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Substring[whole word] check using a string variable

In Python2.7, I am trying the following:
>>> import re
>>> text='0.0.0.0/0 172.36.128.214'
>>> far_end_ip="172.36.128.214"
>>>
>>>
>>> chk=re.search(r"\b172.36.128.214\b",text)
>>> chk
<_sre.SRE_Match object at 0x0000000002349578>
>>> chk=re.search(r"\b172.36.128.21\b",text)
>>> chk
>>> chk=re.search(r"\b"+far_end_ip+"\b",text)
>>>
>>> chk
>>>
Q:how can i make the search work when using the variable far_end_ip

Two issues:
You need to write the last bit of the string as a regex literal or escape the backslash: ... + r"\b"
You should escape the dots in the text to find: ... + re.escape(far_end_ip)
So:
re.search(r"\b" + re.escape(far_end_ip) + r"\b",text)
See also "How to use a variable inside a regular expression?".

How do I split the following string?

I have the following string where I need to extract only the first digits from it.
string = '50.2000\xc2\xb0 E'
How do I extract 50.2000 from string?

If the number can be followed by any kind of character, try using a regex:
>>> import re
>>> r = re.compile(r'(\d+\.\d+)')
>>> r.match('50.2000\xc2\xb0 E').group(1)
'50.2000'

mystring = '50.2000\xc2\xb0 E'
print mystring.split("\xc2", 1)[0]
Output
50.2000

If you just wanted to split the first digits, just slice the string:
start = 10 #start at the 10th digit
print mystring[start:]
Demo:
>>> my_string = 'abcasdkljf23u109842398470ujw{}{\\][\\['
>>> start = 10
>>> print(my_string[start:])
23u109842398470ujw{}{\][\[
You can, split the string at the first \:
>>> s = r'50.2000\xc2\xb0 E'
>>> s.split('\\', 1)
['50.2000', 'xc2\\xb0 E']

You could solve this using a regular expression:
In [1]: import re
In [2]: string = '50.2000\xc2\xb0 E'
In [3]: m = re.match('^([0-9]+\.?[0-9]*)', string)
In [4]: m.group(0)
Out[4]: '50.2000'

Python Regular expression repeat

I have a string like this
--x123-09827--x456-9908872--x789-267504
I am trying to get all value like
123:09827
456:9908872
789:267504
I've tried (--x([0-9]+)-([0-9])+)+
but it only gives me last pair result, I am testing it through python
>>> import re
>>> x = "--x123-09827--x456-9908872--x789-267504"
>>> p = "(--x([0-9]+)-([0-9]+))+"
>>> re.match(p,x)
>>> re.match(p,x).groups()
('--x789-267504', '789', '267504')
How should I write with nested repeat pattern?
Thanks a lot!
David

Code it like this:
x = "--x123-09827--x456-9908872--x789-267504"
p = "--x(?:[0-9]+)-(?:[0-9]+)"
print re.findall(p,x)

Just use the .findall method instead, it makes the expression simpler.
>>> import re
>>> x = "--x123-09827--x456-9908872--x789-267504"
>>> r = re.compile(r"--x(\d+)-(\d+)")
>>> r.findall(x)
[('123', '09827'), ('456', '9908872'), ('789', '267504')]
You can also use .finditer which might be helpful for longer strings.
>>> [m.groups() for m in r.finditer(x)]
[('123', '09827'), ('456', '9908872'), ('789', '267504')]

Use re.finditer or re.findall. Then you don't need the extra pair of parentheses that wrap the entire expression. For example,
>>> import re
>>> x = "--x123-09827--x456-9908872--x789-267504"
>>> p = "--x([0-9]+)-([0-9]+)"
>>> for m in re.finditer(p,x):
>>> print '{0} {1}'.format(m.group(1),m.group(2))

try this
p='--x([0-9]+)-([0-9]+)'
re.findall(p,x)

No need to use regex :
>>> "--x123-09827--x456-9908872--x789-267504".replace('--x',' ').replace('-',':').strip()
'123:09827 456:9908872 789:267504'

You don't need regular expressions for this. Here is a simple one-liner, non-regex solution:
>>> input = "--x123-09827--x456-9908872--x789-267504"
>>> [ x.replace("-", ":") for x in input.split("--x")[1:] ]
['123:09827', '456:9908872', '789:267504']
If this is an exercise on regex, here is a solution that uses the repetition (technically), though the findall(...) solution may be preferred:
>>> import re
>>> input = "--x123-09827--x456-9908872--x789-267504"
>>> regex = '--x(.+)'
>>> [ x.replace("-", ":") for x in re.match(regex*3, input).groups() ]
['123:09827', '456:9908872', '789:267504']

How to get integer values from a string in Python?

Suppose I had a string
string1 = "498results should get"
Now I need to get only integer values from the string like 498. Here I don't want to use list slicing because the integer values may increase like these examples:
string2 = "49867results should get"
string3 = "497543results should get"
So I want to get only integer values out from the string exactly in the same order. I mean like 498,49867,497543 from string1,string2,string3 respectively.
Can anyone let me know how to do this in a one or two lines?

>>> import re
>>> string1 = "498results should get"
>>> int(re.search(r'\d+', string1).group())
498
If there are multiple integers in the string:
>>> map(int, re.findall(r'\d+', string1))
[498]

An answer taken from ChristopheD here: https://stackoverflow.com/a/2500023/1225603
r = "456results string789"
s = ''.join(x for x in r if x.isdigit())
print int(s)
456789

Here's your one-liner, without using any regular expressions, which can get expensive at times:
>>> ''.join(filter(str.isdigit, "1234GAgade5312djdl0"))
returns:
'123453120'

if you have multiple sets of numbers then this is another option
>>> import re
>>> print(re.findall('\d+', 'xyz123abc456def789'))
['123', '456', '789']
its no good for floating point number strings though.

Iterator version
>>> import re
>>> string1 = "498results should get"
>>> [int(x.group()) for x in re.finditer(r'\d+', string1)]
[498]

>>> import itertools
>>> int(''.join(itertools.takewhile(lambda s: s.isdigit(), string1)))

With python 3.6, these two lines return a list (may be empty)
>>[int(x) for x in re.findall('\d+', your_string)]
Similar to
>>list(map(int, re.findall('\d+', your_string))

this approach uses list comprehension, just pass the string as argument to the function and it will return a list of integers in that string.
def getIntegers(string):
numbers = [int(x) for x in string.split() if x.isnumeric()]
return numbers
Like this
print(getIntegers('this text contains some numbers like 3 5 and 7'))
Output
[3, 5, 7]

def function(string):
final = ''
for i in string:
try:
final += str(int(i))
except ValueError:
return int(final)
print(function("4983results should get"))

Another option is to remove the trailing the letters using rstrip and string.ascii_lowercase (to get the letters):
import string
out = [int(s.replace(' ','').rstrip(string.ascii_lowercase)) for s in strings]
Output:
[498, 49867, 497543]

integerstring=""
string1 = "498results should get"
for i in string1:
if i.isdigit()==True
integerstring=integerstring+i
print(integerstring)

How do I coalesce a sequence of identical characters into just one?

Suppose I have this:
My---sun--is------very-big---.
I want to replace all multiple hyphens with just one hyphen.

import re
astr='My---sun--is------very-big---.'
print(re.sub('-+','-',astr))
# My-sun-is-very-big-.

If you want to replace any run of consecutive characters, you can use
>>> import re
>>> a = "AA---BC++++DDDD-EE$$$$FF"
>>> print(re.sub(r"(.)\1+",r"\1",a))
A-BC+D-E$F
If you only want to coalesce non-word-characters, use
>>> print(re.sub(r"(\W)\1+",r"\1",a))
AA-BC+DDDD-EE$FF
If it's really just hyphens, I recommend unutbu's solution.

If you really only want to coalesce hyphens, use the other suggestions. Otherwise you can write your own function, something like this:
>>> def coalesce(x):
... n = []
... for c in x:
... if not n or c != n[-1]:
... n.append(c)
... return ''.join(n)
...
>>> coalesce('My---sun--is------very-big---.')
'My-sun-is-very-big-.'
>>> coalesce('aaabbbccc')
'abc'

As usual, there's a nice itertools solution, using groupby:
>>> from itertools import groupby
>>> s = 'aaaaa----bbb-----cccc----d-d-d'
>>> ''.join(key for key, group in groupby(s))
'a-b-c-d-d-d'

How about:
>>> import re
>>> re.sub("-+", "-", "My---sun--is------very-big---.")
'My-sun-is-very-big-.'
the regular expression "-+" will look for 1 or more "-".

re.sub('-+', '-', "My---sun--is------very-big---")

How about an alternate without the re module:
'-'.join(filter(lambda w: len(w) > 0, 'My---sun--is------very-big---.'.split("-")))
Or going with Tim and FogleBird's previous suggestion, here's a more general method:
def coalesce_factory(x):
return lambda sent: x.join(filter(lambda w: len(w) > 0, sent.split(x)))
hyphen_coalesce = coalesce_factory("-")
hyphen_coalesce('My---sun--is------very-big---.')
Though personally, I would use the re module first :)
mcpeterson

Another simple solution is the String object's replace function.
while '--' in astr:
astr = astr.replace('--','-')

if you don't want to use regular expressions:
my_string = my_string.split('-')
my_string = filter(None, my_string)
my_string = '-'.join(my_string)

I have
my_str = 'a, b,,,,, c, , , d'
I want
'a,b,c,d'
compress all the blanks (the "replace" bit), then split on the comma, then if not None join with a comma in between:
my_str_2 = ','.join([i for i in my_str.replace(" ", "").split(',') if i])