Checking if string inside quotes?

[Michael Yanowitz]

Hello:

If I have a long string (such as a Python file).
I search for a sub-string in that string and find it.
Is there a way to determine if that found sub-string is
inside single-quotes or double-quotes or not inside any quotes?
If so how?

Thanks in advance:
Michael Yanowitz

 

[half.italian]

Hello:

If I have a long string (such as a Python file).
I search for a sub-string in that string and find it.
Is there a way to determine if that found sub-string is
inside single-quotes or double-quotes or not inside any quotes?
If so how?

Thanks in advance:
Michael Yanowitz

I think the .find() method returns the index of the found string. You
could check one char before and then one char after the length of the
string to see. I don't use regular expressions much, but I'm sure
that's a more elegant approach.

This will work. You'll get in index error if you find the string at
the very end of the file.

s = """
foo
"bar"
"""
findme = "foo"
index = s.find(findme)

if s[index-1] == "'" and s[index+len(findme)] == "'":
print "single quoted"
elif s[index-1] == "\"" and s[index+len(findme)] == "\"":
print "double quoted"
else:
print "unquoted"

~Sean

 

[castironpi]

Thanks, but it is a little more complicated than that,
the string could be deep in quotes.

The problem is in string substitution.
Suppose I have a dictionary with MY_IP : "172.18.51.33"

I need to replace all instances of MY_IP with "172.18.51.33"
in the file.
It is easy in cases such as:
if (MY_IP == "127.0.0.1"):

But suppose I encounter:"
("(size==23) and (MY_IP==127.0.0.1)")

In this case I do not want:
("(size==23) and ("172.18.51.33"==127.0.0.1)")
but:
("(size==23) and (172.18.51.33==127.0.0.1)")
without the internal quotes.
How can I do this?
I presumed that I would have to check to see if the string
was already in quotes and if so remove the quotes. But not
sure how to do that?
Or is there an easier way?

Thanks in advance:
Michael Yanowitz

-----Original Message-----
From: [email protected]

[mailto[email protected]]On Behalf
Of (e-mail address removed)
Sent: Wednesday, May 09, 2007 5:12 PM
To: (e-mail address removed)
Subject: Re: Checking if string inside quotes?

Hello:

If I have a long string (such as a Python file).
I search for a sub-string in that string and find it.
Is there a way to determine if that found sub-string is
inside single-quotes or double-quotes or not inside any quotes?
If so how?

Thanks in advance:
Michael Yanowitz

I think the .find() method returns the index of the found string. You
could check one char before and then one char after the length of the
string to see. I don't use regular expressions much, but I'm sure
that's a more elegant approach.

This will work. You'll get in index error if you find the string at
the very end of the file.

s = """
foo
"bar"
"""
findme = "foo"
index = s.find(findme)

if s[index-1] == "'" and s[index+len(findme)] == "'":
print "single quoted"
elif s[index-1] == "\"" and s[index+len(findme)] == "\"":
print "double quoted"
else:
print "unquoted"

~Sean

--http://mail.python.org/mailman/listinfo/python-list

In "nearby" quotes or in quotes at all?
import re
a='abc"def"ghijk'
b=re.sub( r'([\'"])[^\1]*\1', '', a )
b.replace( 'ghi', 'the string' )
#fb: 'abcthe stringjk'
edit()

Here, you get the entire file -in b-, strings omitted entirely, so you
can't write it back.

I've used `tokenize' to parse a file, but you don't get precisely your
original back. Untokenize rearrages your spacings. Equivalent
semantically, so if you want to compile immedately afterwords, you're
alright with that. Short example:
from tokenize import *
import token
from StringIO import StringIO
a= StringIO( 'abc "defghi" ghi jk' )
from collections import deque
b= deque()
for g in generate_tokens( a.readline ):
if g[0]== token.NAME and g[1]== 'ghi':
b.append( ( token.STRING, '"uchoose"' ) )
else:
b.append( g )

untokenize( b )
#fb: 'abc "defghi""uchoose"jk '
edit()
acb

 

[half.italian]

Thanks, but it is a little more complicated than that,
the string could be deep in quotes.

The problem is in string substitution.
Suppose I have a dictionary with MY_IP : "172.18.51.33"

I need to replace all instances of MY_IP with "172.18.51.33"
in the file.
It is easy in cases such as:
if (MY_IP == "127.0.0.1"):

But suppose I encounter:"
("(size==23) and (MY_IP==127.0.0.1)")

In this case I do not want:
("(size==23) and ("172.18.51.33"==127.0.0.1)")
but:
("(size==23) and (172.18.51.33==127.0.0.1)")
without the internal quotes.
How can I do this?
I presumed that I would have to check to see if the string
was already in quotes and if so remove the quotes. But not
sure how to do that?
Or is there an easier way?

Thanks in advance:
Michael Yanowitz

-----Original Message-----
From: [email protected]

[mailto[email protected]]On Behalf
Of (e-mail address removed)
Sent: Wednesday, May 09, 2007 5:12 PM
To: (e-mail address removed)
Subject: Re: Checking if string inside quotes?

Hello:

If I have a long string (such as a Python file).
I search for a sub-string in that string and find it.
Is there a way to determine if that found sub-string is
inside single-quotes or double-quotes or not inside any quotes?
If so how?

Thanks in advance:
Michael Yanowitz

I think the .find() method returns the index of the found string. You
could check one char before and then one char after the length of the
string to see. I don't use regular expressions much, but I'm sure
that's a more elegant approach.

This will work. You'll get in index error if you find the string at
the very end of the file.

s = """
foo
"bar"
"""
findme = "foo"
index = s.find(findme)

if s[index-1] == "'" and s[index+len(findme)] == "'":
print "single quoted"
elif s[index-1] == "\"" and s[index+len(findme)] == "\"":
print "double quoted"
else:
print "unquoted"

~Sean

--http://mail.python.org/mailman/listinfo/python-list

In that case I suppose you'd have to read the file line by line and if
you find your string in the line then search for the indexes of any
matching quotes. If you find matching quotes, see if your word lies
within any of the quote indexes.

#!/usr/bin/env python

file = open("file", 'r')
findme= "foo"
for j, line in enumerate(file):
found = line.find(findme)
if found != -1:
quotecount = line.count("'")
quoteindexes = []
start = 0
for i in xrange(quotecount):
i = line.find("'", start)
quoteindexes.append(i)
start = i+1

f = False
for i in xrange(len(quoteindexes)/2):
if findme in
line[quoteindexes.pop(0):quoteindexes.pop(0)]:
f = True
print "Found %s on line %s: Single-Quoted" % (findme, j
+1)
if not f:
print "Found %s on line %s: Not quoted" % (findme, j+1)


It's not pretty but it works.

~Sean

 

[castironpi]

Thanks, but it is a little more complicated than that,
the string could be deep in quotes.

The problem is in string substitution.
Suppose I have a dictionary with MY_IP : "172.18.51.33"

I need to replace all instances of MY_IP with "172.18.51.33"
in the file.
It is easy in cases such as:
if (MY_IP == "127.0.0.1"):

But suppose I encounter:"
("(size==23) and (MY_IP==127.0.0.1)")

In this case I do not want:
("(size==23) and ("172.18.51.33"==127.0.0.1)")
but:
("(size==23) and (172.18.51.33==127.0.0.1)")
without the internal quotes.
How can I do this?
I presumed that I would have to check to see if the string
was already in quotes and if so remove the quotes. But not
sure how to do that?
Or is there an easier way?

Thanks in advance:
Michael Yanowitz

-----Original Message-----
From: [email protected]

[mailto[email protected]]On Behalf
Of (e-mail address removed)
Sent: Wednesday, May 09, 2007 5:12 PM
To: (e-mail address removed)
Subject: Re: Checking if string inside quotes?

I think the .find() method returns the index of the found string. You
could check one char before and then one char after the length of the
string to see. I don't use regular expressions much, but I'm sure
that's a more elegant approach.

This will work. You'll get in index error if you find the string at
the very end of the file.

s = """
foo
"bar"
"""
findme = "foo"
index = s.find(findme)

if s[index-1] == "'" and s[index+len(findme)] == "'":
print "single quoted"
elif s[index-1] == "\"" and s[index+len(findme)] == "\"":
print "double quoted"
else:
print "unquoted"

~Sean

--http://mail.python.org/mailman/listinfo/python-list

In that case I suppose you'd have to read the file line by line and if
you find your string in the line then search for the indexes of any
matching quotes. If you find matching quotes, see if your word lies
within any of the quote indexes.

#!/usr/bin/env python

file = open("file", 'r')
findme= "foo"
for j, line in enumerate(file):
found = line.find(findme)
if found != -1:
quotecount = line.count("'")
quoteindexes = []
start = 0
for i in xrange(quotecount):
i = line.find("'", start)
quoteindexes.append(i)
start = i+1

f = False
for i in xrange(len(quoteindexes)/2):
if findme in
line[quoteindexes.pop(0):quoteindexes.pop(0)]:
f = True
print "Found %s on line %s: Single-Quoted" % (findme, j
+1)
if not f:
print "Found %s on line %s: Not quoted" % (findme, j+1)

It's not pretty but it works.

~Sean

This approach omits double-quoted strings, escaped single-quotes "'a
\'b' my tag", triple-quoted strings, as well as multi-line strings of
any type.

Depends what constraints you can sacrifice. Maybe character-at-a-
time, or manually untokenize the solution above. For generic input,
use mine.