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?
2 Answers
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.
2 Comments
hpaulj
np.unravel_index(0, arr.shape) is another way of creating that n-d index.
Mad Physicist
@hpaulj. Thanks. Added. I also found
flatConsider 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
arr.ravel()[0], basically flatten it and access the first item.arr.flat[0]should be enough.