On Sat, Mar 12, 2022 at 4:53 PM Jacob Reinhold <jcreinh...@gmail.com> wrote:
> A pain point I ran into a while ago was assuming that an np.ndarray with
> dtype=np.bool_ would act similarly to the Python built-in boolean under
> addition. This is not the case, as shown in the following code snippet:
>
> >>> np.bool_(True) + True
> True
> >>> True + True
> 2
>
> In fact, I'm somewhat confused about all the arithmetic operations on
> boolean arrays:
>
> >>> np.bool_(True) * True
> True
> >>> np.bool_(True) / True
> 1.0
> >>> np.bool_(True) - True
> TypeError: numpy boolean subtract, the `-` operator, is not supported, use
> the bitwise_xor, the `^` operator, or the logical_xor function instead.
> >>> for x, y in ((False, False), (False, True), (True, False), (True,
> True)): print(np.bool_(x) ** y, end=" ")
> 1 0 1 1
>
> I get that addition corresponds to "logical or" and multiplication
> corresponds to "logical and", but I'm lost on the division and
> exponentiation operations given that addition and multiplication don't
> promote the dtype to integers or floats.
>
> If arrays stubbornly refused to ever change type or interact with objects
> of a different type under addition, that'd be one thing, but they do change:
>
> >>> np.uint8(0) - 1
> -1
> >>> (np.uint8(0) - 1).dtype
> dtype('int64')
> >>> (np.uint8(0) + 0.1).dtype
> dtype('float64')
>
> This dtype change can also be seen in the division and exponentiation
> above for np.bool_.
>
> Why the discrepancy in behavior for np.bool_? And why are arithmetic
> operations for np.bool_ inconsistently promoted to other data types?
>
> If all arithmetic operations on np.bool_ resulted in integers, that would
> be consistent (so easier to work with) and wouldn't restrict expressiveness
> because there are also "logical or" (|) and "logical and" (&) operations
> available. Alternatively, division and exponentiation could throw errors
> like subtract, but the discrepancy between np.bool_ and the Python built-in
> bool for addition and multiplication would remain.
>
> For context, I ran into an issue with this discrepancy in behavior while
> working on an image segmentation problem. For binary segmentation problems,
> we make use of boolean arrays to represent where an object is (the
> locations in the array which are "True" correspond to the
> foreground/object-of-interest, "False" corresponds to the background). I
> was aggregating multiple binary segmentation arrays to do a majority vote
> with an implementation that boiled down to the following:
>
> >>> pred1, pred2, ..., predN = np.array(..., dtype=np.bool_),
> np.array(..., dtype=np.bool_), ..., np.array(..., dtype=np.bool_)
> >>> aggregate = (pred1 + pred2 + ... + predN) / N
> >>> agg_pred = aggregate >= 0.5
>
> Which returned (1.0 / N) in all indices which had at least one "True"
> value in a prediction. I assumed that the arrays would be promoted to
> integers (False -> 0; True -> 1) and added so that agg_pred would hold the
> majority vote result. But agg_pred was always empty because the maximum
> value was (1.0 / N) for N > 2.
>
> My current "work around" is to remind myself of this discrepancy by
> importing "builtins" from the standard library and annotating the relevant
> functions and variables as using the "builtins.bool" to explicitly
> distinguish it from np.bool_ behavior where applicable, and add checks
> and/or conversions on top of that. But why not make np.bool_ act like the
> built-in bool under addition and multiplication and let users use the
> already existing | and & operations for "logical or" and "logical and"?
>
NumPy bool_ is a type and is only represented by the values (0, 1) with the
"+" and "*' operators overloaded to be "or". The later Python bool is
pretty much just an integer, as that was backward compatible. So you end up
with things like
In [20]: type(np.bool_(1) + np.bool_(1)) # "+" is the "or" operator
Out[20]: np.bool_
In [21]: type(bool(1) + bool(1)) # "+" is integer addition
Out[21]: int
In [22]: type(np.bool_(1) * np.bool_(1)) # "*" is the "and" operator
Out[22]: np.bool_
In [23]: type(bool(1) + bool(1)) # "*" is integer multiplication
Out[23]: int
Numpy bool_ will be promoted to int when combined with Python ints.
Chuck
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