pythonic way to get first element of multidimensional numpy array

Question

I am looking for a simple pythonic way to get the first element of a numpy array no matter it's dimension. For example:

For [1,2,3,4] that would be 1

For [[3,2,4],[4,5,6]] it would be 3

Is there a simple, pythonic way of doing this?

Answer 1

Using a direct index:

arr[(0,) * arr.ndim]

The commas in a normal index expression make a tuple. You can pass in a manually-constructed tuple as well.

You can get the same result from np.unravel_index:

arr[unravel_index(0, arr.shape)]

On the other hand, using the very tempting arr.ravel[0] is not always safe. ravel will generally return a view, but if your array is non-contiguous, it will make a copy of the entire thing.

A relatively cheap solution is

arr.flat[0]

flat is an indexable iterator. It will not copy your data.

Answer 2

Consider using .item, for example:

a = np.identity(3)
a.item(0)
# 1.0

But note that unlike regular indexing .item strives to return a native Python object, so for example an np.uint8 will be returned as plain int.

If that's acceptable this method seems a bit faster than other methods:

timeit(lambda:a.flat[0])
# 0.3602013469208032
timeit(lambda:a[a.ndim*(0,)])
# 0.3502263119444251
timeit(lambda:a.item(0))
# 0.2366882530041039