I have a string as below
"Server: myserver.mysite.com\r\nAddress: 111.122.133.144\r\n\r\nName: myserver.mysite.com\r\nAddress: 123.144.412.111\r\nAliases: alias1.myserver.mysite.com\r\n\t myserver.mysite.com\r\n\r\n"
I'm currently struggling to write a function in python that will find all aliases and put them in a list. So basically, I need a list that will be ['alias1.myserver.mysite.com', 'myserver.mysite.com']
I tried the following code
pattern = '(?<=Aliases: )([\S*]+)'
name = re.findall(pattern, mystring)
but it only matches the first alias and not both of them.
Any ideas on this?
Greatly appreciated!
Try the following:
import re
s = "Server: myserver.mysite.com\r\nAddress: 111.122.133.144\r\n\r\nName: myserver.mysite.com\r\nAddress: 123.144.412.111\r\nAliases: alias1.myserver.mysite.com\r\n\t myserver.mysite.com\r\n\r\n"
l = re.findall(r'\S+', s.split('Aliases: ')[1])
print(l)
Prints:
['alias1.myserver.mysite.com', 'myserver.mysite.com']
Explanation
First we split the string into two pieces and keep the second piece with s.split('Aliases: ')[1]. This evaluates to the part of the string that follows 'Aliases: '.
Next we use findall with the regaular expression:
\S+
This matches all consecutive strings of one or more non-space characters.
But this can be more simply done in this case without using a regex:
s = "Server: myserver.mysite.com\r\nAddress: 111.122.133.144\r\n\r\nName: myserver.mysite.com\r\nAddress: 123.144.412.111\r\nAliases: alias1.myserver.mysite.com\r\n\t myserver.mysite.com\r\n\r\n"
l = s.split('Aliases: ')[1].split()
print(l)
Try this :
import re
regex = re.compile(r'[\n\r\t]')
t="Server: myserver.mysite.com\r\nAddress: 111.122.133.144\r\n\r\nName: myserver.mysite.com\r\nAddress: 123.144.412.111\r\nAliases: alias1.myserver.mysite.com\r\n\t myserver.mysite.com\r\n\r\n"
t = regex.sub(" ", t)
t = t.split("Aliases:")[1].strip().split()
print(t)
Related
Suppose I have a string like test-123.
I want to test whether it matches a pattern like test-<number>, where <number> means one or more digit symbols.
I tried this code:
import re
correct_string = 'test-251'
wrong_string = 'test-123x'
regex = re.compile(r'test-\d+')
if regex.match(correct_string):
print 'Matching correct string.'
if regex.match(wrong_string):
print 'Matching wrong_string.'
How can I make it so that only the correct_string matches, and the wrong_string doesn't? I tried using .search instead of .match but it didn't help.
Try with specifying the start and end rules in your regex:
re.compile(r'^test-\d+$')
For exact match regex = r'^(some-regex-here)$'
^ : Start of string
$ : End of string
Since Python 3.4 you can use re.fullmatch to avoid adding ^ and $ to your pattern.
>>> import re
>>> p = re.compile(r'\d{3}')
>>> bool(p.match('1234'))
True
>>> bool(p.fullmatch('1234'))
False
I think It may help you -
import re
pattern = r"test-[0-9]+$"
s = input()
if re.match(pattern,s) :
print('matched')
else :
print('not matched')
You can try re.findall():
import re
correct_string = 'test-251'
if len(re.findall("test-\d+", correct_string)) > 0:
print "Match found"
A pattern such as \btest-\d+\b should do you;
matches = re.search(r'\btest-\d+\', search_string)
Demo
This requires the matching of word boundaries, so prevents other substrings from occuring after your desired match.
I have some sample strings:
s = 'neg(able-23, never-21) s2-1/3'
i = 'amod(Market-8, magical-5) s1'
I've got the problem where I can figure out if the string has 's1' or 's3' using:
word = re.search(r's\d$', s)
But if I want to know if the contains 's2-1/3' in it, it won't work.
Is there a regex expression that can be used so that it works for both cases of 's#' and 's#+?
Thanks!
You can allow the characters "-" and "/" to be captured as well, in addition to just digits. It's hard to tell the exact pattern you're going for here, but something like this would capture "s2-1/3" from your example:
import re
s = "neg(able-23, never-21) s2-1/3"
word = re.search(r"s\d[-/\d]*$", s)
I'm guessing that maybe you would want to extract that with some expression, such as:
(s\d+)-?(.*)$
Demo 1
or:
(s\d+)-?([0-9]+)?\/?([0-9]+)?$
Demo 2
Test
import re
expression = r"(s\d+)-?(.*)$"
string = """
neg(able-23, never-21) s211-12/31
neg(able-23, never-21) s2-1/3
amod(Market-8, magical-5) s1
"""
print(re.findall(expression, string, re.M))
Output
[('s211', '12/31'), ('s2', '1/3'), ('s1', '')]
How can i get word example from such string:
str = "http://test-example:123/wd/hub"
I write something like that
print(str[10:str.rfind(':')])
but it doesn't work right, if string will be like
"http://tests-example:123/wd/hub"
You can use this regex to capture the value preceded by - and followed by : using lookarounds
(?<=-).+(?=:)
Regex Demo
Python code,
import re
str = "http://test-example:123/wd/hub"
print(re.search(r'(?<=-).+(?=:)', str).group())
Outputs,
example
Non-regex way to get the same is using these two splits,
str = "http://test-example:123/wd/hub"
print(str.split(':')[1].split('-')[1])
Prints,
example
You can use following non-regex because you know example is a 7 letter word:
s.split('-')[1][:7]
For any arbitrary word, that would change to:
s.split('-')[1].split(':')[0]
many ways
using splitting:
example_str = str.split('-')[-1].split(':')[0]
This is fragile, and could break if there are more hyphens or colons in the string.
using regex:
import re
pattern = re.compile(r'-(.*):')
example_str = pattern.search(str).group(1)
This still expects a particular format, but is more easily adaptable (if you know how to write regexes).
I am not sure why do you want to get a particular word from a string. I guess you wanted to see if this word is available in given string.
if that is the case, below code can be used.
import re
str1 = "http://tests-example:123/wd/hub"
matched = re.findall('example',str1)
Split on the -, and then on :
s = "http://test-example:123/wd/hub"
print(s.split('-')[1].split(':')[0])
#example
using re
import re
text = "http://test-example:123/wd/hub"
m = re.search('(?<=-).+(?=:)', text)
if m:
print(m.group())
Python strings has built-in function find:
a="http://test-example:123/wd/hub"
b="http://test-exaaaample:123/wd/hub"
print(a.find('example'))
print(b.find('example'))
will return:
12
-1
It is the index of found substring. If it equals to -1, the substring is not found in string. You can also use in keyword:
'example' in 'http://test-example:123/wd/hub'
True
Suppose I have a string like test-123.
I want to test whether it matches a pattern like test-<number>, where <number> means one or more digit symbols.
I tried this code:
import re
correct_string = 'test-251'
wrong_string = 'test-123x'
regex = re.compile(r'test-\d+')
if regex.match(correct_string):
print 'Matching correct string.'
if regex.match(wrong_string):
print 'Matching wrong_string.'
How can I make it so that only the correct_string matches, and the wrong_string doesn't? I tried using .search instead of .match but it didn't help.
Try with specifying the start and end rules in your regex:
re.compile(r'^test-\d+$')
For exact match regex = r'^(some-regex-here)$'
^ : Start of string
$ : End of string
Since Python 3.4 you can use re.fullmatch to avoid adding ^ and $ to your pattern.
>>> import re
>>> p = re.compile(r'\d{3}')
>>> bool(p.match('1234'))
True
>>> bool(p.fullmatch('1234'))
False
I think It may help you -
import re
pattern = r"test-[0-9]+$"
s = input()
if re.match(pattern,s) :
print('matched')
else :
print('not matched')
You can try re.findall():
import re
correct_string = 'test-251'
if len(re.findall("test-\d+", correct_string)) > 0:
print "Match found"
A pattern such as \btest-\d+\b should do you;
matches = re.search(r'\btest-\d+\', search_string)
Demo
This requires the matching of word boundaries, so prevents other substrings from occuring after your desired match.
I am trying to write code that will take a string and remove specific data from it. I know that the data will look like the line below, and I only need the data within the " " marks, not the marks themselves.
inputString = 'type="NN" span="123..145" confidence="1.0" '
Is there a way to take a Substring of a string within two characters to know the start and stop points?
You can extract all the text between pairs of " characters using regular expressions:
import re
inputString='type="NN" span="123..145" confidence="1.0" '
pat=re.compile('"([^"]*)"')
while True:
mat=pat.search(inputString)
if mat is None:
break
strings.append(mat.group(1))
inputString=inputString[mat.end():]
print strings
or, easier:
import re
inputString='type="NN" span="123..145" confidence="1.0" '
strings=re.findall('"([^"]*)"', inputString)
print strings
Output for both versions:
['NN', '123..145', '1.0']
fields = inputString.split('"')
print fields[1], fields[3], fields[5]
You could split the string at each space to get a list of 'key="value"' substrings and then use regular expressions to parse the substrings.
Using your input string:
>>> input_string = 'type="NN" span="123..145" confidence="1.0" '
>>> input_string_split = input_string.split()
>>> print input_string_split
[ 'type="NN"', 'span="123..145"', 'confidence="1.0"' ]
Then use regular expressions:
>>> import re
>>> pattern = r'"([^"]+)"'
>>> for substring in input_string_split:
match_obj = search(pattern, substring)
print match_obj.group(1)
NN
123..145
1.0
The regular expression '"([^"]+)"' matches anything within quotation marks (provided there is at least one character). The round brackets indicate the bit of the regular expression that you are interested in.