There was one more error. (smiles ;) Fixed below
def all_equal_array_list(alist) -> bool:
if alist[1:] == alist[:-1]:
return True
return False
def all_equal(alist) -> bool:
if len(alist) == 0 or all(i == alist[0] for i in alist[1:]):
return True
return False
There were some issues with my test. After sending the email, I thought
those times couldn't be real. Here are better results:
all_equal(tlst) times = [6.719925760501064e-05] seconds.
all_equal_array_list(tlst) times = [4.2184268278069794e-06] seconds.
import timeit
def all_equal(alist)
On 08Jan2019 15:28, Dan Sommers <2qdxy4rzwzuui...@potatochowder.com> wrote:
>>>>> a = [1, 1, 1, 1, 1]
>>>>> a[1:] == a[:-1]
>>True
>>>>> a == a[::-1]
>>True
>>
>>>>> a = [1, 2, 3, 4, 3, 2, 1]
>>>>> a[1:] == a[:-1]
>>False
>>>>> a == a[::-1]
>>True
On 08Jan2019 15:28, Dan Sommers <2qdxy4rzwzuui...@potatochowder.com> wrote:
>>> a = [1, 1, 1, 1, 1]
>>> a[1:] == a[:-1]
True
>>> a == a[::-1]
True
>>> a = [1, 2, 3, 4, 3, 2, 1]
>>> a[1:] == a[:-1]
False
>>> a == a[::-1]
True
Looks like Peter's pretty clever after
On 1/8/19 2:31 PM, Tobiah wrote:> On 1/8/19 9:20 AM, Alister wrote:
>> On Tue, 08 Jan 2019 17:15:17 +, Alister wrote:
>>
>>> On Tue, 08 Jan 2019 16:48:58 +0200, Serhiy Storchaka wrote:
>>>
08.01.19 11:07, Peter Otten пише:
> Bob van der Poel wrote:
>
>> I need to see if all
On 1/8/19 9:20 AM, Alister wrote:
On Tue, 08 Jan 2019 17:15:17 +, Alister wrote:
On Tue, 08 Jan 2019 16:48:58 +0200, Serhiy Storchaka wrote:
08.01.19 11:07, Peter Otten пише:
Bob van der Poel wrote:
I need to see if all the items in my list are the same. I was using
set()
for this,
On Tue, Jan 08, 2019 at 01:05:09PM +1100, Chris Angelico wrote:
> No, because the iteration would never happen. If the list is empty,
> a[1:] is also empty, and i==a[0] will never be evaluated. So it is
> safe. (But I agree that it's not instantly obvious.)
Oh yeah, you're right. I overthought it
On Tue, 08 Jan 2019 17:15:17 +, Alister wrote:
> On Tue, 08 Jan 2019 16:48:58 +0200, Serhiy Storchaka wrote:
>
>> 08.01.19 11:07, Peter Otten пише:
>>> Bob van der Poel wrote:
>>>
I need to see if all the items in my list are the same. I was using
set()
for this, but that
On Tue, 08 Jan 2019 16:48:58 +0200, Serhiy Storchaka wrote:
> 08.01.19 11:07, Peter Otten пише:
>> Bob van der Poel wrote:
>>
>>> I need to see if all the items in my list are the same. I was using
>>> set()
>>> for this, but that doesn't work if items are themselves lists. So,
>>> assuming that
Thanks guys for the help on this. As it turns out I have learned new
commands as a result of the question: all() and any() will work perfectly!
The other solutions, as always, are enlightening. And, no, Chris, this is
not homework :)
On Tue, Jan 8, 2019 at 9:31 AM Neil Cerutti wrote:
> On
On 2019-01-08, MRAB wrote:
> On 2019-01-08 00:47, i...@koeln.ccc.de wrote:
>> You might do something like
>>
>> if len(a) == 0 or all(i == a[0] for i in a[1:]):
>>
> You don't need to check the length of the list because if the list is
> empty, 'all' will return True anyway.
Neat! I
08.01.19 11:07, Peter Otten пише:
Bob van der Poel wrote:
I need to see if all the items in my list are the same. I was using set()
for this, but that doesn't work if items are themselves lists. So,
assuming that a is a list of some things, the best I've been able to come
up with it:
if
Bob van der Poel wrote:
> I need to see if all the items in my list are the same. I was using set()
> for this, but that doesn't work if items are themselves lists. So,
> assuming that a is a list of some things, the best I've been able to come
> up with it:
>
> if a.count( targ ) == len(a):
Tim Chase wrote:
> def all_equal(iterable):
> i = iter(iterable)
> first = next(i)
> return all(x == first for x in i)
>
> It's undefined for an empty list (well, it throws a StopIteration
> but you can special-case that), but should hand the cases with
> 1 element and 2+ elements
On Tue, Jan 8, 2019 at 1:03 PM Skip Montanaro wrote:
>
> >
> > > if len(a) == 0 or all(i == a[0] for i in a[1:]):
> > >
> > You don't need to check the length of the list because if the list is
> > empty, 'all' will return True anyway.
> >
>
> Given the structure of the expression passed as
>
> > if len(a) == 0 or all(i == a[0] for i in a[1:]):
> >
> You don't need to check the length of the list because if the list is
> empty, 'all' will return True anyway.
>
Given the structure of the expression passed as an argument to all(), won't
you get an IndexError if a is empty without
On 2019-01-08 00:47, i...@koeln.ccc.de wrote:
You might do something like
if len(a) == 0 or all(i == a[0] for i in a[1:]):
You don't need to check the length of the list because if the list is
empty, 'all' will return True anyway.
This should be linear complexity and short circuiting
On Tue, Jan 8, 2019 at 12:26 PM Tim Chase wrote:
> def all_equal(iterable):
> i = iter(iterable)
> first = next(i)
> return all(x == first for x in i)
>
> And I even like how nicely it reads :-)
Yes, there's something beautiful about writing "first = next" :-)
ChrisA
--
On 2019-01-07 17:14, Bob van der Poel wrote:
> I need to see if all the items in my list are the same. I was using
> set() for this, but that doesn't work if items are themselves
> lists. So, assuming that a is a list of some things, the best I've
> been able to come up with it:
>
> if
On 07Jan2019 17:14, bvdp wrote:
I need to see if all the items in my list are the same. I was using set()
for this, but that doesn't work if items are themselves lists. So, assuming
that a is a list of some things, the best I've been able to come up with it:
if a.count( targ ) == len(a):
You might do something like
if len(a) == 0 or all(i == a[0] for i in a[1:]):
This should be linear complexity and short circuiting and in general it
doesn't get much better than this. Though I wouldn't bet there isn't a
better (faster/clearer/more readable) solution.
On Mon, Jan 07, 2019 at
On 2019-01-08 00:14, Bob van der Poel wrote:
I need to see if all the items in my list are the same. I was using set()
for this, but that doesn't work if items are themselves lists. So, assuming
that a is a list of some things, the best I've been able to come up with it:
if a.count( targ )
I need to see if all the items in my list are the same. I was using set()
for this, but that doesn't work if items are themselves lists. So, assuming
that a is a list of some things, the best I've been able to come up with it:
if a.count( targ ) == len(a):
I'm somewhat afraid that this won't
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