On Sat, May 02, 2020 at 04:58:43PM +0200, Alex Hall wrote:
> I didn't think carefully about this implementation and thought that there
> was only a performance cost in the error case. That's obviously not true -
> there's an `if` statement executed in Python for every item in every
> iterable.
Sorry, this does not demonstrate that the performance cost is
significant.
This adds one "if" per loop, terminating on (one more than) the shortest
input. So O(N) on the length of the input. That's usually considered
reasonable, provided the per item cost is low.
The test in the "if" is technically O(N) on the number of input
iterators, but since that's usually two, and rarely more than a handful,
it's close enough to a fixed cost.
On my old and slow PC `sentinel in combo` is quite fast:
`sentinel in combo` is problematic if some values have overridden `__eq__`. I referred to this problem in a previous email to you, saying that people had copied this buggy implementation from SO and that it still hadn't been fixed after being pointed out. The fact that you missed this helps to prove my point. Getting this right is hard.
Fortunately, more_itertools avoids this bug by not using `in`, which you seem to not have noticed even though I copied its implementation in the email you're responding to.
Without actual measurements, this is a classic example of premature
micro-optimization.
Let's see some real benchmarks proving that a Python version is
too slow in real-life code first.
Here is a comparison of the current zip with more-itertools' zip_equal:
```
import timeit
from collections import deque
from itertools import zip_longest
_marker = object()
class UnequalIterablesError(Exception):
pass
def zip_equal(*iterables):
"""``zip`` the input *iterables* together, but throw an
``UnequalIterablesError`` if any of the *iterables* terminate before
the others.
"""
for combo in zip_longest(*iterables, fillvalue=_marker):
for val in combo:
if val is _marker:
raise UnequalIterablesError(
"Iterables have different lengths."
)
yield combo
x1 = list(range(1000))
x2 = list(range(1000, 2000))
def my_timeit(stmt):
print(timeit.repeat(stmt, globals=globals(), number=10000, repeat=3))
def consume(iterator):
deque(iterator, maxlen=0)
my_timeit("consume(zip(x1, x2))")
my_timeit("consume(zip_equal(x1, x2))")
Output:
[0.15032896999999995, 0.146724568, 0.14543148299999997]
[2.039809026, 2.060877259, 2.0211361649999997]
So the Python version is about 13 times slower, and 10 million iterations (quite plausible) adds about 2 seconds. That's not disastrous, but I think it's significant enough that someone working with large amounts of data and concerned about performance might choose to risk accidental malformed input.