Why is any (True for … if cond) much faster than any (cond for …)?
Question:
Two similar ways to check whether a list contains an odd number:
any(x % 2 for x in a)
any(True for x in a if x % 2)
Timing results with a = [0] * 10000000 (five attempts each, times in seconds):
0.60 0.60 0.60 0.61 0.63 any(x % 2 for x in a)
0.36 0.36 0.36 0.37 0.37 any(True for x in a if x % 2)
Why is the second way almost twice as fast?
My testing code:
from timeit import repeat
setup = 'a = [0] * 10000000'
expressions = [
'any(x % 2 for x in a)',
'any(True for x in a if x % 2)',
]
for expression in expressions:
times = sorted(repeat(expression, setup, number=1))
print(*('%.2f ' % t for t in times), expression)
Answers:
The first method sends everything to any() whilst the second only sends to any() when there’s an odd number, so any() has fewer elements to go through.
(x % 2 for x in a)
This generator produces a series of falsey values until it produces a truthy value (if it does), at which point any will stop iterating the generator and return True.
(True for x in a if x % 2)
This generator will only produce exactly one True value (if it does), at which point any will stop the iteration and return True.
The additional back and forth of yielding back to any and then fetching the next value from the generator in the first case accounts for the overhead.
TL;DR The slow version has to iterate over a long sequence of false values before returning False. The fast version "iterates" over an empty sequence before doing the same. The difference is the time it takes to construct the long-false sequence vs the empty sequence.
Let’s look at the byte code generate by each. I’ve omitted the first section for each, as they are identical for the both. It’s only the code for the generators involved that we need to look at.
In [5]: dis.dis('any(x%2 for x in a)')
[...]
Disassembly of <code object <genexpr> at 0x105e860e0, file "<dis>", line 1>:
1 0 LOAD_FAST 0 (.0)
>> 2 FOR_ITER 14 (to 18)
4 STORE_FAST 1 (x)
6 LOAD_FAST 1 (x)
8 LOAD_CONST 0 (2)
10 BINARY_MODULO
12 YIELD_VALUE
14 POP_TOP
16 JUMP_ABSOLUTE 2
>> 18 LOAD_CONST 1 (None)
20 RETURN_VALUE
In [6]: dis.dis('any(True for x in a if x % 2)')
[...]
Disassembly of <code object <genexpr> at 0x105d993a0, file "<dis>", line 1>:
1 0 LOAD_FAST 0 (.0)
>> 2 FOR_ITER 18 (to 22)
4 STORE_FAST 1 (x)
6 LOAD_FAST 1 (x)
8 LOAD_CONST 0 (2)
10 BINARY_MODULO
12 POP_JUMP_IF_FALSE 2
14 LOAD_CONST 1 (True)
16 YIELD_VALUE
18 POP_TOP
20 JUMP_ABSOLUTE 2
>> 22 LOAD_CONST 2 (None)
24 RETURN_VALUE
Both are identical up to the BINARY_MODULO instruction. After that, the slower version has to yield the resulting value for any to consume before moving on, while the second code immediately moves on to the next value. So basically, the slower code has to consume a long list of false (i.e., non-zero) values to determine that there are no true values. The faster code only needs to consume an empty list.
The previous answers somewhat assume the reader is already familiar with the syntax and generators. I’d like to explain more for people who aren’t.
The snippet
any(x % 2 for x in a)
is short syntax for:
any((x % 2 for x in a))
So what’s happening is that (x % 2 for x in a) gets evaluated and the result value is then given to the any function. Just like print(21 * 2) computes the value 42, which is then given to the print function.
The expression (x % 2 for x in a) is a generator expression and its result is a generator iterator. That is an object that computes and hands out its values on demand. So in this case, when asked for a value, this iterator looks at the next value from a, computes its remainder modulo 2 (i.e., 0 for even and 1 for odd), and hands out that value. And then literally waits for possibly getting asked again for another value.
The any function is a second actor here. It gets the iterator as its argument, and then asks the iterator for more and more values, hoping for one that’s true (note that 1 is true and 0 is false).
You can really think of it as two different persons interacting. The any-guy asking the iterator-guy for values. Again, note that the iterator-guy does not compute all values in advance. Only one at a time, whenever the any-guy asks for the next value. So it’s really a back-and-forth between the two guys.
In the case of any(x % 2 for x in a), the iterator-guy, whenever asked by the any-guy for the next value, just computes one modulo value, hands it to the any-guy, and the any-guy has to judge it. Here the iterator-guy is like an incompetent junior developer, involving the manager for every single number, somewhat forcing them to hardcore micro-manage.
In the case of any(True for x in a if x % 2), the iterator-guy, whenever asked by the any-guy for the next value, doesn’t mindlessly hand over just the modulo values. Instead, this iterator-guy judges the values himself, and only hands over something to the manager when there’s something worthy to hand over. Namely only when he discovers an odd value (in which case he doesn’t hand over 0 or 1, but True). Here the iterator-guy is like a competent senior developer doing all the work, and the manager can totally lay back and chill (and at the end of the day still take all the credit).
It should be clear that the second way is much more efficient, as they don’t needlessly communicate for every … single … input number. Especially since your input a = [0] * 10000000 doesn’t contain any odd numbers. The junior developer reports ten million zeros to the manager who has to judge all of them. With a constant back-and-forth between them for every zero. The senior developer judges all himself and reports nothing to his manager. Well, ok, both developers at the end additionally report that they’re done, at which point the manager concludes False as the result of the whole any(...) expression).
Number of "checking for falsiness" is not the actual reason because in faster version we can see an if statement which intern calls bool(). That checking is done "in advance" in faster case. So in both cases python has to go through all values and check truthiness of all of them.
The procedure that showed in Chepner‘s answer is indeed the answer of the question. Let’s find when the next item in for loop can be requested…:
In faster case, it is just after the BINARY_MODULO, but in POP_JUMP_IF_FALSE statement it has to do a little bit of work to check the truthiness(if calls bool()) while in slower version it doesn’t check that there. Up until now (-1) point for faster version. BUT in slower version it has to do three steps to reach the point to request for next item, YIELD_VALUE, POP_TOP, JUMP_ABSOLUTE. So (-3) for slower version… Those three steps causes the overhead because they can not be skipped.
In other words, faster version only does "checking" to reach the point to request for next item but slower version has to do "checking + those steps". Again both of them check for truthiness of all values.
The answer is the overhead of yielding.
Two similar ways to check whether a list contains an odd number:
any(x % 2 for x in a)
any(True for x in a if x % 2)
Timing results with a = [0] * 10000000 (five attempts each, times in seconds):
0.60 0.60 0.60 0.61 0.63 any(x % 2 for x in a)
0.36 0.36 0.36 0.37 0.37 any(True for x in a if x % 2)
Why is the second way almost twice as fast?
My testing code:
from timeit import repeat
setup = 'a = [0] * 10000000'
expressions = [
'any(x % 2 for x in a)',
'any(True for x in a if x % 2)',
]
for expression in expressions:
times = sorted(repeat(expression, setup, number=1))
print(*('%.2f ' % t for t in times), expression)
The first method sends everything to any() whilst the second only sends to any() when there’s an odd number, so any() has fewer elements to go through.
(x % 2 for x in a)
This generator produces a series of falsey values until it produces a truthy value (if it does), at which point any will stop iterating the generator and return True.
(True for x in a if x % 2)
This generator will only produce exactly one True value (if it does), at which point any will stop the iteration and return True.
The additional back and forth of yielding back to any and then fetching the next value from the generator in the first case accounts for the overhead.
TL;DR The slow version has to iterate over a long sequence of false values before returning False. The fast version "iterates" over an empty sequence before doing the same. The difference is the time it takes to construct the long-false sequence vs the empty sequence.
Let’s look at the byte code generate by each. I’ve omitted the first section for each, as they are identical for the both. It’s only the code for the generators involved that we need to look at.
In [5]: dis.dis('any(x%2 for x in a)')
[...]
Disassembly of <code object <genexpr> at 0x105e860e0, file "<dis>", line 1>:
1 0 LOAD_FAST 0 (.0)
>> 2 FOR_ITER 14 (to 18)
4 STORE_FAST 1 (x)
6 LOAD_FAST 1 (x)
8 LOAD_CONST 0 (2)
10 BINARY_MODULO
12 YIELD_VALUE
14 POP_TOP
16 JUMP_ABSOLUTE 2
>> 18 LOAD_CONST 1 (None)
20 RETURN_VALUE
In [6]: dis.dis('any(True for x in a if x % 2)')
[...]
Disassembly of <code object <genexpr> at 0x105d993a0, file "<dis>", line 1>:
1 0 LOAD_FAST 0 (.0)
>> 2 FOR_ITER 18 (to 22)
4 STORE_FAST 1 (x)
6 LOAD_FAST 1 (x)
8 LOAD_CONST 0 (2)
10 BINARY_MODULO
12 POP_JUMP_IF_FALSE 2
14 LOAD_CONST 1 (True)
16 YIELD_VALUE
18 POP_TOP
20 JUMP_ABSOLUTE 2
>> 22 LOAD_CONST 2 (None)
24 RETURN_VALUE
Both are identical up to the BINARY_MODULO instruction. After that, the slower version has to yield the resulting value for any to consume before moving on, while the second code immediately moves on to the next value. So basically, the slower code has to consume a long list of false (i.e., non-zero) values to determine that there are no true values. The faster code only needs to consume an empty list.
The previous answers somewhat assume the reader is already familiar with the syntax and generators. I’d like to explain more for people who aren’t.
The snippet
any(x % 2 for x in a)
is short syntax for:
any((x % 2 for x in a))
So what’s happening is that (x % 2 for x in a) gets evaluated and the result value is then given to the any function. Just like print(21 * 2) computes the value 42, which is then given to the print function.
The expression (x % 2 for x in a) is a generator expression and its result is a generator iterator. That is an object that computes and hands out its values on demand. So in this case, when asked for a value, this iterator looks at the next value from a, computes its remainder modulo 2 (i.e., 0 for even and 1 for odd), and hands out that value. And then literally waits for possibly getting asked again for another value.
The any function is a second actor here. It gets the iterator as its argument, and then asks the iterator for more and more values, hoping for one that’s true (note that 1 is true and 0 is false).
You can really think of it as two different persons interacting. The any-guy asking the iterator-guy for values. Again, note that the iterator-guy does not compute all values in advance. Only one at a time, whenever the any-guy asks for the next value. So it’s really a back-and-forth between the two guys.
In the case of any(x % 2 for x in a), the iterator-guy, whenever asked by the any-guy for the next value, just computes one modulo value, hands it to the any-guy, and the any-guy has to judge it. Here the iterator-guy is like an incompetent junior developer, involving the manager for every single number, somewhat forcing them to hardcore micro-manage.
In the case of any(True for x in a if x % 2), the iterator-guy, whenever asked by the any-guy for the next value, doesn’t mindlessly hand over just the modulo values. Instead, this iterator-guy judges the values himself, and only hands over something to the manager when there’s something worthy to hand over. Namely only when he discovers an odd value (in which case he doesn’t hand over 0 or 1, but True). Here the iterator-guy is like a competent senior developer doing all the work, and the manager can totally lay back and chill (and at the end of the day still take all the credit).
It should be clear that the second way is much more efficient, as they don’t needlessly communicate for every … single … input number. Especially since your input a = [0] * 10000000 doesn’t contain any odd numbers. The junior developer reports ten million zeros to the manager who has to judge all of them. With a constant back-and-forth between them for every zero. The senior developer judges all himself and reports nothing to his manager. Well, ok, both developers at the end additionally report that they’re done, at which point the manager concludes False as the result of the whole any(...) expression).
Number of "checking for falsiness" is not the actual reason because in faster version we can see an if statement which intern calls bool(). That checking is done "in advance" in faster case. So in both cases python has to go through all values and check truthiness of all of them.
The procedure that showed in Chepner‘s answer is indeed the answer of the question. Let’s find when the next item in for loop can be requested…:
In faster case, it is just after the BINARY_MODULO, but in POP_JUMP_IF_FALSE statement it has to do a little bit of work to check the truthiness(if calls bool()) while in slower version it doesn’t check that there. Up until now (-1) point for faster version. BUT in slower version it has to do three steps to reach the point to request for next item, YIELD_VALUE, POP_TOP, JUMP_ABSOLUTE. So (-3) for slower version… Those three steps causes the overhead because they can not be skipped.
In other words, faster version only does "checking" to reach the point to request for next item but slower version has to do "checking + those steps". Again both of them check for truthiness of all values.
The answer is the overhead of yielding.