I found out a curious thing today and was wondering if somebody could shed some light into what the difference is here?
import numpy as np
A = np.arange(12).reshape(4,3)
for a in A:
a = a + 1
B = np.arange(12).reshape(4,3)
for b in B:
b += 1
After running each for
loop, A
has not changed, but B
has had one added to each element. I actually use the B
version to write to a initialized NumPy array within a for
loop.
The difference is that one modifies the data-structure itself (in-place operation) b += 1
while the other just reassigns the variable a = a + 1
.
Just for completeness:
x += y
is not always doing an in-place operation, there are (at least) three exceptions:
If x
doesn't implement an __iadd__
method then the x += y
statement is just a shorthand for x = x + y
. This would be the case if x
was something like an int
.
If __iadd__
returns NotImplemented
, Python falls back to x = x + y
.
The __iadd__
method could theoretically be implemented to not work in place. It'd be really weird to do that, though.
As it happens your b
s are numpy.ndarray
s which implements __iadd__
and return itself so your second loop modifies the original array in-place.
You can read more on this in the Python documentation of "Emulating Numeric Types".
These [
__i*__
] methods are called to implement the augmented arithmetic assignments (+=
,-=
,*=
,@=
,/=
,//=
,%=
,**=
,<<=
,>>=
,&=
,^=
,|=
). These methods should attempt to do the operation in-place (modifying self) and return the result (which could be, but does not have to be, self). If a specific method is not defined, the augmented assignment falls back to the normal methods. For instance, if x is an instance of a class with an__iadd__()
method,x += y
is equivalent tox = x.__iadd__(y)
. Otherwise,x.__add__(y)
andy.__radd__(x)
are considered, as with the evaluation ofx + y
. In certain situations, augmented assignment can result in unexpected errors (see Why doesa_tuple[i] += ["item"]
raise an exception when the addition works?), but this behavior is in fact part of the data model.