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I have a simple regex to grab a string, It works fine with words like HelloWorld without space and how can i grab a words with space or more than 1 word like Hello World
Text File
FAN PS-2 is NOT PRESENT #like this 'NOT PRESENT'
Regex
Value FAN_PS (\S*) # regex
Start
^FAN PS is ${FAN_PS_1}
What should I change in my regex so I can grab more than 1 word?
Thank you.
You have to change your regular expressions. The current expressions \S* match a string of non-white-space characters, so it is normal that you only get the first word.
If you change \S* to [\S ]*, you will get multiple words. You can even change it to the simpler .* if you do not care about certain characters.
Read the python regex reference for information on different character classes.
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I have a text containing a URL that needs to be reworked.
text='dfs:/?url=https://myserver/c12&ofg={"tes":{"id":1812}}'
I need to replace programmatically the id value (in this example 1812, which is unknown before the execution) with a fixed substring (e.g. 189). So the end result must be
'dfs:/?url=https://myserver/c12&ofg={"tes":{"id":189}}'
As I'm programming in Python, I guess that I should use the regular expression (module re) to automatically replace that value between "id": and }} but I couldn't find one that works for this use case.
I assume you are always generating the same URL with that pattern, and the value to 'change' is always in {"id":X}. One way to solve this particular problem is with a positive lookbehind + re.sub replacement.
import re
pattern = re.compile(r"(?<=\"id\":)\d+")
string = "dfs:/?url=https://myserver/c12&ofg={\"tes\":{\"id\":1812}}"
print(pattern.sub("desired_value", string))
Generated output will contain desired_value in place of the 1812. A good explanation of what is happening is done in regex101 but a quick rep of what is happening in the pattern:
Matches any digit one or more times ONLY if behind has "id":, without consuming characters
what about simply splitting the string twice? eg.
my_string = 'dfs:/?url=https://myserver/c12&ofg={"tes":{"id":1812}}'
substring = my_string.split('"id":',1)[1]
substring = substring.split('}}')[0]
print(my_string.replace(substring, "189"))
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Having a regex pattern which can be found several times in a text, how would one chooses and differs between the matches in this case?
choose the first match.
ignore the peripherals.
only one match using index.
Example:
regex_pattern = r"(AB.+?AbB)|(CD.+?CdD)|(EF.+?EfF)"
text = "for finding several CDadjacent CDtagsCdDCdD in a
text this is an ABexampleAbB text"
the first match is CDadjacent CDtagsCdD.
while one might wants to match both:
CDadjacent CDtagsCdDCdD
and. . . . . . . CDtagsCdD
Found the answer faster than I thought.
using '(',')' (round brackets) lets you use parts of the regex as one unit and by adding the '+; (plus sign) you can look for more than one occurrences of a pattern.
for this example this pattern will solve the issue:
r"(AB.+?(AbB)+)|(CD.+?(CdD)+)|(EF.+?(EfF)+)"
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I have the following strings.
string1 = "按照由 GPV 提供的相关报告; 世界卫生组织 WHO 发布的有关研究"
string2 = "\n\n 介绍 INTRODUCTION"
How can I remove the spaces between Chinese characters and English acronyms?
The expected result is:
"按照由GPV提供的相关报告; 世界卫生组织WHO发布的有关研究".
However, the re pattern should not remove the space between 介绍 and INTRODUCTION since there are no Chinese characters on the right side of INTRODUCTION.
If you can use the third-party regex implementation module regex, it supports \p{script} tokens which make this task easy :
\p{Han}+\s+\p{Latin}+\s+\p{Han}+
Python native re's unfortunately doesn't support these.
In order to remove the spaces, use capturing groups to select the surrounding words and refer to those in your replacement pattern :
Match (\p{Han}+)\s+(\p{Latin}+)\s+(\p{Han}+)
Replace by \1\2\3
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Is it possible to remove periods from the middle of a string (sentence), leaving the ending period?
The answers that I have seen, basically strip all of the periods.
Remove periods at the end of sentences in python
If I understand correctly, this should do what you want:
import re
string = 'You can. use this to .remove .extra dots.'
string = re.sub('\.(?!$)', '', string)
It uses regex to replace all dots, except if the dot is at the end of the string. (?!$) is a negative lookahead, so the regex looks for any dot not directly followed by $ (end of line).
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I need to write a regex matching pattern code to either return true if there is one '+' between two words and nothing else. I have written the code to check if there is only one '+' in the string but how will I check it is between two words?
The code is below:
import re
inputStr= "ali+ahmedafaw+"
inputStr2= "hello+world+again"
plus=re.findall(r'[+]', inputStr)
print (plus)
l_plus=len(plus)
print "The length is ",l_plus
if l_plus<=1:
print "True"
else:
print "False"
Actually it depends on what you mean by word. If you mean a word with more than one character, you can simply use [a-zA-Z]+ around the + character. Or other patterns which will match different characters like \w to match word characters.
re.search(r'[a-zA-Z]+\+[a-zA-Z]+', input_str)
But if you just want it doesn't appears at the leading and trailing of your text you can use negative look-around:
re.search(r'(?<!^)\+(?!$)', input_str)