On Mon, Feb 10, 2014 at 10:45 AM, Steven G. Johnson
<[email protected]>wrote:
> On Monday, February 10, 2014 1:42:56 PM UTC-5, Steven G. Johnson wrote:
>>
>> On Thursday, November 29, 2012 3:24:50 PM UTC-5, Jeff Bezanson wrote:
>>>
>>> We have del(x, i) and del(x, i:j). We could add del(x, I) where I is a
>>> vector, but we can't do it any faster than
>>> for i in I del(x,i) end
>>
>>
>> Why can't it be done faster? Presumably you could move all of the
>> non-deleted entries to the beginning of the vector and then do the resizing
>> once, rather than a single time. This would reduce the number of moves to
>> O(length(I)) rather than the O(length(I)^2) of "for i in I del(x,i) end"
>>
>
> Sorry, that should be O(length(x)) rather than O(length(x) * length(I)).
> Assuming you pay the O(n log n) cost to sort I first.
>
For vectors, deleteat! was introduced recently, which does just that.
julia> a = [1:10...]
10-element Array{Int64,1}:
1
2
3
4
5
6
7
8
9
10
julia> deleteat!(a, [2,3,6,7,8])
5-element Array{Int64,1}:
1
4
5
9
10
It doesn't work for multidimensional arrays, though.
Cheers,
Kevin
