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
'())('
Related
I am trying to replace \ with /. However, I'm having no success.
Following is the snapshot of the scenario that I am trying to achieve
string = "//SQL-SERVER/Lacie/City of X/Linservo\171002"
print string.replace("\\","/")
Output:
//SQL-SERVER/Lacie/City of X/Linservoy002
Desired output:
//SQL-SERVER/Lacie/City of X/Linservo/171002
You need to escape "\" with an extra "\".
>>> string = "//SQL-SERVER/Lacie/City of X/Linservo\\171002"
>>> string
'//SQL-SERVER/Lacie/City of X/Linservo\\171002'
>>> print string.replace("\\","/")
//SQL-SERVER/Lacie/City of X/Linservo/171002
string = r"//SQL-SERVER/Lacie/City of X/Linservo\171002"
print string.replace("\\","/")
output
//SQL-SERVER/Lacie/City of X/Linservo/171002
You have errors both in replace function and in string definition.
In your string definition \171 gives char with octal value of 171 – y
In you replace function, backslash escapes quote.
You should escape backslashes
string = "//SQL-SERVER/Lacie/City of X/Linservo\\171002"
string.replace("\\","/")
You can simply use ".replace" in python or if you want you can use regex :
import re
string = r"//SQL-SERVER/Lacie/City of X/Linservo\171002"
pattern=r'[\\]'
replaced_string=re.sub(pattern,"/",string)
print(replaced_string)
Since your original question shows : "X/Linservo\171002" here \171 referring to character encoding so it's replacing \171 to "y". you can try this in python interpreter :
In[2]: print("\171")
y
I have a number of strings from which I am aiming to remove charactars using replace. However, this dosent seem to wake. To give a simplified example, this code:
row = "b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'"
row = row.replace("b'", "").replace("'", "").replace('b"', '').replace('"', '')
print(row.encode('ascii', errors='ignore'))
still ouputs this b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38' wheras I would like it to output James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38. How can I do this?
Edit: Updataed the code with a better example.
You seem to be mistaking single quotes for double quotes. Simple replace 'b:
>>> row = "xyz'b"
>>> row.replace("'b", "")
'xyz'
As an alternative to str.replace, you can simple slice the string to remove the unwanted leading and trailing characters:
>>> row[2:-1]
'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'
In your first .replace, change b' to 'b. Hence your code should be:
>>> row = "xyz'b"
>>> row = row.replace("'b", "").replace("'", "").replace('b"', '').replace('"', '')
# ^ changed here
>>> print(row.encode('ascii', errors='ignore'))
xyz
I am assuming rest of the conditions you have are the part of other task/matches that you didn't mentioned here.
If all you want is to take the string before first ', then you may just do:
row.split("'")[0]
You haven't listed this to remove 'b:
.replace("'b", '')
import ast
row = "b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'"
b_string = ast.literal_eval(row)
print(b_string)
u_string = b_string.decode('utf-8')
print(u_string)
out:
b_string:b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'
u_string: James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38
The real question is how to convert a string to python object.
You get a string which contains an a binary string, to convert it to python's binary string object, you should use eval(). ast.literal_eval() is more safe way to do it.
Now you get a binary string, you can convert it to unicode string which do not start with "b" by using decode()
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
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.