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How to exactly match a word and then replace in Python?

Have a list which contains strings etc. as shown below
strpool = ['fruit,apple:3', '', '[1,abcd, ['fruit,apple'], ['1,kdlld', apple,taste]]']
Wanted to search exactly for the word 'apple' and replace with 'apple:3'
I tried the below code,
print str(strpool).replace("apple","apple:3")
print (re.sub(r"\bapple\b","apple:3",str(strpool)))
But its replacing even the apple:3 as well into apple:3:3 , not just the string apple.
Thought apple:3 would be considered as a string and that doesn't get changed.
Update:
How can I exactly match a string name without any other components attached to it and replace all of them inside a list ?
Tried re.sub but for that need to convert the list to string, instead is there any other way ?

Use a negative lookahead to match a word unless it's followed by :.
print(re.sub(r'\bapple\b(?!:)', 'apple:3', str(strpool))

Replace text between parentheses in python

My string will contain () in it. What I need to do is to change the text between the brackets.
Example string: "B.TECH(CS,IT)".
In my string I need to change the content present inside the brackets to something like this.. B.TECH(ECE,EEE)
What I tried to resolve this problem is as follows..
reg = r'(()([\s\S]*?)())'
a = 'B.TECH(CS,IT)'
re.sub(reg,"(ECE,EEE)",a)
But I got output like this..
'(ECE,EEE)B(ECE,EEE).(ECE,EEE)T(ECE,EEE)E(ECE,EEE)C(ECE,EEE)H(ECE,EEE)((ECE,EEE)C(ECE,EEE)S(ECE,EEE),(ECE,EEE)I(ECE,EEE)T(ECE,EEE))(ECE,EEE)'
Valid output should be like this..
B.TECH(CS,IT)
Where I am missing and how to correctly replace the text.

The problem is that you're using parentheses, which have another meaning in RegEx. They're used as grouping characters, to catch output.
You need to escape the () where you want them as literal tokens. You can escape characters using the backslash character: \(.
Here is an example:
reg = r'\([\s\S]*\)'
a = 'B.TECH(CS,IT)'
re.sub(reg, '(ECE,EEE)', a)
# == 'B.TECH(ECE,EEE)'

The reason your regex does not work is because you are trying to match parentheses, which are considered meta characters in regex. () actually captures a null string, and will attempt to replace it. That's why you get the output that you see.
To fix this, you'll need to escape those parens – something along the lines of
\(...\)
For your particular use case, might I suggest a simpler pattern?
In [268]: re.sub(r'\(.*?\)', '(ECE,EEE)', 'B.TECH(CS,IT)')
Out[268]: 'B.TECH(ECE,EEE)'

Dynamically Removing string with regex python

I am currently having trouble removing the end of strings using regex. I have tried using .partition with unsuccessful results. I am now trying to use regex unsuccessfully. All the strings follow the format of some random words **X*.* Some more words. Where * is a digit and X is a literal X. For Example 21X2.5. Everything after this dynamic string should be removed. I am trying to use re.sub('\d\d\X\d.\d', string). Can someone point me in the right direction with regex and how to split the string?
The expected output should read:
some random words 21X2.5
Thanks!

Use following regex:
re.search("(.*?\d\dX\d\.\d)", "some random words 21X2.5 Some more words").groups()[0]
Output:
'some random words 21X2.5'

Your regex is not correct. The biggest problem is that you need to escape the period. Otherwise, the regex treats the period as a match to any character. To match just that pattern, you can use something like:
re.findall('[\d]{2}X\d\.\d', 'asb12X4.4abc')
[\d]{2} matches a sequence of two integers, X matches the literal X, \d matches a single integer, \. matches the literal ., and \d matches the final integer.
This will match and return only 12X4.4.
It sounds like you instead want to remove everything after the matched expression. To get your desired output, you can do something like:
re.split('(.*?[\d]{2}X\d\.\d)', 'some random words 21X2.5 Some more words')[1]
which will return some random words 21X2.5. This expression pulls everything before and including the matched regex and returns it, discarding the end.
Let me know if this works.

To remove everything after the pattern, i.e do exactly as you say...:
s = re.sub(r'(\d\dX\d\.\d).*', r'\1', s)
Of course, if you mean something else than what you said, something different will be needed! E.g if you want to also remove the pattern itself, not just (as you said) what's after it:
s = re.sub(r'\d\dX\d\.\d.*', r'', s)
and so forth, depending on what, exactly, are your specs!-)

Using Python regex to sanitize input string

I have the string:
text = 'href = "www.google.com" onmouseover = blahblah >'
I want 'href = "www.google.com">'
Currently, my function looks like this:
text = re.sub(r'href = \".*\".*>', 'href = \".*\">', text)
which ends up removing the website link and replacing it with the string '.*' . I think I'm supposed to use ?Pname somehow?, but do not know ho to write it properly so that I get the correct output.

You don't want to substitute in .*, you want to substitute in whatever the first .* matched.
To do that, you need a backreference, like \1.
And this means you need something for the backreference to refer back to—a capture group, like (.*) instead of .*.
More generally, the replacement string is not a regular expression, it's a different kind of thing—basically, it's a template that's all literal characters except for backreferences.* So, you don't want to try to escape the quotes, unless you want literal backslashes in the results.
So:
>>> re.sub(r'href = \"(.*)\".*>', r'href = "\1">', text)
'href = "www.google.com">'
This is explained in more detail in Search and Replace in the Regular Expression HOWTO.
* Or it can be a function which takes each match object and returns a string.

An alternative way to accomplish your goal is to take a substring. No regular expression is needed. The idea is to find the second double-quote character using the string method index().
For a string called input, this expression gives you the position of the second double-quote character:
input.index('"', input.index('"')+1)
If that value is k, write input[:k+1] to extract everything up to and including the second double-quote character.
Try out the following in your Python interpreter.
input = 'href = "www.google.com" onmouseover=hax0rFunction()>'
k = input.index('"', input.index('"')+1)
input[0:k+1]

re.sub repl function returning \1 does not replace the group

I am trying to write a generic replace function for a regex sub operation in Python (trying in both 2 and 3) Where the user can provide a regex pattern and a replacement for the match. This could be just a simple string replacement to replacing using the groups from the match.
In the end, I get from the user a dictionary in this form:
regex_dict = {pattern:replacement}
When I try to replace all the occurrences of a pattern via this command, the replacement works for replacements for a group number, (such as \1) and I call the following operation:
re.sub(pattern, regex_dict[pattern], text)
This works as expected, but I need to do additional stuff when a match is found. Basically, what I try to achieve is as follows:
replace_function(matchobj):
result = regex_dict[matchobj.re]
##
## Do some other things
##
return result
re.sub(pattern, replace_function, text)
I see that this works for normal replacements, but the re.sub does not use the group information to get the match when the function is used.
I also tried to convert the \1 pattern to \g<1>, hoping that the re.sub would understand it, but to no avail.
Am I missing something vital?
Thanks in advance!
Additional notes: I compile the pattern using strings as in bytes, and the replacements are also in bytes. I have non-Latin characters in my pattern, but I read everything in bytes, including the text where the regex substitution will operate on.
EDIT
Just to clarify, I do not know in advance what kind of replacement the user will provide. It could be some combination of normal strings and groups, or just a string replacement.
SOLUTION
replace_function(matchobj):
repl = regex_dict[matchobj.re]
##
## Do some other things
##
return matchobj.expand(repl)
re.sub(pattern, replace_function, text)

I suspect you're after .expand, if you've got a compiled regex object (for instance), you can provide a string to be taken into consideration for the replacements, eg:
import re
text = 'abc'
# This would be your key in the dict
rx = re.compile('a(\w)c')
# This would be the value for the key (the replacement string, eg: `\1\1\1`)
res = rx.match(text).expand(r'\1\1\1')
# bbb