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How to replace comma with space in python list

I want to replace comma to space in the list. How can i do that? Thanks
input:
host_dict['actives'] = list(get_po_bound_ints['result'][1]['portChannels'][mac_to_eth2]['activePorts'].keys())
output:
[{'actives': ['PeerEthernet23',
'PeerEthernet24',
'Ethernet23',
'Ethernet22'],

Replacing commas with empty strings (assuming your list is named my_list):
print(str(my_list).replace(',', ''))

If I understand your question, you have an iterable that you want to convert to a space-separated string. The str.join method does that:
>>> test = ['PeerEthernet23', 'PeerEthernet24', 'Ethernet23', 'Ethernet22']
>>> " ".join(test)
'PeerEthernet23 PeerEthernet24 Ethernet23 Ethernet22'
your script would be
host_dict['actives'] = " ".join(get_po_bound_ints['result'][1]
['portChannels'][mac_to_eth2]['activePorts'].keys())

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()

Splitting string in Python starting from the first numeric character

I need to manage string in Python in this way:
I have this kind of strings with '>=', '=', '<=', '<', '>' in front of them, for example:
'>=1_2_3'
'<2_3_2'
what I want to achieve is splitting the strings to obtain, respectively:
'>=', '1_2_3'
'<', '2_3_2'
basically I need to split them starting from the first numeric character.
There's a way to achieve this result with regular expressions without iterating over the string checking if a character is a number or a '_'?
thank you.

This will do:
re.split(r'(^[^\d]+)', string)[1:]
Example:
>>> re.split(r'(^[^\d]+)', '>=1_2_3')[1:]
['>=', '1_2_3']
>>> re.split(r'(^[^\d]+)', '<2_3_2')[1:]
['<', '2_3_2']

import re
strings = ['>=1_2_3','<2_3_2']
for s in strings:
mat = re.match(r'([^\d]*)(\d.*)', s)
print mat.groups()
Outputs:
('>=', '1_2_3')
('<', '2_3_2')
This just groups everything up until the first digit in one group, then that first digit and everything after into a second.
You can access the individual groups with mat.group(1), mat.group(2)

You can split using this regex:
(?<=[<>=])(?=\d)
RegEx Demo

There's probably a better way but you can split with a capture then join the second two elements:
values = re.split(r'(\d)', '>=1_2_3', maxsplit = 1)
values = [values[0], values[1] + values[2]]

how to split brackets using python abcd[00451.00]

I have tried below code to split but I am unable to split
import re
s = "abcd[00451.00]"
print str(s).strip('[]')
I need output as only number or decimal format 00451.00 this value but I am able to get output as abcd[00451.00

If you know for sure that there will be one opening and closing brackets you can do
s = "abcd[00451.00]"
print s[s.index("[") + 1:s.rindex("]")]
# 00451.00
str.index is used to get the first index of the element [ in the string, where as str.rindex is used to get the last index of the element in ]. Based on those indexes, the string is sliced.
If you want to convert that to a floating point number, then you can use float function, like this
print float(s[s.index("[") + 1:s.rindex("]")])
# 451.0

You should use re.search:
import re
s = "abcd[00451.00]"
>>> print re.search(r'\[([^\]]+)\]', s).group(1)
00451.00

You can first split on the '[' and then strip the resulting list of any ']' chars:
[p.strip(']') for p in s.split('[')]

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'