Develop Reference

Why is the split() returning list objects that are empty? [duplicate]

I have the following file names that exhibit this pattern:
000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...
I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:
time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
But this gives me two extra empty strings in the returned list:
time_info=['', '20111007T084734', '20111008T023142', '']
How do I get only the two time stamp information? i.e. I want:
time_info=['20111007T084734', '20111008T023142']

I'm no Python expert but maybe you could just remove the empty strings from your list?
str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)

Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.
>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')
You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')

If the timestamps are always after the second _ then you can use str.split and str.strip:
>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']

Since this came up on google and for completeness, try using re.findall as an alternative!
This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.
Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.

>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']
or, somewhat more general:
>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']

Python: list strip overkill

I just want to remove the '.SI' in the list but it will overkill by remove any that contain S or I in the list.
ab = ['abc.SI','SIV.SI','ggS.SI']
[x.strip('.SI') for x in ab]
>> ['abc','V','gg']
output which I want is
>> ['abc','SIV','ggS']
any elegant way to do it? prefer not to use for loop as my list is long

Why strip ? you can use .replace():
[x.replace('.SI', '') for x in ab]
Output:
['abc', 'SIV', 'ggS']
(this will remove .SI anywhere, have a look at other answers if you want to remove it only at the end)
The reason strip() doesn't work is explained in the docs:
The chars argument is not a prefix or suffix; rather, all combinations of its values are stripped
So it will strip any character in the string that you pass as an argument.

If you want to remove the substring only from the end, the correct way to achieve this will be:
>>> ab = ['abc.SI','SIV.SI','ggS.SI']
>>> sub_string = '.SI'
# checks the presence of substring at the end
# v
>>> [s[:-len(sub_string)] if s.endswith(sub_string) else s for s in ab]
['abc', 'SIV', 'ggS']
Because str.replace() (as mentioned in TrakJohnson's answer) removes the substring even if it is within the middle of string. For example:
>>> 'ab.SIrt'.replace('.SI', '')
'abrt'

use this [x[:-3] for x in ab].

Use split instead of strip and get the first element:
[x.split('.SI')[0] for x in ab]

Python split a string at an underscore

How do I split a string at the second underscore in Python so that I get something like this
name = this_is_my_name_and_its_cool
split name so I get this ["this_is", "my_name_and_its_cool"]

the following statement will split name into a list of strings
a=name.split("_")
you can combine whatever strings you want using join, in this case using the first two words
b="_".join(a[:2])
c="_".join(a[2:])
maybe you can write a small function that takes as argument the number of words (n) after which you want to split
def func(name, n):
a=name.split("_")
b="_".join(a[:n])
c="_".join(a[n:])
return [b,c]

Assuming that you have a string with multiple instances of the same delimiter and you want to split at the nth delimiter, ignoring the others.
Here's a solution using just split and join, without complicated regular expressions. This might be a bit easier to adapt to other delimiters and particularly other values of n.
def split_at(s, c, n):
words = s.split(c)
return c.join(words[:n]), c.join(words[n:])
Example:
>>> split_at('this_is_my_name_and_its_cool', '_', 2)
('this_is', 'my_name_and_its_cool')

I think you're trying the split the string based on second underscore. If yes, then you used use findall function.
>>> import re
>>> s = "this_is_my_name_and_its_cool"
>>> re.findall(r'^[^_]*_[^_]*|[^_].*$', s)
['this_is', 'my_name_and_its_cool']
>>> [i for i in re.findall(r'^[^_]*_[^_]*|(?!_).*$', s) if i]
['this_is', 'my_name_and_its_cool']

print re.split(r"(^[^_]+_[^_]+)_","this_is_my_name_and_its_cool")
Try this.

Here's a quick & dirty way to do it:
s = 'this_is_my_name_and_its_cool'
i = s.find('_'); i = s.find('_', i+1)
print [s[:i], s[i+1:]]
output
['this_is', 'my_name_and_its_cool']
You could generalize this approach to split on the nth separator by putting the find() into a loop.

How do I strip a string given a list of unwanted characters? Python

Is there a way to pass in a list instead of a char to str.strip() in python? I have been doing it this way:
unwanted = [c for c in '!##$%^&*(FGHJKmn']
s = 'FFFFoFob*&%ar**^'
for u in unwanted:
s = s.strip(u)
print s
Desired output, this output is correct but there should be some sort of a more elegant way than how i'm coding it above:
oFob*&%ar

Strip and friends take a string representing a set of characters, so you can skip the loop:
>>> s = 'FFFFoFob*&%ar**^'
>>> s.strip('!##$%^&*(FGHJKmn')
'oFob*&%ar'
(the downside of this is that things like fn.rstrip(".png") seems to work for many filenames, but doesn't really work)

Since, you are looking to not delete elements from the middle, you can just use.
>>> 'FFFFoFob*&%ar**^'.strip('!##$%^&*(FGHJKmn')
'oFob*&%ar'
Otherwise, Use str.translate().
>>> 'FFFFoFob*&%ar**^'.translate(None, '!##$%^&*(FGHJKmn')
'oobar'

Python regular words cut

I have string: './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
I need string: '27-10-2011 17:07:02'
How can i do this in python?

There are many ways to do this, one way is to use str.partition:
text='./money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
before,_,after = text.partition('[')
print(after[:-1])
# 27-10-2011 17:07:02
Another is to use str.split:
before,after = text.split('[',1)
print(after[:-1])
# 27-10-2011 17:07:02
or str.find and str.rfind:
ind1 = text.find('[')+1
ind2 = text.rfind(']')
print(text[ind1:ind2])
All these methods rely on the desired substring immediately following the first left-bracket [.
The first two methods also rely on the desired substring ending at the next-to-last character in text. The last method (using rfind) searches from the right for the index of the right-bracket, so it is a little more general, and does not depend on quite so many (potential off-by-one) constants.

If your string has always the same structure this is probably the simplest solution:
s = r'./money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
s[s.find("[")+1:s.find("]")]
Update:
After seeing some of the other answers this is a slight improvement:
s[s.find("[")+1:-1]
Exploiting the fact that the closing square bracket is the last character in your string.

If the format is "fixed", you can also use this
>>> s = './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
>>> s[-20:-1:]
'27-10-2011 17:07:02'
>>>

You can also use regular expression:
import re
s = './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
print re.search(r'\[(.*?)\]', s).group(1)

Try with a regex :
import re
re.findall(".*\[(.*)\]", './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]')
>>> ['27-10-2011 17:07:02']

Probably the easiest way(if you know the string will always be in this format
>>> s = './money.log_rotated.27.10.2011_17:15:01:[27-10-2011 17:07:02]'
>>> s[s.index('[') + 1:-1]
'27-10-2011 17:07:02'