Confused about backslashes in regular expressions
Question:
I am confused with the backslash in regular expressions. Within a regex a d means a decimal digit. If you add a backslash in front of the backslash this special meaning gets lost. In the regex-howto one can read:
Perhaps the most important metacharacter is the backslash, [ or [ or \.
So print(re.search('d', 'd')) gives None because d matches any decimal digit but there is none in d.
I now would expect print(re.search('\d', 'd')) to match d but the answer is still None.
Only print(re.search('\d', 'd')) gives as output <_sre.SRE_Match object; span=(0, 2), match='\d'>.
Does someone have an explanation?
Answers:
Python’s own string parsing (partially) comes in your way.
If you want to see what re sees, type
print 'd'
print '\d'
print '\d'
on the Python command prompt. You see that d and \d both result in d, the latter one being taken care by the Python string parser.
If you want to avoid any hassle with these, use raw strings as suggested by the re module documentation: r'\d' will result in \d seen by the RE module.
The confusion is due to the fact that the backslash character re module ever sees your string. For instance, n is converted to a newline character, t is converted to a tab character, etc. To get an actual \ gives a single \.
If you want to see how Python is expanding your string escapes, just print out the string. For example:
s = 'a\btc'
print(s)
If s is part of an aggregate data type, e.g. a list or a tuple, and if you print that aggregate, Python will enclose the string in single quotes and will include the
Once you know how your string is being encoded, you can then think about what the re module will do with it. For instance, if you want to escape re module, you will need to pass \ to re, which means you will need to use \\ in your quoted Python string. The Python string will end up with \ and the re module will treat this as a single literal
An alternative way to include r'ab' is equivalent to "a\b".
An r character before the regular expression in a call to search() specifies that the regular expression is a raw string. This allows backslashes to be used in the regular expression as regular characters rather than in an escape sequence of characters. Let me explain …
Before the re module’s search method processes the strings that are passed to it, the Python interpreter takes an initial pass over the string. If there are backslashes present in a string, the Python interpreter must decide if each is part of a Python escape sequence (e.g. n or t) or not.
Note: at this point Python does not care whether or not ” is a regular expression meta-character.
If the ” is followed by a recognized Python escape character (t,n, etc.), then the backslash and the escape character are replaced with the actual Unicode or 8-bit character. For example, ‘t’ would be replaced with the ASCII character for tab. Otherwise it is passed by and interpreted as a ” character.
Consider the following.
>>> s = 't'
>>> print ("[" + s + "]")
>>> [ ] // an actual tab character after preprocessing
>>> s = 'd'
>>> print ("[" + s + "]")
>>> [d] // 'd' after preprocessing
Sometimes we want to include in a string a character sequence that includes ” without it being interpreted by Python as an escape sequence. To do this we escape the ” with a ”. Now when Python sees ” it replaces the two backslashes with a single ” character.
>>> s = '\t'
>>> print ("[" + s + "]")
>>> [t] // 't' after preprocessing
After the Python interpreter take a pass on both strings, they are passed to the re module’s search method. The search method parses the regular expression string to identify the regular expression’s meta-characters.
Now ” is also a special regular expression meta-character and is interpreted as one UNLESS it is escaped at the time that the re search() method is executed.
Consider the following call.
>>> match = re.search('a\t','a\t') //Match is None
Here, match is None. Why? Lets look at the strings after the Python interpreter makes its pass.
String 1: 'at'
String 2: 'at'
So why is match equal to None? When search() interprets String 1, since it is a regular expression, the backslash is interpreted as a meta-character, not an ordinary character. The backslash in String 2 however is not in a regular expression and has already been processed by the Python interpreter, so it is interpreted as an ordinary character.
So the search() method is looking for ‘a escape-t’ in the string ‘at’ which are not a match.
To fix this we can tell the search() method to not interpret the ” as a meta-character. We can do this by escaping it.
Consider the following call.
>>> match = re.search('a\\t','a\t') // Match contains 'at'
Again, lets look at the strings after the Python interpreter has made its pass.
String 1: 'a\t'
String 2: 'at'
Now when the search() method processes the regular expression, it sees that the second backslash is escaped by the first and should not be considered a meta-character. It therefore interprets the string as ‘at’, which matches String 2.
An alternate way to have search() consider ” as a character is to place an r before the regular expression. This tells the Python interpreter to NOT preprocess the string.
Consider this.
>>> match = re.search(r'a\t','a\t') // match contains 'at'
Here the Python interpreter does not modify the first string but does process the second string. The strings passed to search() are:
String 1: 'a\t'
String 2: 'at'
As in the previous example, search interprets the ” as the single character ” and not a meta-character, thus matches with String 2.
I am confused with the backslash in regular expressions. Within a regex a has a special meaning, e.g. d means a decimal digit. If you add a backslash in front of the backslash this special meaning gets lost. In the regex-howto one can read:
Perhaps the most important metacharacter is the backslash,
. As in Python string literals, the backslash can be followed by various characters to signal various special sequences. It’s also used to escape all the metacharacters so you can still match them in patterns; for example, if you need to match a[or, you can precede them with a backslash to remove their special meaning:[or\.
So print(re.search('d', 'd')) gives None because d matches any decimal digit but there is none in d.
I now would expect print(re.search('\d', 'd')) to match d but the answer is still None.
Only print(re.search('\d', 'd')) gives as output <_sre.SRE_Match object; span=(0, 2), match='\d'>.
Does someone have an explanation?
Python’s own string parsing (partially) comes in your way.
If you want to see what re sees, type
print 'd'
print '\d'
print '\d'
on the Python command prompt. You see that d and \d both result in d, the latter one being taken care by the Python string parser.
If you want to avoid any hassle with these, use raw strings as suggested by the re module documentation: r'\d' will result in \d seen by the RE module.
The confusion is due to the fact that the backslash character is used as an escape at two different levels. First, the Python interpreter itself performs substitutions for before the re module ever sees your string. For instance, n is converted to a newline character, t is converted to a tab character, etc. To get an actual character, you can escape it as well, so \ gives a single character. If the character following the isn’t a recognized escape character, then the is treated like any other character and passed through, but I don’t recommend depending on this. Instead, always escape your characters by doubling them, i.e. \.
If you want to see how Python is expanding your string escapes, just print out the string. For example:
s = 'a\btc'
print(s)
If s is part of an aggregate data type, e.g. a list or a tuple, and if you print that aggregate, Python will enclose the string in single quotes and will include the escapes (in a canonical form), so be aware of how your string is being printed. If you just type a quoted string into the interpreter, it will also display it enclosed in quotes with escapes.
Once you know how your string is being encoded, you can then think about what the re module will do with it. For instance, if you want to escape in a string you pass to the re module, you will need to pass \ to re, which means you will need to use \\ in your quoted Python string. The Python string will end up with \ and the re module will treat this as a single literal character.
An alternative way to include characters in Python strings is to use raw strings, e.g. r'ab' is equivalent to "a\b".
An r character before the regular expression in a call to search() specifies that the regular expression is a raw string. This allows backslashes to be used in the regular expression as regular characters rather than in an escape sequence of characters. Let me explain …
Before the re module’s search method processes the strings that are passed to it, the Python interpreter takes an initial pass over the string. If there are backslashes present in a string, the Python interpreter must decide if each is part of a Python escape sequence (e.g. n or t) or not.
Note: at this point Python does not care whether or not ” is a regular expression meta-character.
If the ” is followed by a recognized Python escape character (t,n, etc.), then the backslash and the escape character are replaced with the actual Unicode or 8-bit character. For example, ‘t’ would be replaced with the ASCII character for tab. Otherwise it is passed by and interpreted as a ” character.
Consider the following.
>>> s = 't'
>>> print ("[" + s + "]")
>>> [ ] // an actual tab character after preprocessing
>>> s = 'd'
>>> print ("[" + s + "]")
>>> [d] // 'd' after preprocessing
Sometimes we want to include in a string a character sequence that includes ” without it being interpreted by Python as an escape sequence. To do this we escape the ” with a ”. Now when Python sees ” it replaces the two backslashes with a single ” character.
>>> s = '\t'
>>> print ("[" + s + "]")
>>> [t] // 't' after preprocessing
After the Python interpreter take a pass on both strings, they are passed to the re module’s search method. The search method parses the regular expression string to identify the regular expression’s meta-characters.
Now ” is also a special regular expression meta-character and is interpreted as one UNLESS it is escaped at the time that the re search() method is executed.
Consider the following call.
>>> match = re.search('a\t','a\t') //Match is None
Here, match is None. Why? Lets look at the strings after the Python interpreter makes its pass.
String 1: 'at'
String 2: 'at'
So why is match equal to None? When search() interprets String 1, since it is a regular expression, the backslash is interpreted as a meta-character, not an ordinary character. The backslash in String 2 however is not in a regular expression and has already been processed by the Python interpreter, so it is interpreted as an ordinary character.
So the search() method is looking for ‘a escape-t’ in the string ‘at’ which are not a match.
To fix this we can tell the search() method to not interpret the ” as a meta-character. We can do this by escaping it.
Consider the following call.
>>> match = re.search('a\\t','a\t') // Match contains 'at'
Again, lets look at the strings after the Python interpreter has made its pass.
String 1: 'a\t'
String 2: 'at'
Now when the search() method processes the regular expression, it sees that the second backslash is escaped by the first and should not be considered a meta-character. It therefore interprets the string as ‘at’, which matches String 2.
An alternate way to have search() consider ” as a character is to place an r before the regular expression. This tells the Python interpreter to NOT preprocess the string.
Consider this.
>>> match = re.search(r'a\t','a\t') // match contains 'at'
Here the Python interpreter does not modify the first string but does process the second string. The strings passed to search() are:
String 1: 'a\t'
String 2: 'at'
As in the previous example, search interprets the ” as the single character ” and not a meta-character, thus matches with String 2.