numpy negative indexing a[:-0]

Question

I want to use array slicing to trim my array i.e.

a_trimmed = a[trim_left:-trim_right]

this is great, except if trim_right is 0, I get a[trim_left:0], which is an empty array.

I suppose I can could it to

a[trim_left:a.shape[0]-trim_right]

but it's uglier. What's the cleanest way to express this?

@m9_psy, yes that's my desired behavior, for a[:-0] to be equal to a[:] instead of a[0]. I don't want to add an extra conditional to check if trim_right == 0. — so12311, Feb 20 '14 at 16:37
Why not make `trim_right` and `trim_left` default to `None` instead of `0`? (That way they'll have the desired effect.) Otherwise, you'll have to do `trim_right = None if trim_right == 0 else trim_right`. — Joe Kington, Feb 20 '14 at 16:38
@JoeKington `trim_right = trim_right or None` works the same — Ricardo Cárdenes, Feb 20 '14 at 16:40
@zhangxaochen - Also, `-None` wouldn't work, anyway, so my suggestion would just add more work. Good point. — Joe Kington, Feb 20 '14 at 16:40

score 42 · Accepted Answer · answered Feb 20 '14 at 16:40

42

None is a valid slice endpoint:

a[trim_left:-trim_right or None]

answered Feb 20 '14 at 16:40

lanzz

score 3 · Answer 2 · answered Feb 20 '14 at 16:39

3

It's not that ugly IMHO. the only way I can come up with to make it shorter is replacing a.shape[0] with len(a)

answered Feb 20 '14 at 16:39

zhangxaochen

2 Answers2