error: bad escape (end of pattern) at position 0 while trying to replace to backslah

Question

I need to replace single backslash in my string. I'm trying to perform '\' but it gives with me double backslash. When i'm performing '\' it throw error bad escape (end of pattern) at position 0

string = 'abrakadabra'
string.replace('a','\\')
#or
re.sub('a','\\','abrakadabra')

In [47]: string.replace('a','\'')
Out[47]: "'br'k'd'br'"

In [48]: string.replace('a','\')
 File "<ipython-input-48-e884682860ae>", line 1
string.replace('a','\')
                       ^

SyntaxError: EOL while scanning string literal

In [49]: string.replace('a','\\')
Out[49]: '\\br\\k\\d\\br\\'

In [50]:

expecting for result: \br\k\d\br\

Run `print(string.replace('a','\\'))` to see there are no double backslashes. — VPfB, May 19 '19 at 08:42

score 11 · Answer 1 · edited Jan 10 '23 at 14:16

11

You should use '\\\\', then you will pass '\\' to re, which is just one escaped backslash.

print(re.sub('a','\\\\','abrakadabra'))
# \br\k\d\br\

edited Jan 10 '23 at 14:16

petezurich

9,280
9
43
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answered May 19 '19 at 08:51

gregk

179
5

7

Or explicitly tell python it's actually raw backslashes `r'\\'` – Zionsof May 19 '19 at 13:50

error: bad escape (end of pattern) at position 0 while trying to replace to backslah

1 Answers1