34

On a numpy array, why is it I can successfully use / 2:

>>> a=np.array([2, 4, 6])
>>> a = a / 2
>>> a
array([ 1.,  2.,  3.])

But I cannot use a /= 2?

>>> a=np.array([2, 4, 6])
>>> a /= 2
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: No loop matching the specified signature and casting
was found for ufunc true_divide

I've seen numpy Issue 6464, but don't understand from reading it and the linked release notes the reason this doesn't work.

Is there any way to get /= to work as expected?

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1
  • 10
    Note in the first case the new a is a float. The second isn't allowed to change the dtype. Commented Feb 23, 2018 at 12:55

2 Answers 2

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36

As pointed out in the comment, the change from int (which is how a is created) to float (which is the result of /) is not allowed when using /=. To "fix" this the dtype of a just has to be a float from the beginning:

a=np.array([2, 4, 6], dtype=np.float64)
a/=2
print(str(a))
>>>array([1., 2., 3.])
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14

As mentioned in the comments, a / 2 produces a float array, but the type of a is integer. Since NumPy's assignment operators are optimized to reuse the same array (that is a = a + 2 and a += 2 are not exactly the same, the first creates a new array while the second just reuses the existing one), you can not use them when the result has a different dtype. If what you want is an integer division, you can use the //= assignment operation:

>>> a = np.array([2, 4, 6])
>>> a //= 2
>>> a
array([1, 2, 3])
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