spilt question

[loial]

I want to split a string so that I always return everything BEFORE the LAST underscore

HELLO_xxxxxxxx.lst # should return HELLO
HELLO_GOODBYE_xxxxxxxx.ls # should return HELLO_GOODBYE

I have tried with rsplit but cannot get it to work.

Any help appreciated

[Walter Hurry]

mystr.rsplit("_",1)[0]

[Fábio Santos]

str.split takes a limit argument. Try your_string.split('_', 1)

--
http://mail.python.org/mailman/listinfo/python-list

[Chris Angelico]

Try with a limit:

>>> "HELLO_GOODBYE_xxxxxxxx.ls".rsplit("_",1)
['HELLO_GOODBYE', 'xxxxxxxx.ls']
>>> "HELLO_GOODBYE_xxxxxxxx.ls".rsplit("_",1)[0]
'HELLO_GOODBYE'

You can easily get docs on it:

>>> help("".rsplit)
Help on built-in function rsplit:

rsplit(...)
S.rsplit(sep=None, maxsplit=-1) -> list of strings

Return a list of the words in S, using sep as the
delimiter string, starting at the end of the string and
working to the front. If maxsplit is given, at most maxsplit
splits are done. If sep is not specified, any whitespace string
is a separator.

ChrisA

[Dave Angel]

The rsplit() method was a good idea; it will work. So exactly what did
you try and what about it did not work. How are you stuck going from
your partial answer to a complete one?

While you're at it, you need to consider at least two more cases

HELLO.txt #should return ""
HELLO_and_____goodbye #should return "HELLO_and____"

--
DaveA

[Tim Chase]

.rsplit takes an optional "how many splits do you want?" parameter
that defaults to giving you all of them. Just ask for one
right-most split:

TESTS = [
("HELLO_xxxxxxx.lst", "HELLO"),
("HELLO_GOODBYE_xxxxx.ls", "HELLO_GOODBYE"),
]

for input, expected in TESTS:
result = input.rsplit('_', 1)[0]
if result == expected:
verdict = "passed"
else:
verdict = "failed"
print "%r -> %r == %r (%s)" % (
input,
result,
expected,
verdict,
)

-tkc

[Ned Batchelder]

>>> "HELLO_GOODBYE_xxxxxxxx.ls".rpartition('_')
('HELLO_GOODBYE', '_', 'xxxxxxxx.ls')
>>> "HELLO_GOODBYE_xxxxxxxx.ls".rsplit('_', 1)
['HELLO_GOODBYE', 'xxxxxxxx.ls']
>>> "HELLO_GOODBYE_xxxxxxxx.ls".rpartition('_')[0]
'HELLO_GOODBYE'
>>> "HELLO_GOODBYE_xxxxxxxx.ls".rsplit('_', 1)[0]
'HELLO_GOODBYE'

For the future, getting help works better if you show what you tried, and what it produced.

--Ned.

[Dave Angel]

On 05/16/2013 11:15 AM, Chris Angelico wrote:
> On Fri, May 17, 2013 at 1:00 AM, loial <jldun...@gmail.com> wrote:

> Try with a limit:
>
>>>> "HELLO_GOODBYE_xxxxxxxx.ls".rsplit("_",1)
> ['HELLO_GOODBYE', 'xxxxxxxx.ls']
>>>> "HELLO_GOODBYE_xxxxxxxx.ls".rsplit("_",1)[0]
> 'HELLO_GOODBYE'
>
> You can easily get docs on it:
>
>>>> help("".rsplit)
> Help on built-in function rsplit:
>
> rsplit(...)
> S.rsplit(sep=None, maxsplit=-1) -> list of strings
>
> Return a list of the words in S, using sep as the
> delimiter string, starting at the end of the string and
> working to the front. If maxsplit is given, at most maxsplit
> splits are done. If sep is not specified, any whitespace string
> is a separator.
>
> ChrisA
>

Now that you've all done his homework for him, see if The OP can spot
the one case where that won't work. It's easy to test for, but still
important to get right (for some definition of right).

--
DaveA