Python re.sub: ignore backreferences in the replacement string

Question

I want to replace a pattern with a string. The string is given in a variable. It might, of course, contains '\1', and it should not be interpreted as a backreference - but simply as \1.

How can I achieve that?

Answer 1

The previous answer using re.escape() would escape too much, and you would get undesirable backslashes in the replacement and the replaced string.

It seems like in Python only the backslash needs escaping in the replacement string, thus something like this could be sufficient:

replacement = replacement.replace("\\", "\\\\")

Example:

import re

x = r'hai! \1 <ops> $1 \' \x \\'
print "want to see: "
print x

print "getting: "
print re.sub(".(.).", x, "###")
print "over escaped: "
print re.sub(".(.).", re.escape(x), "###")
print "could work: "
print re.sub(".(.).", x.replace("\\", "\\\\"), "###")

Output:

want to see: 
hai! \1 <ops> $1 \' \x \\
getting: 
hai! # <ops> $1 \' \x \
over escaped: 
hai\!\ \1\ \<ops\>\ \$1\ \\'\ \x\ \\
could work: 
hai! \1 <ops> $1 \' \x \\