After replacing all word characters in a string with the character '^', using re.sub("\w", "^" , stringorphrase) I'm left with :
>>> '^^^ ^^ ^^^^'
Is there any way to remove the single quotes so it looks cleaner?
>>> ^^^ ^^ ^^^^
Are you sure it's just not how it's displayed in the interactive prompt or something (and there aren't actually apost's in your string)?
If the ' is actually part of the string, and is first/last then either:
string = string.strip("'")
or:
string = string[1:-1] # lop ending characters off
Use the print statement. The quotes aren't actually part of the string.
To remove all occurrences of single quotes:
mystr = some_string_with_single_quotes
answer = mystr.replace("'", '')
To remove single quotes ONLY at the ends of the string:
mystr = some_string_with_single_quotes
answer = mystr.strip("'")
Hope this helps
Related
string = "hi())("
string = string.rstrip("abcdefghijklmnoprstuwxyz")
print(string)
I want to remove every letter from given string using rstrip method, however it does not change the string in the slightest.
Output:
'hi())('
What i Want:
'())('
I know that I can use regex, but I really don't understand why it doesn't work.
Note : It is a part of the Valid Parentheses challenge on code-wars
You have to use lstrip instead of rstrip:
>>> string = "hi())("
>>> string = string.lstrip("abcdefghijklmnoprstuwxyz")
>>> string
'())('
I want to remove all the text before and including */ in a string.
For example, consider:
string = ''' something
other things
etc. */ extra text.
'''
Here I want extra text. as the output.
I tried:
string = re.sub("^(.*)(?=*/)", "", string)
I also tried:
string = re.sub(re.compile(r"^.\*/", re.DOTALL), "", string)
But when I print string, it did not perform the operation I wanted and the whole string is printing.
I suppose you're fine without regular expressions:
string[string.index("*/ ")+3:]
And if you want to strip that newline:
string[string.index("*/ ")+3:].rstrip()
The problem with your first regex is that . does not match newlines as you noticed. With your second one, you were closer but forgot the * that time. This would work:
string = re.sub(re.compile(r"^.*\*/", re.DOTALL), "", string)
You can also just get the part of the string that comes after your "*/":
string = re.search(r"(\*/)(.*)", string, re.DOTALL).group(2)
Update: After doing some research, I found that the pattern (\n|.) to match everything including newlines is inefficient. I've updated the answer to use [\s\S] instead as shown on the answer I linked.
The problem is that . in python regex matches everything except newlines. For a regex solution, you can do the following:
import re
strng = ''' something
other things
etc. */ extra text.
'''
print(re.sub("[\s\S]+\*/", "", strng))
# extra text.
Add in a .strip() if you want to remove that remaining leading whitespace.
to keep text until that symbol you can do:
split_str = string.split(' ')
boundary = split_str.index('*/')
new = ' '.join(split_str[0:boundary])
print(new)
which gives you:
something
other things
etc.
string_list = string.split('*/')[1:]
string = '*/'.join(string_list)
print(string)
gives output as
' extra text. \n'
Basically, I'm asking the user to input a string of text into the console, but the string is very long and includes many line breaks. How would I take the user's string and delete all line breaks to make it a single line of text. My method for acquiring the string is very simple.
string = raw_input("Please enter string: ")
Is there a different way I should be grabbing the string from the user? I'm running Python 2.7.4 on a Mac.
P.S. Clearly I'm a noob, so even if a solution isn't the most efficient, the one that uses the most simple syntax would be appreciated.
How do you enter line breaks with raw_input? But, once you have a string with some characters in it you want to get rid of, just replace them.
>>> mystr = raw_input('please enter string: ')
please enter string: hello world, how do i enter line breaks?
>>> # pressing enter didn't work...
...
>>> mystr
'hello world, how do i enter line breaks?'
>>> mystr.replace(' ', '')
'helloworld,howdoienterlinebreaks?'
>>>
In the example above, I replaced all spaces. The string '\n' represents newlines. And \r represents carriage returns (if you're on windows, you might be getting these and a second replace will handle them for you!).
basically:
# you probably want to use a space ' ' to replace `\n`
mystring = mystring.replace('\n', ' ').replace('\r', '')
Note also, that it is a bad idea to call your variable string, as this shadows the module string. Another name I'd avoid but would love to use sometimes: file. For the same reason.
You can try using string replace:
string = string.replace('\r', '').replace('\n', '')
You can split the string with no separator arg, which will treat consecutive whitespace as a single separator (including newlines and tabs). Then join using a space:
In : " ".join("\n\nsome text \r\n with multiple whitespace".split())
Out: 'some text with multiple whitespace'
https://docs.python.org/2/library/stdtypes.html#str.split
The canonic answer, in Python, would be :
s = ''.join(s.splitlines())
It splits the string into lines (letting Python doing it according to its own best practices). Then you merge it. Two possibilities here:
replace the newline by a whitespace (' '.join())
or without a whitespace (''.join())
updated based on Xbello comment:
string = my_string.rstrip('\r\n')
read more here
Another option is regex:
>>> import re
>>> re.sub("\n|\r", "", "Foo\n\rbar\n\rbaz\n\r")
'Foobarbaz'
If anybody decides to use replace, you should try r'\n' instead '\n'
mystring = mystring.replace(r'\n', ' ').replace(r'\r', '')
A method taking into consideration
additional white characters at the beginning/end of string
additional white characters at the beginning/end of every line
various end-line characters
it takes such a multi-line string which may be messy e.g.
test_str = '\nhej ho \n aaa\r\n a\n '
and produces nice one-line string
>>> ' '.join([line.strip() for line in test_str.strip().splitlines()])
'hej ho aaa a'
UPDATE:
To fix multiple new-line character producing redundant spaces:
' '.join([line.strip() for line in test_str.strip().splitlines() if line.strip()])
This works for the following too
test_str = '\nhej ho \n aaa\r\n\n\n\n\n a\n '
Regular expressions is the fastest way to do this
s='''some kind of
string with a bunch\r of
extra spaces in it'''
re.sub(r'\s(?=\s)','',re.sub(r'\s',' ',s))
result:
'some kind of string with a bunch of extra spaces in it'
The problem with rstrip() is that it does not work in all cases (as I myself have seen few). Instead you can use
text = text.replace("\n"," ")
This will remove all new line '\n' with a space.
You really don't need to remove ALL the signs: lf cr crlf.
# Pythonic:
r'\n', r'\r', r'\r\n'
Some texts must have breaks, but you probably need to join broken lines to keep particular sentences together.
Therefore it is natural that line breaking happens after priod, semicolon, colon, but not after comma.
My code considers above conditions. Works well with texts copied from pdfs.
Enjoy!:
def unbreak_pdf_text(raw_text):
""" the newline careful sign removal tool
Args:
raw_text (str): string containing unwanted newline signs: \\n or \\r or \\r\\n
e.g. imported from OCR or copied from a pdf document.
Returns:
_type_: _description_
"""
pat = re.compile((r"[, \w]\n|[, \w]\r|[, \w]\r\n"))
breaks = re.finditer(pat, raw_text)
processed_text = raw_text
raw_text = None
for i in breaks:
processed_text = processed_text.replace(i.group(), i.group()[0]+" ")
return processed_text
I have a question regarding strip() in Python. I am trying to strip a semi-colon from a string, I know how to do this when the semi-colon is at the end of the string, but how would I do it if it is not the last element, but say the second to last element.
eg:
1;2;3;4;\n
I would like to strip that last semi-colon.
Strip the other characters as well.
>>> '1;2;3;4;\n'.strip('\n;')
'1;2;3;4'
>>> "".join("1;2;3;4;\n".rpartition(";")[::2])
'1;2;3;4\n'
how about replace?
string1='1;2;3;4;\n'
string2=string1.replace(";\n","\n")
>>> string = "1;2;3;4;\n"
>>> string.strip().strip(";")
"1;2;3;4"
This will first strip any leading or trailing white space, and then remove any leading or trailing semicolon.
Try this:
def remove_last(string):
index = string.rfind(';')
if index == -1:
# Semi-colon doesn't exist
return string
return string[:index] + string[index+1:]
This should be able to remove the last semicolon of the line, regardless of what characters come after it.
>>> remove_last('Test')
'Test'
>>> remove_last('Test;abc')
'Testabc'
>>> remove_last(';test;abc;foobar;\n')
';test;abc;foobar\n'
>>> remove_last(';asdf;asdf;asdf;asdf')
';asdf;asdf;asdfasdf'
The other answers provided are probably faster since they're tailored to your specific example, but this one is a bit more flexible.
You could split the string with semi colon and then join the non-empty parts back again using ; as separator
parts = '1;2;3;4;\n'.split(';')
non_empty_parts = []
for s in parts:
if s.strip() != "": non_empty_parts.append(s.strip())
print "".join(non_empty_parts, ';')
If you only want to use the strip function this is one method:
Using slice notation, you can limit the strip() function's scope to one part of the string and append the "\n" on at the end:
# create a var for later
str = "1;2;3;4;\n"
# format and assign to newstr
newstr = str[:8].strip(';') + str[8:]
Using the rfind() method(similar to Micheal0x2a's solution) you can make the statement applicable to many strings:
# create a var for later
str = "1;2;3;4;\n"
# format and assign to newstr
newstr = str[:str.rfind(';') + 1 ].strip(';') + str[str.rfind(';') + 1:]
re.sub(r';(\W*$)', r'\1', '1;2;3;4;\n') -> '1;2;3;4\n'
i've the following code:
import re
key = re.escape('#one #two #some #tests #are #done')
print(key)
key = key.split()
print(key)
and the following output:
\#one\ \#two\ \#some\ \#tests\ \#are\ \#done
['\\#one\\', '\\#two\\', '\\#some\\', '\\#tests\\', '\\#are\\', '\\#done']
How come the backslashes are duplicated? I just want them once in my list, because i would like to use this list in a regular expression.
Thanks in advance! John
There is only one backslash each, but when printing the repr of the strings, they are duplicated (escaped) - just as you would need to duplicate them when using a string to build a regex. So everything is fine.
For example:
>>> len("\\")
1
>>> len("\\n")
2
>>> len("\n")
1
>>> print "\\n"
\n
>>> print "\n"
>>>
The \ character is an escape character, that is a character that changes the meaning of the subsequent character[s]. For example the "n" character is simply an "n". But if you escape it like "\n" it becomes the "newline" character. So, if you need to use a \ literal, you need to escape it with... itself: \\
The backslashes are not duplicated. To realize this, try to do:
for element in key:
print element
And you will see this output:
\#one\
\#two\
\#some\
\#tests\
\#are\
\#done
When you have printed whole list, the python used representation where strings are printed not as they are, but they are printed as python expression (notice the quotes "", they are not in the strings)
To actually encode string containing backslash, you need to duplicate that backslash. That is it.
When you convert a list to a string (e.g. to print it), it calls repr on each object contained in the list. That's why you get the quotes and extra backslashes in your second line of output. Try this:
s = "\\a string with an escaped backslash"
print s # prints: \a string with an escaped backslash
print repr(s) # prints: '\\a string with an escaped backslash'
The repr call puts quotes around the string, and shows the backslash escapes.