Are all items in list the same?

[Karen Shaeffer]

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) -> bool:

    if len(alist) == 0 or all(i == alist[0] for i in alist[1:]):

        return True

    return False


def all_equal_array_list(alist) -> bool:

    if alist == alist[:-1]:

        return True

    return False


tlst = [42 for _j in range(5000)]


def timeit_all_equal() -> bool:

    return all_equal(tlst)


def timeit_all_equal_array_list() -> bool:

    return all_equal_array_list(tlst)


if __name__ == '__main__':

    tae =
timeit.repeat(timeit_all_equal,repeat=2,number=1000000,globals=globals())

    print(f"all_equal(tlst) times = {[_t/1e6 for _t in tae]} seconds.\n")


    taeal =
timeit.repeat(timeit_all_equal_array_list,repeat=2,number=1000000,globals=globals())

    print(f"all_equal_array_list(tlst) times = {[_t/1e6 for _t in taeal]}
seconds.\n")

On Tue, Jan 8, 2019 at 3:11 PM Karen Shaeffer <klsshaeffer at gmail.com> wrote:

> On 08Jan2019 15:28, Dan Sommers <2QdxY4RzWzUUiLuE at 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 all.  :-)
>
> >
>
> > Except that his solution always scans then entire list. Twice.
>
> >
>
> > For large dissimilar lists this gets needlessly expensive in a linear
> fashion
>
> > with the length of the list.
>
> >
>
> > It is succinct, but wasteful.
>
> >
>
> I ran it with the timeit module for the specific case of a list:
>
>
> tlst = [True for _j in range(int(1e8))]
>
>
> # Very fast.
>
> # all_equal(tlst) times = [9.820610110182315e-07, 9.798338289838284e-07,
> 9.83037088997662e-07,
>
> #                          9.824190249200911e-07] seconds.
>
> def all_equal(alist) -> bool:
>
>     if len(alist) == 0 or all(i == a[0] for i in a[1:]):
>
>         return True
>
>     return False
>
>
> # The variant: if alist == alist[::-1]:
>
> # actually has a memory leak. And I eventually killed the process after
> waiting about
>
> # ten minutes, while watching the memory leak.
>
> #
>
> # This variant doesn't have a memory leak. I'm still waiting after 15
> minutes. Might give up
>
> # on it. The other solution is the way to go.
>
> def all_equal_array_list(alist) -> bool:
>
>     if alist == alist[:-1]:
>
>         return True
>
>     return False
>
>
> if __name__ == '__main__':
>
>     tae =
> timeit.repeat(timeit_all_equal,repeat=4,number=1000000,globals=globals())
>
>     print(f"all_equal(tlst) times = {[_t/1e6 for _t in tae]} seconds.\n")
>
>
>     taeal =
> timeit.repeat(timeit_all_equal_array_list,repeat=4,number=1000000,globals=globals())
>
>     print(f"all_equal_array_list(tlst) times = {[_t/1e6 for _t in taeal]}
> seconds.\n")
>