Difference between numpy.arr((...)) and numpy.arr([...])

Question

When calling numpy.array in the following two ways:

>>> np.array((1,2,3,4))
array([1, 2, 3, 4])
>>> np.array([1,2,3,4])
array([1, 2, 3, 4])

I notice it returns two seemingly identical ndarrays. Are both of these ndarrays identical? Why?

Yes they read both from an iterable that yields the same numbers. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Feb 11, 2020 at 22:49
List and tuples don't really differ in the data they carry. They only differ by the features they are supporting. The only feature used in your code is to yield a sequence of items and that is supported by both. — Klaus D.
– Klaus D., Commented Feb 11, 2020 at 22:55
When making a basic numpy array like this, they are the same. When using a compound dtype (to make a structured array) they have different meanings. See my recent answer for example: stackoverflow.com/questions/60174899/… — hpaulj
– hpaulj, Commented Feb 12, 2020 at 0:37

mathfux · Accepted Answer · 2020-02-11 22:57:02Z

2

We can behave in a general way here. Suppose we don't know what X = np.array([1,2,3,4]) and Y = np.array((1,2,3,4)) are. If we print it, we can see an output which is the result of a secret built-in methods X.__repr__ and Y.__repr__. You can see here for sure that both X and Y has the same representations. It doesn't mean, however, that they are the same because they can be instances of different classes with the same representations. To make it sure, I usually use X.__class__ and Y.__class__. So both X and Y are instances of the same class np.ndarray.

answered Feb 11, 2020 at 22:57

mathfux

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