About Patterns and the backslash \ character

Goal: find pattern with \ in it

First thought: use \ to escape \

x = re.search("a.\\y", "ab\y")

blows up. Why? The pattern string is scanned twice -- first by the Python interpreter, next by the pattern matcher, so:

"a.\\y" -- seen by Python interpreter; processed to "a.\y"
"a.\y" -- seen by the Python pattern matcher; ERROR as \y is invalid

So try this:

x = re.search("a.\\\\y", "ab\y")

That works. Note "\y" in the string is a backslash followed by a "y". That's because the Python interpreter looks for \ as part of an escape sequence. If it is, the \ and following character are processed into the special character. If not, it's treated as a backslash.

That's why printing x gives you span=(0,4)

Note in the match, x has match='ab\\y'; that's because the single \ prints as 2.

We can see this as: 

x = 'ab\y'

displaying x shows:

'ab\\y'

but len(x) gives 4. Also, note if you do print(x), you get

'ab\y'

Now, look at 

y = 'ab\t'

and \t is an escape sequence (a tab). Then if you display y, you get

'ab\t'

with len(y) is 3 as the \t is a tab (hex 0x9).

There's another way -- using the "raw string" notation. That suppresses the processing by the Python interpreter.

x = r'ab\y'

is the same as

x = 'ab\\y'

Now watch this:

y = 'ab\t'

Now len(y) is 4 as the interpretation of \ as an escape is suppressed.