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 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")
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