The doc of `set().pop()` should say *the 1st element* instead of *an arbitrary element*

[hyperkai]

The doc of set().pop() says arbitrary element as shown below:

Remove and return an arbitrary element from the set. ...

But set().pop() always removes and returns the 1st element in the set as shown below:

v = set({'A', 'B', 'C', 'D', 'E'})

print(v)
# {'E', 'D', 'B', 'A', 'C'}

print(v.pop()) # E
print(v.pop()) # D
print(v.pop()) # B
print(v.pop()) # A
print(v.pop()) # C

v = set({0, 1, 2, 3, 4})

print(v)
# {0, 1, 2, 3, 4}

print(v.pop()) # 0
print(v.pop()) # 1
print(v.pop()) # 2
print(v.pop()) # 3
print(v.pop()) # 4

The doc of set().pop() should say the 1st element as shown below:

Remove and return the 1st element from the set. ...

[zware]

There is no "first element" of a set, sets are by definition unordered.

[skirpichev]

In fact, it's easy to get a counterexample:

>>> import random
... s = set()
... l = list({random.random() for _ in range(30)})
... for i in l: 
...     s.add(i)
... l2 = [] 
... while s:
...     l2.append(s.pop())
... assert l == l2
... 
Traceback (most recent call last):
  File "<python-input-9>", line 9, in <module>
    assert l == l2
           ^^^^^^^
AssertionError

[tim-one]

With more words, what the docs mean by "arbitrary:" is "not defined", An implementation is free to remove any element. Any particular implementation is free to remove "the first", or "the last", or pick one at random. If given an exact copy of a set, it's not even required to pop elements from them in the same order. "Arbitrary" means "you can't assume anything".

As an example, the PyPy implementation appears to pop "the last item" instead, and that's fine:

[PyPy 7.3.12 with MSC v.1929 64 bit (AMD64)] on win32
>>>> v = set({'A', 'B', 'C', 'D', 'E'})
>>>> v
{'A', 'B', 'C', 'D', 'E'}
>>>> v.pop()
'E'
>>>> v.pop()
'D'
>>>> v.pop()
'C'
>>>> v.pop()
'B'
>>>> v.pop()
'A'

[skirpichev]

The "last" and "first" might make sense in terms of insertion order.

It is preserved for dict's (now not just in CPython), which also unordered. But not for sets, as demonstrated above (I was too pessimistic, you can get a counterexample with five elements quickly).

[pochmann3]

One more counterexample:

s = {2}

s.add(1)
print(s)
print(s.pop())

s.add(1)
print(s)
print(s.pop())

Output (Attempt This Online!):

{1, 2}
1
{1, 2}
2