Python re "bogus escape error" - python

I've been messing around with the python re modules .search method. cur is the input from a Tkinter entry widget. Whenever I enter a "\" into the entry widget, it throws this error. I'm not all to sure what the error is or how to deal with it. Any insight would be much appreciated.
cur is a string
tup[0] is also a string
Snippet:
se = re.search(cur, tup[0], flags=re.IGNORECASE)
The error:
Exception in Tkinter callback
Traceback (most recent call last):
File "C:\Python26\Lib\Tkinter.py", line 1410, in __call__
return self.func(*args)
File "C:\Python26\Suite\quidgets7.py", line 2874, in quick_links_results
self.quick_links_results_s()
File "C:\Python26\Suite\quidgets7.py", line 2893, in quick_links_results_s
se = re.search(cur, tup[0], flags=re.IGNORECASE)
File "C:\Python26\Lib\re.py", line 142, in search
return _compile(pattern, flags).search(string)
File "C:\Python26\Lib\re.py", line 245, in _compile
raise error, v # invalid expression
error: bogus escape (end of line)

"bogus escape (end of line)" means that your pattern ends with a backslash. This has nothing to do with Tkinter. You can duplicate the error pretty easily in an interactive shell:
>>> import re
>>> pattern="foobar\\"
>>> re.search(pattern, "foobar")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/re.py", line 142, in search
return _compile(pattern, flags).search(string)
File "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/re.py", line 241, in _compile
raise error, v # invalid expression
sre_constants.error: bogus escape (end of line)
The solution? Make sure your pattern doesn't end with a single backslash.

The solution to this issue is to use a raw string as the replacement text. The following won't work:
re.sub('this', 'This \\', 'this is a text')
It will throw the error: bogus escape (end of line)
But the following will work just fine:
re.sub('this', r'This \\', 'this is a text')
Now, the question is how do you convert a string generated during program runtime into a raw string in Python. You can find a solution for this here. But I prefer using a simpler method to do this:
def raw_string(s):
if isinstance(s, str):
s = s.encode('string-escape')
elif isinstance(s, unicode):
s = s.encode('unicode-escape')
return s
The above method can convert only ascii and unicode strings into raw strings. Well, this has been working great for me till date :)

If you are trying to search for "cur" in "tup[0]" you should do this through "try:... except:..." block to catch invalid pattern:
try :
se = re.search(cur, tup[0], flags=re.IGNORECASE)
except re.error, e:
# print to stdout or any status widget in your gui
print "Your search pattern is not valid."
# Some details for error:
print e
# Or some other code for default action.

The first parameter to re is the pattern to search for, thus if 'cur' contains a backslash at the end of the line, it'll be an invalid escape sequence. You've probably swapped your arguments around (I don't know what tup[0] is, but is it your pattern?) and it should be like this
se = re.search(tup[0], cur, flags=re.IGNORECASE)
As you very rarely use user input as a pattern (unless you're doing a regular expression search mechanism, in which case you might want to show the error instead).
HTH.
EDIT:
The error it is reporting is that you're using an escape character before the end of line (which is what bogus escape (end of line) means), that is your pattern ends with a backslash, which is not a valid pattern. Escape character (backslash) must be followed by another character, which removes or adds special meaning to that character (not sure exactly how python does it, posix makes groups by adding escape to parentheses, perl removes the group effect by escaping it). That is \* matches a literal asterix, whereas * matches the preceding character 0 or more times.

Related

How to fix bad escape regex error (python re)

I've been messing around with re.sub() to see how I would change the format from Y-m-d to M/d/y. To perform the test, I defined the starting variable: current_date = "2012-05-26"
I would try to achieve to convert that date to 05/26/2012.
I tried to achieve this without using DateTime but with regex. I used re.sub as below:
formatted_date = re.sub(r"\d{2,4}-\d{1,2}-\d{1,2}", r"[^a-zA-Z]\d{1,2}/\d{1,2}/\d{2,4}", current_date)
The first regex is to match the original format of Y-M-D and the second Regex is to try to convert it to the format that I want it to be. I got the following error:
Traceback (most recent call last):
File "C:\Users\ghub4\AppData\Local\Programs\Python\Python39\lib\sre_parse.py", line 1039, in parse_template
this = chr(ESCAPES[this][1])
KeyError: '\\d'
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "c:\Users\ghub4\OneDrive\Desktop\test_sub.py", line 5, in <module>
formatted_date = re.sub(r"\d{2,4}-\d{1,2}-\d{1,2}", r"[^a-zA-Z]\d{1,2}/\d{1,2}/\d{2,4}", current_date)
File "C:\Users\ghub4\AppData\Local\Programs\Python\Python39\lib\re.py", line 210, in sub
return _compile(pattern, flags).sub(repl, string, count)
File "C:\Users\ghub4\AppData\Local\Programs\Python\Python39\lib\re.py", line 327, in _subx
template = _compile_repl(template, pattern)
File "C:\Users\ghub4\AppData\Local\Programs\Python\Python39\lib\re.py", line 318, in _compile_repl
return sre_parse.parse_template(repl, pattern)
File "C:\Users\ghub4\AppData\Local\Programs\Python\Python39\lib\sre_parse.py", line 1042, in parse_template
raise s.error('bad escape %s' % this, len(this))
re.error: bad escape \d at position 9
Full Code:
import re
current_date = "2012-05-26"
formatted_date = re.sub(r"\d{2,4}-\d{1,2}-\d{1,2}", r"[^a-zA-Z]\d{1,2}/\d{1,2}/\d{2,4}", current_date)
print(formatted_date)
I've traced the error to potential the second regex but I'm unsure where position 9 is and how to fix the error. Another reason why I'm not sure how to fix it is due to the first error where it stated a keyerror raised by \\d. I'm sure that when the regex is interpret somewhere in the code, it is taking the \d as \\d instead which Im also not sure how to prevent that. I'm also pretty sure that the second regex may backfire on me and I am working on a solution on that after this question is posted. How would I be able to correct these errors?
The replacement string for a regex is not a regex in itself, rather it is a string which may contain references to groups captured by the original regex. In your case, you want to capture the year, month and day and then output them in the result string. You do that with () around the values you want to capture, and then refer to the groups by \1, \2, and \3 in the replacement string, with the numbers being assigned in order of the groups being captured. So for your code, you want:
formatted_date = re.sub(r"(\d{2,4})-(\d{1,2})-(\d{1,2})", r"\2/\3/\1", current_date)
Try and group your digits (If you goal is testing then position 9 is your first \d in your second regex-check - It is an invalid group reference):
formatted_date = re.sub(r"(\d{2,4})-(\d{1,2})-(\d{1,2})",r"\2/\3/\1",current_date)

What is the RegEx pattern for 24-06-2015 10:15:45: Aditya Krishnakant:?

What is the RegEx pattern for 24-06-2015 10:15:45: Aditya Krishnakant:
If you look at the whatsapp chat transcript, it looks like a mess. The purpose of this code is to print messages sent by a person in a new line (for better readability). This is my code
import re
f = open("wa_chat.txt", "r")
match = re.findall(r'(\d{2})\:(\d{2})\:(\d{4})\s(\d{2})\:(\d{2})\:(\d{2})\:\s(\w)\s(\w)\:', f)
for content in match:
print(f.readlines(), '\n')
f.close()
I am getting the following error message:
Traceback (most recent call last):
File "whatsapp.py", line 4, in <module>
match = re.findall(r'(\d{2})\:(\d{2})\:(\d{4})\s(\d{2})\:(\d{2})\:(\d{2})\:\s(\w)\s(\w)\:', f)
File "/usr/lib/python2.7/re.py", line 177, in findall
return_compile(pattern, flags).findall(string)
TypeError: expected string or buffer
Where am I going wrong?
For some reason you're putting \: where - should be. Also, instead of \s you can be more specific and just use a space. You can be more specific with those kinds of things because you know exactly what the format is. Your other big problem is that you're only using \w, which only matches one alphanumeric character, when you should use \w+, matching the whole word. Lastly, your actual error is coming from the fact that you're passing in a file object instead of the string containing its contents, i.e. f.read(). Here's some code that should work:
import re
f = open("wa_chat.txt", 'r')
match = re.findall(r'(\d{2})-(\d{2})-(\d{4}) (\d{2}):(\d{2}):(\d{2}): (\w+) (\w+):', f.read())
print match #or do whatever you want with it
Note that match will be a list of tuples since you wanted to use grouping.

pyPEG2 parsing of newlines

I'm trying to use pyPEG2 to translate MoinMoin markup to Markdown, and I need to pay attention to newlines in certain cases. However, I can't even get my newline parsing tests to work. I'm new to pyPEG and my Python is rusty. Please bear with me.
Here's the code:
#!/usr/local/bin/python3
from pypeg2 import *
import re
class Newline(List):
grammar = re.compile(r'\n')
parse("\n", Newline)
parse("""
""", Newline)
This results in:
Traceback (most recent call last):
File "./pyPegNewlineTest.py", line 7, in <module>
parse("\n", Newline)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/pypeg2/__init__.py", line 667, in parse
t, r = parser.parse(text, thing)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/pypeg2/__init__.py", line 794, in parse
raise r
File "<string>", line 2
^
SyntaxError: expecting match on \n
It's as if pypeg is inserting an empty line after the \n.
Trying other options such as
grammar = re.compile(r'\n', re.MULTILINE)
grammar = re.compile(r'\r\n|\r|\n', re.MULTILINE)
grammar = contiguous(re.compile(r'\r\n|\r|\n', re.MULTILINE))
and various combinations of those don't change the error message (although I don't think I tried all combinations). Changing Newline to subclass str instead of List doesn't change the error either.
Update
I have figured out that pypeg is stripping the newline before parsing it:
#!/usr/local/bin/python3
from pypeg2 import *
import re
class Newline(str):
grammar = contiguous(re.compile(r'a'))
parse("\na", Newline)
parse("""
a""", Newline)
print("Success, of a sort.")
Running this results in:
Success, of a sort.
If I override the Newline's parse method I don't even see the newline. The first thing it gets is the "a". This is consistent with what I'm seeing elsewhere. pypeg strips all leading whitespace, even when you specify contiguous.
So, that's what's happening. Not sure what to do about it.
Yes by default pypeg remove the whitespaces including the newlines.
This is easly configurable by setting the optional whitespace argument in the parse() function, e.g. in:
parse("\na", Newline, whitespace=re.compile(r"[ \t\r]"))
Doing so spaces and tabs will still be skipped, but not newlines \n.
With this example the parser now correctly find the syntax error:
SyntaxError: expecting match on a

Doctest Involving Escape Characters

Have a function fix(), as a helper function to an output function which writes strings to a text file.
def fix(line):
"""
returns the corrected line, with all apostrophes prefixed by an escape character
>>> fix('DOUG\'S')
'DOUG\\\'S'
"""
if '\'' in line:
return line.replace('\'', '\\\'')
return line
Turning on doctests, I get the following error:
Failed example:
fix('DOUG'S')
Exception raised:
Traceback (most recent call last):
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/doctest.py", line 1254, in __run
compileflags, 1) in test.globs
File "<doctest convert.fix[0]>", line 1
fix('DOUG'S')
^
No matter what combination of \ and 's I use, the doctest doesn't seem to to want to work, even though the function itself works perfectly. Have a suspicion that it is a result of the doctest being in a block comment, but any tips to resolve this.
Is this what you want?:
def fix(line):
r"""
returns the corrected line, with all apostrophes prefixed by an escape character
>>> fix("DOUG\'S")
"DOUG\\'S"
>>> fix("DOUG'S") == r"DOUG\'S"
True
>>> fix("DOUG'S")
"DOUG\\'S"
"""
return line.replace("'", r"\'")
import doctest
doctest.testmod()
raw strings are your friend...
First, this is what happens if you actually call your function in the interactive interpreter:
>>> fix("Doug's")
"Doug\\'s"
Note that you don't need to escape single quotes in double-quoted strings, and that Python does not do this in the representation of the resulting string – only the back slash gets escaped.
This means the correct docstring should be (untested!)
"""
returns the corrected line, with all apostrophes prefixed by an escape character
>>> fix("DOUG'S")
"DOUG\\\\'S"
"""
I'd use a raw string literal for this docstring to make this more readable:
r"""
returns the corrected line, with all apostrophes prefixed by an escape character
>>> fix("DOUG'S")
"DOUG\\'S"
"""

Substitute parenthesis for their regular expression

I'm trying to copy a file,
>>> originalFile = '/Users/alvinspivey/Documents/workspace/Image_PCA/spectra_text/HIS/jean paul test 1 - Copy (2)/bean-1-aa.txt'
>>> copyFile = os.system('cp '+originalFile+' '+NewTmpFile)
But must first replace the spaces and parenthesis before the open function will work:
/Users/alvinspivey/Documents/workspace/Image_PCA/spectra_text/HIS/jean\ paul\ test\ 1\ -\ Copy\ \(2\)/bean-1-aa.txt
spaces ' ' --> '\ '
parenthesis '(' --> '\(' etc.
Replacing the spaces work:
>>> originalFile = re.sub(r'\s',r'\ ', os.path.join(root,file))
but parenthesis return an error:
>>> originalFile = re.sub(r'(',r'\(', originalFile)
Traceback (most recent call last):
File "", line 1, in
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/re.py", line 151, in sub
return _compile(pattern, flags).sub(repl, string, count)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/re.py", line 244, in _compile
raise error, v # invalid expression
sre_constants.error: unbalanced parenthesis
Am I replacing parenthesis correctly?
Also, when using re.escape() for this, the file is not returned correctly. So it is not an alternative.
( has special meaning in regular expressions (grouping), you have to escape it:
originalFile = re.sub(r'\(',r'\(', originalFile)
or, since you don't use regex features for the replacement:
originalFile = re.sub(r'\(','\(', originalFile)
The regular expression r'(' is translated as start a capturing group. Which is why Python is complaining.
If all you are doing is replacing spaces and parenthesis then maybe just string.replace will do ?
Alternatively, if you avoid calling a shell (os.system) to do the copy, you don't need to worry about escaping spaces and other special characters,
import shutil
originalFile = '/Users/alvinspivey/Documents/workspace/Image_PCA/spectra_text/HIS/jean paul test 1 - Copy (2)/bean-1-aa.txt'
newTmpFile = '/whatever.txt'
shutil.copy(originalFile, newTmpFile)
Use shutil.copy to copy files, rather than calling the system.
Use subprocess rather than os.system - it avoids calling into the shell, so doesn't need the quoting.

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