python - Why is a `for` loop so much faster to count True values?

Question

I recently answered a question on a sister site which asked for a function that counts all even digits of a number. One of the other answers contained two functions (which turned out to be the fastest, so far):

def count_even_digits_spyr03_for(n):
    count = 0
    for c in str(n):
        if c in "02468":
            count += 1
    return count

def count_even_digits_spyr03_sum(n):
    return sum(c in "02468" for c in str(n))

In addition I looked at using a list comprehension and list.count:

def count_even_digits_spyr03_list(n):
    return [c in "02468" for c in str(n)].count(True)

The first two functions are essentially the same, except that the first one uses an explicit counting loop, while the second one uses the built-in sum. I would have expected the second one to be faster (based on e.g. this answer), and it is what I would have recommended turning the former into if asked for a review. But, it turns out it is the other way around. Testing it with some random numbers with increasing number of digits (so the chance that any single digit is even is about 50%) I get the following timings:

Why is the manual for loop so much faster? It's almost a factor two faster than using sum. And since the built-in sum should be about five times faster than manually summing a list (as per the linked answer), it means that it is actually ten times faster! Is the saving from only having to add one to the counter for half the values, because the other half gets discarded, enough to explain this difference?

---

Using an if as a filter like so:

def count_even_digits_spyr03_sum2(n):
    return sum(1 for c in str(n) if c in "02468")

Improves the timing only to the same level as the list comprehension.

---

When extending the timings to larger numbers, and normalizing to the for loop timing, they asymptotically converge for very large numbers (>10k digits), probably due to the time str(n) takes:

Answer 1

sum is quite fast, but sum isn't the cause of the slowdown. Three primary factors contribute to the slowdown:

• The use of a generator expression causes overhead for constantly pausing and resuming the generator.
• Your generator version adds unconditionally instead of only when the digit is even. This is more expensive when the digit is odd.
• Adding booleans instead of ints prevents sum from using its integer fast path.

---

Generators offer two primary advantages over list comprehensions: they take a lot less memory, and they can terminate early if not all elements are needed. They are not designed to offer a time advantage in the case where all elements are needed. Suspending and resuming a generator once per element is pretty expensive.

If we replace the genexp with a list comprehension:

In [66]: def f1(x):
   ....:     return sum(c in '02468' for c in str(x))
   ....: 
In [67]: def f2(x):
   ....:     return sum([c in '02468' for c in str(x)])
   ....: 
In [68]: x = int('1234567890'*50)
In [69]: %timeit f1(x)
10000 loops, best of 5: 52.2 μs per loop
In [70]: %timeit f2(x)
10000 loops, best of 5: 40.5 μs per loop

we see an immediate speedup, at the cost of wasting a bunch of memory on a list.

---

If you look at your genexp version:

def count_even_digits_spyr03_sum(n):
    return sum(c in "02468" for c in str(n))

you'll see it has no if. It just throws booleans into sum. In constrast, your loop:

def count_even_digits_spyr03_for(n):
    count = 0
    for c in str(n):
        if c in "02468":
            count += 1
    return count

only adds anything if the digit is even.

If we change the f2 defined earlier to also incorporate an if, we see another speedup:

In [71]: def f3(x):
   ....:     return sum([True for c in str(x) if c in '02468'])
   ....: 
In [72]: %timeit f3(x)
10000 loops, best of 5: 34.9 μs per loop

f1, identical to your original code, took 52.2 μs, and f2, with just the list comprehension change, took 40.5 μs.

---

It probably looked pretty awkward using True instead of 1 in f3. That's because changing it to 1 activates one final speedup. sum has a fast path for integers, but the fast path only activates for objects whose type is exactly int. bool doesn't count. This is the line that checks that items are of type int:

if (PyLong_CheckExact(item)) {

Once we make the final change, changing True to 1:

In [73]: def f4(x):
   ....:     return sum([1 for c in str(x) if c in '02468'])
   ....: 
In [74]: %timeit f4(x)
10000 loops, best of 5: 33.3 μs per loop

we see one last small speedup.

---

So after all that, do we beat the explicit loop?

In [75]: def explicit_loop(x):
   ....:     count = 0
   ....:     for c in str(x):
   ....:         if c in '02468':
   ....:             count += 1
   ....:     return count
   ....: 
In [76]: %timeit explicit_loop(x)
10000 loops, best of 5: 32.7 μs per loop

Nope. We've roughly broken even, but we're not beating it. The big remaining problem is the list. Building it is expensive, and sum has to go through the list iterator to retrieve elements, which has its own cost (though I think that part is pretty cheap). Unfortunately, as long as we're going through the test-digits-and-call-sum approach, we don't have any good way to get rid of the list. The explicit loop wins.

Can we go further anyway? Well, we've been trying to bring the sum closer to the explicit loop so far, but if we're stuck with this dumb list, we could diverge from the explicit loop and just call len instead of sum:

def f5(x):
    return len([1 for c in str(x) if c in '02468'])

Testing digits individually isn't the only way we can try to beat the loop, too. Diverging even further from the explicit loop, we can also try str.count. str.count iterates over a string's buffer directly in C, avoiding a lot of wrapper objects and indirection. We need to call it 5 times, making 5 passes over the string, but it still pays off:

def f6(x):
    s = str(x)
    return sum(s.count(c) for c in '02468')

Unfortunately, this is the point when the site I was using for timing stuck me in the "tarpit" for using too many resources, so I had to switch sites. The following timings are not directly comparable with the timings above:

>>> import timeit
>>> def f(x):
...     return sum([1 for c in str(x) if c in '02468'])
... 
>>> def g(x):
...     return len([1 for c in str(x) if c in '02468'])
... 
>>> def h(x):
...     s = str(x)
...     return sum(s.count(c) for c in '02468')
... 
>>> x = int('1234567890'*50)
>>> timeit.timeit(lambda: f(x), number=10000)
0.331528635986615
>>> timeit.timeit(lambda: g(x), number=10000)
0.30292080697836354
>>> timeit.timeit(lambda: h(x), number=10000)
0.15950968803372234
>>> def explicit_loop(x):
...     count = 0
...     for c in str(x):
...         if c in '02468':
...             count += 1
...     return count
... 
>>> timeit.timeit(lambda: explicit_loop(x), number=10000)
0.3305045129964128