How to match literal backslashes read from a text file using regular expressions?

[cricfan]

I'm parsing a text file to extract word definitions. For example the
input text file contains the following content:

di.va.gate \'di_--v*-.ga_-t\ vb
pas.sim \'pas-*m\ adv : here and there : THROUGHOUT

I am trying to obtain words between two literal backslashes (\ .. \). I
am not able to match words between two literal backslashes using the
regxp - re.compile(r'\\[^\\]*\\').

Here is my sample script:

import re;

#slashPattern = re.compile(re.e scape(r'\\[^\\]*\\'));
pattern = r'\\[^\\]*\\'
slashPattern = re.compile(patt ern);

fdr = file( "parseinput",'r ');
line = fdr.readline();

while (line != ""):
if (slashPattern.m atch(line)):
print line.rstrip() + " <-- matches pattern " + pattern
else:
print line.rstrip() + " <-- DOES not match pattern " +
pattern
line = fdr.readline();
print;
----------
The output

C:\home\krishna \lang\python>py thon wsparsetest.py
python wsparsetest.py
di.va.gate \'di_--v*-.ga_-t\ vb <-- DOES not match
pattern \\[^\\]*\\
pas.sim \'pas-*m\ adv : here and there : THROUGHOUT <-- DOES not match
pattern \\[^\\]*\\
-----------

What should I be doing to match those literal backslashes?

Thanks

[John Machin]

cr*****@gmail.c om wrote:

I'm parsing a text file to extract word definitions. For example the
input text file contains the following content:

di.va.gate \'di_--v*-.ga_-t\ vb
pas.sim \'pas-*m\ adv : here and there : THROUGHOUT

I am trying to obtain words between two literal backslashes (\ .. \). I
am not able to match words between two literal backslashes using the
regxp - re.compile(r'\\[^\\]*\\').

Here is my sample script:

import re;
Lose the semicolons ...

#slashPattern = re.compile(re.e scape(r'\\[^\\]*\\'));
pattern = r'\\[^\\]*\\'
slashPattern = re.compile(patt ern);

fdr = file( "parseinput",'r ');
line = fdr.readline();

You should upgrade so that you have a modern Python and a modern
tutor[ial] -- then you will be writing:

for line in fdr:
do_something_wi th(line)

while (line != ""):
Lose the extraneous parentheses ...
if (slashPattern.m atch(line)):
Your main problem is that you should be using the search() method, not
the match() method. Read the section on this topic in the re docs!!

import re
pat = re.compile(r'\\[^\\]*\\')
pat.match(r'abc d \xyz\ pqr')
pat.search(r'ab cd \xyz\ pqr')

<_sre.SRE_Mat ch object at 0x00AE8988>

print line.rstrip() + " <-- matches pattern " + pattern
else:
print line.rstrip() + " <-- DOES not match pattern " +
pattern
line = fdr.readline();
print;
----------
The output

C:\home\krishna \lang\python>py thon wsparsetest.py
python wsparsetest.py
di.va.gate \'di_--v*-.ga_-t\ vb <-- DOES not match
pattern \\[^\\]*\\
pas.sim \'pas-*m\ adv : here and there : THROUGHOUT <-- DOES not match
pattern \\[^\\]*\\
-----------

What should I be doing to match those literal backslashes?

Thanks

[George Sakkis]

This should give you an idea of how to go about it (needs python 2.3 or
newer):
import re
slashPattern = re.compile(r'\\ (.*?)\\')

for i,line in enumerate(file( "parseinput ")):
print "line", i+1,
match = slashPattern.se arch(line)
if match:
print "matched:", match.group(1)
else:
print "did not match"

#===== output =============== ========

line 1 matched: 'di_--v*-.ga_-t
line 2 matched: 'pas-*m

#============== =============== =======
George