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Python string.rstrip() doesn't strip specified characters

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
'())('

Remove Characters from string with replace not working

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()

Converting regex whitespace characters from list into string

So i want to convert regex whitespaces into a string for example
list1 = ["Hello","\s","my","\s","name","\s","is"]
And I want to convert it to a string like
"Hello my name is"
Can anyone please help.
But also if there was characters such as
"\t"
how would i do this?

list = ["Hello","\s","my","\s","name","\s","is"]
str1 = ''.join(list).replace("\s"," ")
Output :
>>> str1
'Hello my name is'
Update :
If you have something like this list1 = ["Hello","\s","my","\s","name","\t","is"] then you can use multiple replace
>>> str1 = ''.join(list).replace("\s"," ").replace("\t"," ")
>>> str1
'Hello my name is'
or if it's only \t
str1 = ''.join(list).replace("\t","anystring")

I would highly recommend using the join string function mentioned in one of the earlier answers, as it is less verbose. However, if you absolutely needed to use regex in order to complete the task, here's the answer:
import re
list1 = ["Hello","\s","my","\s","name","\s","is"]
list_str = ''.join(list1)
updated_str = re.split('\\\s', list_str)
updated_str = ' '.join(updated_str)
print(updated_str)
Output is:
'Hello my name is'
In order to use raw string notation, replace the 5th line of code with the one below:
updated_str = re.split(r'\\s', list_str)
Both will have the same output result.

You don't even need regular expressions for that:
s = ' '.join([item for item in list if item is not '\s'])
Please note that list is an invalid name for a variable in python as it conflicts with the list function.

Split string by hyphen

I have a strings in the format of feet'-inches" (i.e. 18'-6") and I want to split it so that the values of the feet and inches are separated.
I have tried:
re.split(r'\s|-', `18'-6`)
but it still returns 18'-6.
Desired output: [18,6] or similar
Thanks!

Just split normally replacing the ':
s="18'-6"
a, b = s.replace("'","").split("-")
print(a,b)
If you have both " and ' one must be escaped so just split and slice up to the second last character:
s = "18'-6\""
a, b = s.split("-")
print(a[:-1], b[:-1])
18 6

You can use
import re
p = re.compile(ur'[-\'"]')
test_str = u"18'-6\""
print filter(None,re.split(p, test_str))
Output:
[u'18', u'6']
Ideone demo

A list comprehension will do the trick:
In [13]: [int(i[:-1]) for i in re.split(r'\s|-', "18'-6\"")]
Out[13]: [18, 6]
This assumes that your string is of the format feet(int)'-inches(int)", and you are trying to get the actual ints back, not just numbers in string format.

The built-in split method can take an argument that will cause it to split at the specified point.
"18'-16\"".replace("'", "").replace("\"", "").split("-")
A one-liner. :)

Strip in Python

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'