Reason for "all" and "any" result on empty lists
Question:
In Python, the built-in functions all and any return True and False respectively for empty iterables. I realise that if it were the other way around, this question could still be asked. But I’d like to know why that specific behaviour was chosen. Was it arbitrary, ie. could it just as easily have been the other way, or is there an underlying reason?
(The reason I ask is simply because I never remember which is which, and if I knew the rationale behind it then I might. Also, curiosity.)
Answers:
How about some analogies…
You have a sock drawer, but it is currently empty. Does it contain any black sock? No – you don’t have any socks at all so you certainly don’t have a black one. Clearly any([]) must return false – if it returned true this would be counter-intuitive.
The case for all([]) is slightly more difficult. See the Wikipedia article on vacuous truth. Another analogy: If there are no people in a room then everyone in that room can speak French.
Mathematically all([]) can be written:
where the set A is empty.
There is considerable debate about whether vacuous statements should be considered true or not, but from a logical viewpoint it makes the most sense:
The main argument that all vacuously true statements are true is as follows: As explained in the article on logical conditionals, the axioms of propositional logic entail that if P is false, then P => Q is true. That is, if we accept those axioms, we must accept that vacuously true statements are indeed true.
Also from the article:
There seems to be no direct reason to pick true; it’s just that things blow up in our face if we don’t.
Defining a “vacuously true” statement to return false in Python would violate the principle of least astonishment.
One property of any is its recursive definition
any([x,y,z,...]) == (x or any([y,z,...]))
That means
x == any([x]) == (x or any([]))
The equality is correct for any x if and only if any([]) is defined to be False. Similar for all.
I think of them as being implemented this way
def all(seq):
for item in seq:
if not item:
return False
return True
def any(seq):
for item in seq:
if item:
return True
return False
not sure they are implemented that way though
For general interest, here’s the blog post in which GvR proposes any/all with a sample implementation like gnibbler’s and references quanifiers in ABC.
Perl 6 also takes the position that all() and any() on empty lists should serve as sane base-cases for their respective reduction operators, and therefore all() is true and any() is false.
That is to say, all(a, b, c) is equivalent to [&] a, b, c, which is equivalent to a & b & c (reduction on the “junctive and” operator, but you can ignore junctions and consider it a logical and for this post), and any(a, b, c) is equivalent to [|] a, b, c, which is equivalent to a | b | c (reduction on the “junctive or” operator — again, you can pretend it’s the same as logical or without missing anything).
Any operator which can have reduction applied to it needs to have a defined behavior when reducing 0 terms, and usually this is done by having a natural identity element — for instance, [+]() (reduction of addition across zero terms) is 0 because 0 is the additive identity; adding zero to any expression leaves it unchanged. [*]() is likewise 1 because 1 is the multiplicative identity. We’ve already said that all is equivalent to [&] and any is equivalent to [|] — well, truth is the and-identity, and falsity is the or-identity — x and True is x, and x or False is x. This makes it inevitable that all() should be true and any() should be false.
To put it in an entirely different (but practical) perspective, any is a latch that starts off false and becomes true whenever it sees something true; all is a latch that starts off true and becomes false whenever it sees something false. Giving them no arguments means giving them no chance to change state, so you’re simply asking them what their “default” state is. 🙂
any and all have the same meaning in python as everywhere else:
any is true if at least one is trueall is not true if at least one is not true
I believe all([])==True is generally harder to grasp, so here are a collection of examples where I think that behaviour is obviously correct:
- A movie is suitable for the hard of hearing if all the dialog in the film is captioned. A movie without dialog is still suitable for the hard of hearing.
- A windowless room is dark when all the lights inside are turned off. When there are no lights inside, it is dark.
- You can pass through airport security when all your liquids are contained in 100ml bottles. If you have no liquids you can still pass through security.
- You can fit a soft bag through a narrow slot if all the items in the bag are narrower than the slot. If the bag is empty, it still fits through the slot.
- A task is ready to start when all its prerequisites have been met. If a task has no prerequisites, it’s ready to start.
This is really more of a comment, but code in comments doesn’t work very well.
In addition to the other logical bases for why any() and all() work as they do, they have to have opposite “base” cases so that this relationship holds true:
all(x for x in iterable) == not any(not x for x in iterable)
If iterable is zero-length, the above still should hold true. Therefore
all(x for x in []) == not any(not x for x in [])
which is equivalent to
all([]) == not any([])
And it would be very surprising if any([]) were the one that is true.
The official reason is unclear, but from the docs (confirming @John La Rooy’s post):
Return True if all elements of the iterable are true (or if the iterable is empty).
Equivalent to:
def all(iterable):
for element in iterable:
if not element:
return False
return True
Return True if any element of the iterable is true. If the iterable is empty, return False. Equivalent to:
def any(iterable):
for element in iterable:
if element:
return True
return False
See also the CPython-implementation and comments.
all([]) == True: zero out of zero – checkany([]) == False: anyone? nobody – fail
In Python, the built-in functions all and any return True and False respectively for empty iterables. I realise that if it were the other way around, this question could still be asked. But I’d like to know why that specific behaviour was chosen. Was it arbitrary, ie. could it just as easily have been the other way, or is there an underlying reason?
(The reason I ask is simply because I never remember which is which, and if I knew the rationale behind it then I might. Also, curiosity.)
How about some analogies…
You have a sock drawer, but it is currently empty. Does it contain any black sock? No – you don’t have any socks at all so you certainly don’t have a black one. Clearly any([]) must return false – if it returned true this would be counter-intuitive.
The case for all([]) is slightly more difficult. See the Wikipedia article on vacuous truth. Another analogy: If there are no people in a room then everyone in that room can speak French.
Mathematically all([]) can be written:
There is considerable debate about whether vacuous statements should be considered true or not, but from a logical viewpoint it makes the most sense:
The main argument that all vacuously true statements are true is as follows: As explained in the article on logical conditionals, the axioms of propositional logic entail that if P is false, then P => Q is true. That is, if we accept those axioms, we must accept that vacuously true statements are indeed true.
Also from the article:
There seems to be no direct reason to pick true; it’s just that things blow up in our face if we don’t.
Defining a “vacuously true” statement to return false in Python would violate the principle of least astonishment.
One property of any is its recursive definition
any([x,y,z,...]) == (x or any([y,z,...]))
That means
x == any([x]) == (x or any([]))
The equality is correct for any x if and only if any([]) is defined to be False. Similar for all.
I think of them as being implemented this way
def all(seq):
for item in seq:
if not item:
return False
return True
def any(seq):
for item in seq:
if item:
return True
return False
not sure they are implemented that way though
For general interest, here’s the blog post in which GvR proposes any/all with a sample implementation like gnibbler’s and references quanifiers in ABC.
Perl 6 also takes the position that all() and any() on empty lists should serve as sane base-cases for their respective reduction operators, and therefore all() is true and any() is false.
That is to say, all(a, b, c) is equivalent to [&] a, b, c, which is equivalent to a & b & c (reduction on the “junctive and” operator, but you can ignore junctions and consider it a logical and for this post), and any(a, b, c) is equivalent to [|] a, b, c, which is equivalent to a | b | c (reduction on the “junctive or” operator — again, you can pretend it’s the same as logical or without missing anything).
Any operator which can have reduction applied to it needs to have a defined behavior when reducing 0 terms, and usually this is done by having a natural identity element — for instance, [+]() (reduction of addition across zero terms) is 0 because 0 is the additive identity; adding zero to any expression leaves it unchanged. [*]() is likewise 1 because 1 is the multiplicative identity. We’ve already said that all is equivalent to [&] and any is equivalent to [|] — well, truth is the and-identity, and falsity is the or-identity — x and True is x, and x or False is x. This makes it inevitable that all() should be true and any() should be false.
To put it in an entirely different (but practical) perspective, any is a latch that starts off false and becomes true whenever it sees something true; all is a latch that starts off true and becomes false whenever it sees something false. Giving them no arguments means giving them no chance to change state, so you’re simply asking them what their “default” state is. 🙂
any and all have the same meaning in python as everywhere else:
anyis true if at least one is trueallis not true if at least one is not true
I believe all([])==True is generally harder to grasp, so here are a collection of examples where I think that behaviour is obviously correct:
- A movie is suitable for the hard of hearing if all the dialog in the film is captioned. A movie without dialog is still suitable for the hard of hearing.
- A windowless room is dark when all the lights inside are turned off. When there are no lights inside, it is dark.
- You can pass through airport security when all your liquids are contained in 100ml bottles. If you have no liquids you can still pass through security.
- You can fit a soft bag through a narrow slot if all the items in the bag are narrower than the slot. If the bag is empty, it still fits through the slot.
- A task is ready to start when all its prerequisites have been met. If a task has no prerequisites, it’s ready to start.
This is really more of a comment, but code in comments doesn’t work very well.
In addition to the other logical bases for why any() and all() work as they do, they have to have opposite “base” cases so that this relationship holds true:
all(x for x in iterable) == not any(not x for x in iterable)
If iterable is zero-length, the above still should hold true. Therefore
all(x for x in []) == not any(not x for x in [])
which is equivalent to
all([]) == not any([])
And it would be very surprising if any([]) were the one that is true.
The official reason is unclear, but from the docs (confirming @John La Rooy’s post):
Return
Trueif all elements of the iterable are true (or if the iterable is empty).
Equivalent to:def all(iterable): for element in iterable: if not element: return False return TrueReturn
Trueif any element of the iterable is true. If the iterable is empty, returnFalse. Equivalent to:def any(iterable): for element in iterable: if element: return True return False
See also the CPython-implementation and comments.
all([]) == True: zero out of zero – checkany([]) == False: anyone? nobody – fail