Steven D'Aprano:
> I believe that it is considered an abuse of doctest to write a function
> with 28 lines of code and 19 tests
I agree with you. Most of my functions/methods have more compact
doctests (putting things on many lines has the advantage of letting
you locate the failing tests in a sim
On 9 Feb, 15:20, Steven D'Aprano <[EMAIL PROTECTED]
cybersource.com.au> wrote:
> On Sat, 09 Feb 2008 05:44:27 -0800, bearophileHUGS wrote:
> > def issubseq(sub, items):
> > """issubseq(sub, items): return true if the sequence 'sub' is a
> > contiguous subsequence of the 'items' sequence.
>
> [s
On Sat, 09 Feb 2008 05:44:27 -0800, bearophileHUGS wrote:
> def issubseq(sub, items):
> """issubseq(sub, items): return true if the sequence 'sub' is a
> contiguous subsequence of the 'items' sequence.
[snip]
A stylistic note for you...
I believe that it is considered an abuse of doctest to
Steven D'Aprano:
> What you're trying to do is the equivalent of substring searching. There
> are simple algorithms that do this, and there are fast algorithms, but
> the simple ones aren't fast and the fast ones aren't simple.
> However, for your use case, the simple algorithm is probably fast eno
On Fri, 08 Feb 2008 16:39:55 +, Matthew_WARREN wrote:
> Hallo,
>
>
> I need to search list a for the sequence of list b
>
> First I went
a=[1,2,34,4,5,6]
b=[2,3,4]
a in b
> False
You have the order of a and b mixed up there. As you have it, you are
searching b for a.
>
[EMAIL PROTECTED] wrote:
> Hallo,
>
>
> I need to search list a for the sequence of list b
>
I guess most pythonic would be sth like this:
[code]
def naive_seq_match(sequence, pattern):
"""Serches for pattern in sequence. If pattern is found
returns match start index, else returns No
On Feb 8, 5:06 pm, Paul Hankin <[EMAIL PROTECTED]> wrote:
> [EMAIL PROTECTED] wrote:
> > I need to search list a for the sequence of list b
>
> def list_contains(a, b):
> return any(a[i:i+len(b)] == b for i in range(len(a) - len(b) + 1))
>
> list_contains(range(1, 7), [2, 3, 4])
>
> --
> Paul H
[EMAIL PROTECTED] wrote:
> I need to search list a for the sequence of list b
def list_contains(a, b):
return any(a[i:i+len(b)] == b for i in range(len(a) - len(b) + 1))
list_contains(range(1, 7), [2, 3, 4])
--
Paul Hankin
--
http://mail.python.org/mailman/listinfo/python-list
Hallo,
I need to search list a for the sequence of list b
First I went
>>> a=[1,2,34,4,5,6]
>>> b=[2,3,4]
>>> a in b
False
So
''.join([ v.__str__() for v in b ]) in ''.join([ v.__str__() for v in a ])
>>> s=SomeObject()
>>> a=[1,2,3,[s,3,[4,s,9]],s,4]
>>> b=[3,[s,3,[4,s,9]],s,4]
>>> ''.join([
