I have list of demangled-function names like _Z6__comp7StudentS_
_Z4SortiSt6vectorI7StudentSaIS0_EE. I read wiki and found out that it follows some sort of defined structure. _Z is mangled Symbol followed by a number and then the function name of that length.
So I wanted to retrieve that function name using regex. I only come close to _Z(?:\d)(?<function_name>[a-z_A-Z]){\1}. But referring \1 won't work because its string, right? Is there a single regex pattern solution to this.
You can use 2 capture groups, and get the part of the string using the position of capture group 2
import re
pattern = r"_Z(\d+)([a-z_A-Z]+)"
s = "_Z4SortiSt6vectorI7StudentSaIS0_EE"
m = re.search(pattern, s)
if m:
print(m.group(2)[0: int(m.group(1))])
Output
Sort
Using _Z6__comp7StudentS_ will return __comp
Related
How can I check whether regex pattern contains a named capturing group? I want to decide whether to use re.findall or re.finditer based on the form of the regex.
Use the following approach:
pat = '.*(?P<word>\w+\d+\b).+' # sample pattern
has_named_group = bool(re.search(r'\(\?P<\w+>[^)]+\)', pat))
This can also be a function:
def has_named_group(pat):
return bool(re.search(r'\(\?P<\w+>[^)]+\)', pat))
You can use Pattern.groupindex
A dictionary mapping any symbolic group names defined by (?P) to
group numbers. The dictionary is empty if no symbolic groups were used
in the pattern.
For example
import re
pattern = re.compile('(?P<mygroup>.*)')
if pattern.groupindex:
print("The pattern contains a named capturing group")
else:
print("The pattern does not contain a named capturing group")
Output
The pattern contains a named capturing group
I have a string, I have to get digits only from that string.
url = "www.mylocalurl.com/edit/1987"
Now from that string, I need to get 1987 only.
I have been trying this approach,
id = [int(i) for i in url.split() if i.isdigit()]
But I am getting [] list only.
You can use regex and get the digit alone in the list.
import re
url = "www.mylocalurl.com/edit/1987"
digit = re.findall(r'\d+', url)
output:
['1987']
Replace all non-digits with blank (effectively "deleting" them):
import re
num = re.sub('\D', '', url)
See live demo.
You aren't getting anything because by default the .split() method splits a sentence up where there are spaces. Since you are trying to split a hyperlink that has no spaces, it is not splitting anything up. What you can do is called a capture using regex. For example:
import re
url = "www.mylocalurl.com/edit/1987"
regex = r'(\d+)'
numbers = re.search(regex, url)
captured = numbers.groups()[0]
If you do not what what regular expressions are, the code is basically saying. Using the regex string defined as r'(\d+)' which basically means capture any digits, search through the url. Then in the captured we have the first captured group which is 1987.
If you don't want to use this, then you can use your .split() method but this time provide a split using / as the separator. For example `url.split('/').
I am having a string as follows:
A5697[2:10] = {ravi, rageev, raghav, smith};
I want the content after "A5697[2:10] =". So, my output should be:
{ravi, rageev, raghav, smith};
This is my code:
print(re.search(r'(?<=A\d+\[.*\] =\s).*', line).group())
But, this is giving error:
sre_constants.error: look-behind requires fixed-width pattern
Can anyone help to solve this issue? I would prefer to use regex.
You can try re.sub , like below, Since you have given only one data point. I am assuming all the other data points are following the similar pattern.
import re
text = "A5697[2:10] = {ravi, rageev, raghav, smith}"
re.sub(r'(A\d+\[\d+:\d+\]\s+=\s+)(.+)', r'\2', text)
returns,
'{ravi, rageev, raghav, smith}'
re.sub : substitutes the entire match as given as regex with the 2nd capturing group. The second capturing group captures every thing after '= '.
Simply replace the bits you don't want:
print re.sub(r'A\d[^=]*= *','',line)
See demo here: https://rextester.com/NSG17655
I am a newbee in python. I am trying to pull data (XXXX) out from a text with a pattern PDB:XXXX. The XXXX varies, but it is exactly what I want.
Since the data all contain PDB:, I use re.findall() to search and get this pattern. But this only gave me a list of PDB:. How can I get it to include the XXXX???
this is my code:
text = 'blah...........
PDB:AAAA
blah...........
blah...........
PDB:BBBB'
etc.
r = re.findall("PDB:",text)
and the output gave me:
['PDB:', 'PDB:']
My desired output should be something like
['AAAA', 'BBBB']
You need to use """ to quote multi-line strings in Python. Also, to get a specific subset of the matched pattern, you need to use capture groups (the parentheses in my regular expression below).
import re
text = """blah...........
PDB:AAAA
blah...........
blah...........
PDB:BBBB"""
results = re.findall(r"PDB:(.*)", text)
print results #['AAAA', 'BBBB']
I am trying to create a function which will take 2 parameters. A word with wildcards in it like "*arn*val" and a file name containing a dictionary. It returns a list of all words that match the word like ["carnival"].
My code works fine for anything with only one "*" in it, however any more and I'm stumped as to how to do it.
Just searching for the wildcard string in the file was returning nothing.
Here is my code:
dictionary_file = open(dictionary_filename, 'r')
dictionary = dictionary_file.read()
dictionary_file.close()
dictionary = dictionary.split()
alphabet = ["a","b","c","d","e","f","g","h","i",
"j","k","l","m","n","o","p","q","r",
"s","t","u","v","w","x","y","z"]
new_list = []
for letter in alphabet:
if wildcard.replace("*", letter) in dictionary:
new_list += [wildcard.replace("*", letter)]
return new_list
The parameters parameters: First is the wildcard string (wildcard), and second is the dictionary file name (dictionary_filename).
Most answers on this site were about Regex, which I have no knowledge of.
Your particular error is that .replace replaces all occurrences e.g., "*arn*val" -> "CarnCval" or "IarnIval". You want different letters here. You could use the second nested loop over the alphabet (or use itertools.product() to generate all possible letter pairs) to fix it but a simpler way is to use regular expressions:
import re
# each `*` corresponds to an ascii lowercase letter
pattern = re.escape(wildcard).replace("\\*", "[a-z]")
matches = list(filter(re.compile(pattern+"$").match, known_words))
Note: it doesn't support escaping * in the wildcard.
If input wildcards are file patterns then you could use fnmatch module to filter words:
import fnmatch
matches = fnmatch.filter(known_words, wildcard)