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
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 want to take the string 0.71331, 52.25378 and return 0.71331,52.25378 - i.e. just look for a digit, a comma, a space and a digit, and strip out the space.
This is my current code:
coords = '0.71331, 52.25378'
coord_re = re.sub("(\d), (\d)", "\1,\2", coords)
print coord_re
But this gives me 0.7133,2.25378. What am I doing wrong?
You should be using raw strings for regex, try the following:
coord_re = re.sub(r"(\d), (\d)", r"\1,\2", coords)
With your current code, the backslashes in your replacement string are escaping the digits, so you are replacing all matches the equivalent of chr(1) + "," + chr(2):
>>> '\1,\2'
'\x01,\x02'
>>> print '\1,\2'
,
>>> print r'\1,\2' # this is what you actually want
\1,\2
Any time you want to leave the backslash in the string, use the r prefix, or escape each backslash (\\1,\\2).
Python interprets the \1 as a character with ASCII value 1, and passes that to sub.
Use raw strings, in which Python doesn't interpret the \.
coord_re = re.sub(r"(\d), (\d)", r"\1,\2", coords)
This is covered right in the beginning of the re documentation, should you need more info.
a="D:/R_SVN/hostworkspace/middleware/projects/module/com.ofss.fc.module.ac/src/com/ofss/fc/app\ac\service\writeoffrecovery\ext\WriteoffRecoveryApplicationServiceExtExecutor.java"
b=a.replace('\','/')
print b
Error:
b=a.replace('\','/')
SyntaxError: EOL while scanning string literal
As "Backslash notation" is used for "Escape character", you have to add \\ instead of \
a.replace('\\','/')
You have to escape the backslash, because it is a special character:
b=a.replace('\\','/')
In strings \ is escape character e.g if there are two \ like \\ then first one is escape character.
in b=a.replace('\','/') '\' is read as escape character. so you can replace it with \\. In this case first \ will be escaped and second one will perform operation on string a.
code:
>>> a="D:/R_SVN/hostworkspace/middleware/projects/module/com.ofss.fc.module.ac/src/com/ofss/fc/app\ac\service\writeoffrecovery\ext\WriteoffRecoveryApplicationServiceExtExecutor.java"
>>> b=a.replace('\\','/')
>>> print b
D:/R_SVN/hostworkspace/middleware/projects/module/com.ofss.fc.module.ac/src/com/ofss/fc/appc/service/writeoffrecovery/ext/WriteoffRecoveryApplicationServiceExtExecutor.java
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
This question already has answers here:
How can I put an actual backslash in a string literal (not use it for an escape sequence)?
(4 answers)
Closed 7 months ago.
How do I escape a backslash and a single quote or double quote in Python?
For example:
Long string = '''some 'long' string \' and \" some 'escaped' strings'''
value_to_change = re.compile(A EXPRESION TO REPRESENT \' and \")
modified = re.sub(value_to_change, 'thevalue', Long_string)
## Desired Output
modified = '''some 'long' string thevalue and thevalue some 'escaped' strings'''
How you did it
If your "long string" is read from a file (as you mention in a comment) then your question is misleading. Since you obviously don't fully understand how escaping works, the question as you wrote it down probably is different from the question you really have.
If these are the contents of your file (51 bytes as shown + maybe one or two end-of-line characters):
some 'long' string \' and \" some 'escaped' strings
then this is what it will look like in python:
>>> s1 = open('data.txt', 'r').read().strip()
>>> s1
'some \'long\' string \\\' and \\" some \'escaped\' strings'
>>> print s1
some 'long' string \' and \" some 'escaped' strings
What you wrote in the question will produce:
>>> s2 = '''some 'long' string \' and \" some 'escaped' strings'''
>>> s2
'some \'long\' string \' and " some \'escaped\' strings'
>>> print s2
some 'long' string ' and " some 'escaped' strings
>>> len(s)
49
Do you see the difference?
There are no backslashes in s2 because they have special meaning when you use them to write down strings in Python. They have no special meaning when you read them from a file.
If you want to write down a string that afterwards has a backslash in it, you have to protect the backslash you enter. You have to keep Python from thinking it has special meaning. You do that by escaping it - with a backslash.
One way to do this is to use backslashes, but often the easier and less confusing way is to use raw strings:
>>> s3 = r'''some 'long' string \' and \" some 'escaped' strings'''
'some \'long\' string \\\' and \\" some \'escaped\' strings'
>>> print s3
some 'long' string \' and \" some 'escaped' strings
>>> s1 == s3
True
How you meant it
The above was only to show you that your question was confusing.
The actual answer is a bit harder - when you are working with regular expressions, the backslash takes on yet another layer of special meaning. If you want to safely get a backslash through string escaping and through regex escaping to the actual regex, you have to write down multiple backslashes accordingly.
Furthermore, rules for putting single quotes (') in single-quoted raw strings (r'') are a bit tricky as well, so I will use a raw string with triple single-quotes (r'''''').
>>> print re.sub(r'''\\['"]''', 'thevalue', s1)
some 'long' string thevalue and thevalue some 'escaped' strings
The two backslashes stay two backslashes throughout string escaping and then become only one backslash without special meaning through regex escaping. In total, the regex says:
"match one backslash followed by either a single-quote or a double-quote."
How it should be done
Now for the pièce de résistance: The previous is really a good demonstration of what jwz meant1. If you forget about regex (and know about raw strings), the solution becomes much more obvious:
>>> print s1.replace(r'\"', 'thevalue').replace(r"\'", 'thevalue')
some 'long' string thevalue and thevalue some 'escaped' strings
1 Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems.
The problem is that in your string \' and \" get converted to ' and ", so on your example as-is, you won't be able to match only \' without matching the single quotes around long.
But my understanding is that this data comes from a file so assuming you have your_file.txt containing
some 'long' string \' and \" some 'escaped' strings
you can replace \' and \" with following code:
import re
from_file = open("your_file.txt", "r").read()
print(re.sub("\\\\(\"|')", "thevalue", from_file))
Note the four slashes. Since this is a string \ gets converted to \ (as this is an escaped character). Then in the regular expression, the remaining \ gets again converted to \, as this is also regular experssion escaped character. Result will match a single slash and one of the " and ' quotes.
is this what you want?
import re
Long_string = "some long string \' and \" some escaped strings"
value_to_change = re.compile( "'|\"" )
modified = re.sub(value_to_change , 'thevalue' , Long_string )
print modified
I try this to print a single backslash (Python 3):
single_backslash_str = r'\ '[0]
print('single_backslash_str') #output: \
print('repr(single_backslash_str)') #output: '\\'
Hope this will help!
Keep in mind, all these strings are exactly the same:
Long_string = '''some long string \' and \" some escaped strings'''
Long_string = '''some long string ' and " some escaped strings'''
Long_string = """some long string ' and " some escaped strings"""
Long_string = 'some long string \' and \" some escaped strings'
Long_string = "some long string \' and \" some escaped strings"
Long_string = 'some long string \' and " some escaped strings'
Long_string = "some long string ' and \" some escaped strings"
There is no backslash character in any of them. So the regex you're looking for doesn't need to match a backslash and a quote, just a quote:
modified = re.sub("['\"]", 'thevalue', Long_string)
BTW: You also don't have to compile the regex before you use it, re.sub will accept a string regex as well as a compiled one.