For example:
a = (r'''\n1''')
b = (r'''
2''')
print(a)
print(b)
The output of this example is this:
\n1
2
Meaning that even if b is supposed to be a raw string, it does not seem to work like one, why is this?
I also checked:
if '\n' in b:
print('yes')
The output of this is yes meaning that b is a string, and indeed has \n string inside of it.
In the raw string syntax, escape sequences have no special meaning (apart from a backslash before a quote). The characters \ plus n form two characters in a raw string literal, unlike a regular string literal, where those two characters are replaced by a newline character.
An actual newline character, on the other hand, is not an escape sequence. It is just a newline character, and is included in the string as such.
Compare this to using 1 versus \x31; the latter is an escape sequence for the ASCII codepoint for the digit 1. In a regular string literal, both would give you the character 1, in a raw string literal, the escape sequence would not be interpreted:
>>> print('1\x31')
11
>>> print(r'1\x31')
1\x31
All this has nothing to do with parentheses. The parentheses do not alter the behaviour of a r'''...''' raw string. The exact same thing happens when you remove the parentheses:
>>> a = r'''\n1'''
>>> a
'\\n1'
>>> print(a)
\n1
>>> b = r'''
... 2'''
>>> b
'\n2'
>>> print(b)
2
Related
I wrote this piece of code to count the number of characters in the string a\nb\x1f\000d
#CLASS TASK-VI
ctr=0
str1="a\nb\x1f\000d"
for i in range(len(str1)):
ctr=ctr+1
print("Number of characters in the string str1 is: ",ctr)
It returns Number of characters in the string str1 is: 6
Can someone explain this? Thanks in advance.
There are 6 characters in the string:
a
\n, which resolves to the single 'newline' character (also referred to as 'line feed')
b
\x1f, which is a hex escape sequence. \x means that the following two characters (in hexidecimal) will make up a number (in this case, 1f -> 31), and to use the character whose code is that number. Character number 31 happens to be an ASCII control character, known as 'unit separator'.
\000 is an octal escape sequence, which is the same as above but in base 8. In this case, the code it refers to is 0, which is the null character
d
Backslashes are a special control character that 'escape' the following character. Certain 'escape sequences' have special effects - here you see \n, \x, and \0, for example, though there are plenty more if you feel like looking them up. In python, you can make a string not process escape sequences by declaring it as a "raw string", which you do by starting the string with r" instead of just ":
>>> len("a\nb\x1f\000d")
6
>>> len(r"a\nb\x1f\000d")
13
You can also use a double-backslash \\ to escape a backslash, thus preventing it from escaping something else.
>>> len("a\\nb\\x1f\\000d")
13
If you want the backslashes to be counted as characters, make them double backslashes, otherwise they represent escape characters.
>>> str1 = "a\\nb\\x1f\\000d"
>>> len(str1)
13
str1= "a\nb\x1f\000d"
print(f"Number of characters in the string str1 is {len(str1)}")
# 6
str1= r"a\nb\x1f\000d"
print(f"Number of characters in the string str1 is {len(str1)}")
#13
If you source the string out of python then each character will be treated as a string.
For example, make a file named "content.txt". Inside that file type/paste any string you want with any special character, read it with python then it will be treated as a string.
i.e.
with open("content.txt") as fl:
content = fl.read()
print(len(content))
print(content)
I have tried the following code:
strReFindString = u"\\begin{minipage}"+"(.*?)"
strReFindString += u"\\end{minipage}"
lst = re.findall(strReFindString, strBuffer, re.DOTALL)
But it always returns empty list.
How can I do?
Thanks all.
As #BrenBarn said, u"\\b" parses as \b; and \b is not a valid regexp escape, so findall treats it as b (literal b). u"\\\\b" is \\b, which regexp understands as \b (literal backslash, literal b). You can prevent escape-parsing in the string using raw strings, ur"\\b" is equal to u"\\\\b":
ur"\\b" == u"\\\\b"
# => True
Python recommends using raw strings when defining regular expressions in the re module. From the Python documentation:
Regular expressions use the backslash character ('\') to indicate special forms or to allow special characters to be used without invoking their special meaning. This collides with Python’s usage of the same character for the same purpose in string literals; for example, to match a literal backslash, one might have to write '\\' as the pattern string, because the regular expression must be \, and each backslash must be expressed as \ inside a regular Python string literal.
However, in many cases this is not necessary, and you get the same result whether you use a raw string or not:
$ ipython
In [1]: import re
In [2]: m = re.search("\s(\d)\s", "a 3 c")
In [3]: m.groups()
Out[3]: ('3',)
In [4]: m = re.search(r"\s(\d)\s", "a 3 c")
In [5]: m.groups()
Out[5]: ('3',)
Yet, in some cases this is not the case:
In [6]: m = re.search("\s(.)\1\s", "a 33 c")
In [7]: m.groups()
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-12-84a8d9c174e2> in <module>()
----> 1 m.groups()
AttributeError: 'NoneType' object has no attribute 'groups'
In [8]: m = re.search(r"\s(.)\1\s", "a 33 c")
In [9]: m.groups()
Out[9]: ('3',)
And you must escape the special characters when not using a raw string:
In [10]: m = re.search("\\s(.)\\1\\s", "a 33 c")
In [11]: m.groups()
Out[11]: ('3',)
My question is why do the non-escaped, non-raw regex strings work at all with special characters (as in command [2] above)?
The example above works because \s and \d are not escape sequences in python. According to the docs:
Unlike Standard C, all unrecognized escape sequences are left in the string unchanged, i.e., the backslash is left in the string.
But it's best to just use raw strings and not worry about what is or isn't a python escape, or worry about changing it later if you change the regex.
It is because \s and \d are not escape sequences:
>>> print('\s')
\s
>>> print('\d')
\d
>>>
So, they are treated literally as \s and \d. \1 however is an escape sequence:
>>> print('\1')
☺
>>>
This means that it is being interpreted as ☺ instead of \1.
For a complete list of Python's escape sequences, see String and Bytes literals in the documentation.
I wanna replace all the chars which occur more than one time,I used Python's re.sub and my regex looks like this data=re.sub('(.)\1+','##',data), But nothing happened...
Here is my Text:
Text
※※※※※※※※※※※※※※※※※Chapter One※※※※※※※※※※※※※※※※※※
This is the begining...
You need to use raw string here, 1 is interpreted as octal and then its ASCII value present at its integer equivalent is used in the string.
>>> '\1'
'\x01'
>>> chr(01)
'\x01'
>>> '\101'
'A'
>>> chr(0101)
'A'
Use raw string to fix this:
>>> '(.)\1+'
'(.)\x01+'
>>> r'(.)\1+' #Note the `r`
'(.)\\1+'
Use a raw string, so the regex engine interprets backslashes instead of the Python parser. Just put an r in front of the string:
data=re.sub(r'(.)\1+', '##', data)
^ this r is the important bit
Otherwise, \1 is interpreted as character value 1 instead of a backreference.
Can anyone explain why example 1 below works, when the r prefix is not used?
I thought the r prefix must be used whenever escape sequences are used.
Example 2 and example 3 demonstrate this.
# example 1
import re
print (re.sub('\s+', ' ', 'hello there there'))
# prints 'hello there there' - not expected as r prefix is not used
# example 2
import re
print (re.sub(r'(\b\w+)(\s+\1\b)+', r'\1', 'hello there there'))
# prints 'hello there' - as expected as r prefix is used
# example 3
import re
print (re.sub('(\b\w+)(\s+\1\b)+', '\1', 'hello there there'))
# prints 'hello there there' - as expected as r prefix is not used
Because \ begin escape sequences only when they are valid escape sequences.
>>> '\n'
'\n'
>>> r'\n'
'\\n'
>>> print '\n'
>>> print r'\n'
\n
>>> '\s'
'\\s'
>>> r'\s'
'\\s'
>>> print '\s'
\s
>>> print r'\s'
\s
Unless an 'r' or 'R' prefix is present, escape sequences in strings are interpreted according to rules similar to those used by Standard C. The recognized escape sequences are:
Escape Sequence Meaning Notes
\newline Ignored
\\ Backslash (\)
\' Single quote (')
\" Double quote (")
\a ASCII Bell (BEL)
\b ASCII Backspace (BS)
\f ASCII Formfeed (FF)
\n ASCII Linefeed (LF)
\N{name} Character named name in the Unicode database (Unicode only)
\r ASCII Carriage Return (CR)
\t ASCII Horizontal Tab (TAB)
\uxxxx Character with 16-bit hex value xxxx (Unicode only)
\Uxxxxxxxx Character with 32-bit hex value xxxxxxxx (Unicode only)
\v ASCII Vertical Tab (VT)
\ooo Character with octal value ooo
\xhh Character with hex value hh
Never rely on raw strings for path literals, as raw strings have some rather peculiar inner workings, known to have bitten people in the ass:
When an "r" or "R" prefix is present, a character following a backslash is included in the string without change, and all backslashes are left in the string. For example, the string literal r"\n" consists of two characters: a backslash and a lowercase "n". String quotes can be escaped with a backslash, but the backslash remains in the string; for example, r"\"" is a valid string literal consisting of two characters: a backslash and a double quote; r"\" is not a valid string literal (even a raw string cannot end in an odd number of backslashes). Specifically, a raw string cannot end in a single backslash (since the backslash would escape the following quote character). Note also that a single backslash followed by a newline is interpreted as those two characters as part of the string, not as a line continuation.
To better illustrate this last point:
>>> r'\'
SyntaxError: EOL while scanning string literal
>>> r'\''
"\\'"
>>> '\'
SyntaxError: EOL while scanning string literal
>>> '\''
"'"
>>>
>>> r'\\'
'\\\\'
>>> '\\'
'\\'
>>> print r'\\'
\\
>>> print r'\'
SyntaxError: EOL while scanning string literal
>>> print '\\'
\
the 'r' means the the following is a "raw string", ie. backslash characters are treated literally instead of signifying special treatment of the following character.
http://docs.python.org/reference/lexical_analysis.html#literals
so '\n' is a single newline
and r'\n' is two characters - a backslash and the letter 'n'
another way to write it would be '\\n' because the first backslash escapes the second
an equivalent way of writing this
print (re.sub(r'(\b\w+)(\s+\1\b)+', r'\1', 'hello there there'))
is
print (re.sub('(\\b\\w+)(\\s+\\1\\b)+', '\\1', 'hello there there'))
Because of the way Python treats characters that are not valid escape characters, not all of those double backslashes are necessary - eg '\s'=='\\s' however the same is not true for '\b' and '\\b'. My preference is to be explicit and double all the backslashes.
Not all sequences involving backslashes are escape sequences. \t and \f are, for example, but \s is not. In a non-raw string literal, any \ that is not part of an escape sequence is seen as just another \:
>>> "\s"
'\\s'
>>> "\t"
'\t'
\b is an escape sequence, however, so example 3 fails. (And yes, some people consider this behaviour rather unfortunate.)
Try that:
a = '\''
'
a = r'\''
\'
a = "\'"
'
a = r"\'"
\'
Check below example:
print r"123\n123"
#outputs>>>
123\n123
print "123\n123"
#outputs>>>
123
123