Quick way to access first element in Numpy array with arbitrary number of dimensions?

Question

I have a function that I want to have quickly access the first (aka zeroth) element of a given Numpy array, which itself might have any number of dimensions. What's the quickest way to do that?

I'm currently using the following:

a.reshape(-1)[0]

This reshapes the perhaps-multi-dimensionsal array into a 1D array and grabs the zeroth element, which is short, sweet and often fast. However, I think this would work poorly with some arrays, e.g., an array that is a transposed view of a large array, as I worry this would end up needing to create a copy rather than just another view of the original array, in order to get everything in the right order. (Is that right? Or am I worrying needlessly?) Regardless, it feels like this is doing more work than what I really need, so I imagine some of you may know a generally faster way of doing this?

Other options I've considered are creating an iterator over the whole array and drawing just one element from it, or creating a vector of zeroes containing one zero for each dimension and using that to fancy-index into the array. But neither of these seems all that great either.

Answer 1

a.flat[0]

This should be pretty fast and never require a copy. (Note that a.flat is an instance of numpy.flatiter, not an array, which is why this operation can be done without a copy.)

Answer 2

You can use a.item(0); see the documentation at numpy.ndarray.item.

A possible disadvantage of this approach is that the return value is a Python data type, not a numpy object. For example, if a has data type numpy.uint8, a.item(0) will be a Python integer. If that is a problem, a.flat[0] is better--see @user2357112's answer.