".*\[(.*)\]", './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]')
>>> ['27-10-2011 17:07:02']
Probably the easiest way(if you know the string will always be in this format
>>> s = './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
>>> s[s.index('[') + 1:-1]
'27-10-2011 17:07:02'