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Remove brackets and number inside from string Python

I've seen a lot of examples on how to remove brackets from a string in Python, but I've not seen any that allow me to remove the brackets and a number inside of the brackets from that string.
For example, suppose I've got a string such as "abc[1]". How can I remove the "[1]" from the string to return just "abc"?
I've tried the following:
stringTest = "abc[1]"
stringTestWithoutBrackets = str(stringTest).strip('[]')
but this only outputs the string without the final bracket
abc[1
I've also tried with a wildcard option:
stringTest = "abc[1]"
stringTestWithoutBrackets = str(stringTest).strip('[\w+\]')
but this also outputs the string without the final bracket
abc[1

You could use regular expressions for that, but I think the easiest way would be to use split:
>>> stringTest = "abc[1][2][3]"
>>> stringTest.split('[', maxsplit=1)[0]
'abc'

You can use regex but you need to use it with the re module:
re.sub(r'\[\d+\]', '', stringTest)
If the [<number>] part is always at the end of the string you can also strip via:
stringTest.rstrip('[0123456789]')
Though the latter version might strip beyond the [ if the previous character is in the strip list too. For example in "abc1[5]" the "1" would be stripped as well.

Assuming your string has the format "text[number]" and you only want to keep the "text", then you could do:
stringTest = "abc[1]"
bracketBegin = stringTest.find('[')
stringTestWithoutBrackets = stringTest[:bracketBegin]

Turn string into a list and remove carriage returns (Python)

I have a string like this:
['过\r\n啤酒\r\n小心\r\n照顾\r\n锻炼\r\n过去\r\n忘记\r\n哭\r\n包\r\n个子\r\n瘦\r\n选择\r\n奶奶\r\n突然\r\n节目\r\n']
How do I remove all of the "\r\n", and then turn the string into a list like so:
[过, 啤酒, 小心, 照顾, 过去, etc...]

str.split removes all whitespace; this includes \r and \n:
A = ['过\r\n啤酒\r\n小心\r\n照顾\r\n锻炼\r\n过去\r\n忘记\r\n哭\r\n包\r\n个子\r\n瘦\r\n选择\r\n奶奶\r\n突然\r\n节目\r\n']
res = A[0].split()
print(res)
['过', '啤酒', '小心', '照顾', '锻炼', '过去', '忘记', '哭', '包', '个子', '瘦', '选择', '奶奶', '突然', '节目']
As described in the str.split docs:
If sep is not specified or is None, a different splitting
algorithm is applied: runs of consecutive whitespace are regarded as a
single separator, and the result will contain no empty strings at the
start or end if the string has leading or trailing whitespace.

To limit the split to \r\n you can use .splitlines():
>>> li=['过\r\n啤酒\r\n小心\r\n照顾\r\n锻炼\r\n过去\r\n忘记\r\n哭\r\n包\r\n个子\r\n瘦\r\n选择\r\n奶奶\r\n突然\r\n节目\r\n']
>>> li[0].splitlines()
['过', '啤酒', '小心', '照顾', '锻炼', '过去', '忘记', '哭', '包', '个子', '瘦', '选择', '奶奶', '突然', '节目']

Try this:
s = "['过\r\n啤酒\r\n小心\r\n照顾\r\n锻炼\r\n过去\r\n忘记\r\n哭\r\n包\r\n个子\r\n瘦\r\n选择\r\n奶奶\r\n突然\r\n节目\r\n']"
s = s.replace('\r\n', ',').replace("'", '')
print(s)
Output:
[过,啤酒,小心,照顾,锻炼,过去,忘记,哭,包,个子,瘦,选择,奶奶,突然,节目,]
This first replace replaces the \r\n and the second one replaces the single quote from the string as you expected as the output.

Stripping Hex code from a plain text file in Python [duplicate]

I have a string. How do I remove all text after a certain character? (In this case ...)
The text after will ... change so I that's why I want to remove all characters after a certain one.

Split on your separator at most once, and take the first piece:
sep = '...'
stripped = text.split(sep, 1)[0]
You didn't say what should happen if the separator isn't present. Both this and Alex's solution will return the entire string in that case.

Assuming your separator is '...', but it can be any string.
text = 'some string... this part will be removed.'
head, sep, tail = text.partition('...')
>>> print head
some string
If the separator is not found, head will contain all of the original string.
The partition function was added in Python 2.5.
S.partition(sep) -> (head, sep, tail)
Searches for the separator sep in S, and returns the part before it,
the separator itself, and the part after it. If the separator is not
found, returns S and two empty strings.

If you want to remove everything after the last occurrence of separator in a string I find this works well:
<separator>.join(string_to_split.split(<separator>)[:-1])
For example, if string_to_split is a path like root/location/child/too_far.exe and you only want the folder path, you can split by "/".join(string_to_split.split("/")[:-1]) and you'll get
root/location/child

Without a regular expression (which I assume is what you want):
def remafterellipsis(text):
where_ellipsis = text.find('...')
if where_ellipsis == -1:
return text
return text[:where_ellipsis + 3]
or, with a regular expression:
import re
def remwithre(text, there=re.compile(re.escape('...')+'.*')):
return there.sub('', text)

import re
test = "This is a test...we should not be able to see this"
res = re.sub(r'\.\.\..*',"",test)
print(res)
Output: "This is a test"

The method find will return the character position in a string. Then, if you want remove every thing from the character, do this:
mystring = "123⋯567"
mystring[ 0 : mystring.index("⋯")]
>> '123'
If you want to keep the character, add 1 to the character position.

From a file:
import re
sep = '...'
with open("requirements.txt") as file_in:
lines = []
for line in file_in:
res = line.split(sep, 1)[0]
print(res)

This is in python 3.7 working to me
In my case I need to remove after dot in my string variable fees
fees = 45.05
split_string = fees.split(".", 1)
substring = split_string[0]
print(substring)

Yet another way to remove all characters after the last occurrence of a character in a string (assume that you want to remove all characters after the final '/').
path = 'I/only/want/the/containing/directory/not/the/file.txt'
while path[-1] != '/':
path = path[:-1]

another easy way using re will be
import re, clr
text = 'some string... this part will be removed.'
text= re.search(r'(\A.*)\.\.\..+',url,re.DOTALL|re.IGNORECASE).group(1)
// text = some string

Understanding string method strip

After initializing a variable x with the content shown in below, I applied strip with a parameter. The result of strip is unexpected. As I'm trying to strip "ios_static_analyzer/", "rity/ios_static_analyzer/" is getting striped.
Kindly help me know why is it so.
>>> print x
/Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer/
>>> print x.strip()
/Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer/
>>> print x.strip('/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer
>>> print x.strip('ios_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/secu
>>> print x.strip('analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_
>>> print x.strip('_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static
>>> print x.strip('static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/io
>>> print x.strip('_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/io
>>> print x.strip('s_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/io
>>> print x.strip('os_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/secu

Quoting from str.strip docs
Return a copy of the string with the leading and trailing characters
removed. The chars argument is a string specifying the set of
characters to be removed. If omitted or None, the chars argument
defaults to removing whitespace. The chars argument is not a prefix or
suffix; rather, all combinations of its values are stripped:
So, it removes all the characters in the parameter, from both the sides of the string.
For example,
my_str = "abcd"
print my_str.strip("da") # bc
Note: You can think of it like this, it stops removing the characters from the string when it finds a character which is not found in the input parameter string.
To actually, remove the particular string, you should use str.replace
x = "/Users/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer/"
print x.replace('analyzer/', '')
# /Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_
But replace will remove the matches everywhere,
x = "abcd1abcd2abcd"
print x.replace('abcd', '') # 12
But if you want to remove words only at the beginning and ending of the string, you can use RegEx, like this
import re
pattern = re.compile("^{0}|{0}$".format("abcd"))
x = "abcd1abcd2abcd"
print pattern.sub("", x) # 1abcd2

What you need, I think, is replace:
>>> x.replace('ios_static_analyzer/','')
'/Users/msecurity/Desktop/testspace/Hy5_Workspace/security/'
string.replace(s, old, new[, maxreplace])
Return a copy of string s with all occurrences of substring old replaced by new.
So you can replace your string with nothing and get the desired output.

Python x.strip(s) remove from the begginning or the end of the string x any character appearing in s ! So s is just a set of characters, not a string being matched for substring.

string.strip removes a set of characters given as an argument. The chars argument is not a prefix or suffix; rather, all combinations of its values are stripped.

strip does not remove the string given as argument from the object; it removes the characters in the argument.
In this case, strip sees the string s_static_analyzer/ as an iterable of characters that needs to be stripped.

How to delete some characters from a string by matching certain character in python

i am trying to delete certain portion of a string if a match found in the string as below
string = 'Newyork, NY'
I want to delete all the characters after the comma from the string including comma, if comma is present in the string
Can anyone let me now how to do this .

Use .split():
string = string.split(',', 1)[0]
We split the string on the comma once, to save python the work of splitting on more commas.
Alternatively, you can use .partition():
string = string.partition(',')[0]
Demo:
>>> 'Newyork, NY'.split(',', 1)[0]
'Newyork'
>>> 'Newyork, NY'.partition(',')[0]
'Newyork'
.partition() is the faster method:
>>> import timeit
>>> timeit.timeit("'one, two'.split(',', 1)[0]")
0.52929401397705078
>>> timeit.timeit("'one, two'.partition(',')[0]")
0.26499605178833008

You can split the string with the delimiter ",":
string.split(",")[0]
Example:
'Newyork, NY'.split(",") # ['Newyork', ' NY']
'Newyork, NY'.split(",")[0] # 'Newyork'

Try this :
s = "this, is"
m = s.index(',')
l = s[:m]

A fwe options:
string[:string.index(",")]
This will raise a ValueError if , cannot be found in the string. Here, we find the position of the character with .index then use slicing.
string.split(",")[0]
The split function will give you a list of the substrings that were separated by ,, and you just take the first element of the list. This will work even if , is not present in the string (as there'd be nothing to split in that case, we'd have string.split(...) == [string])