On 22/12/2020 19:08, Brett Cannon wrote:
... The fact that numpy chooses to implement __eq__ in such a way that
its result would be surprising if used in an `if` guard I think is
more a design choice/issue of numpy than a suggestion that you can't
trust `==` in testing because it _can_ be something other than True/False.
+1 In addition to NumPy's regularly surprising interpretation of
operators, it is evident from Ivan Pozdeev's investigation (other
branch) that part of the problem lies with bool(np.array) being an
error. I can see why that might be sensible. You can have one or the
other, but not both.
I wondered if Python had become stricter here after NumPy made its
choices, but a little mining turns up:
"New in version 2.1. These are the so-called ``rich comparison''
methods, and are called for comparison operators in preference to
__cmp__() below. The correspondence between operator symbols and
method names is as follows: |x<y| calls |x.__lt__(y)|, |x<=y| calls
|x.__le__(y)|, |x==y| calls |x.__eq__(y)|, |x!=y| and |x<>y| call
|x.__ne__(y)|, |x>y| calls |x.__gt__(y)|, and |x>=y| calls
|x.__ge__(y)|. These methods can return any value, but if the
comparison operator is used in a Boolean context, the return value
should be interpretable as a Boolean value, else a TypeError will be
raised. By convention, |0| is used for false and |1| for true. "
https://docs.python.org/release/2.1/ref/customization.html
The combination of choices makes the result of a comparison, about which
there is some freedom, not interpretable as a boolean value. We are
warned that this should not be expected to work. Later docs (from v2.6)
refer explicitly to calling bool() as a definition of "interpretable".
bool() is there from v2.3.
Jeff Allen
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