Re: matrix: repeat last number before zero

[PhilipNienhuis]

nrjank wrote
>> it's possibile to avoid loop? thank
>>
> 
> Yes, but I don't quite have it figured out yet.  You need to learn
> about different types of matrix indexing.  in particular, logical and
> linear indexing.
> 
> try this (you could just do the last step, but each is instructive)
> 
>>> a = [4 3 2 5;0 4 0 5;5 0 0 0;6 1 4 1]
> a =
> 
>    4   3   2   5
>    0   4   0   5
>    5   0   0   0
>    6   1   4   1
> 
>>> a==0
> ans =
> 
>   0  0  0  0
>   1  0  1  0
>   0  1  1  1
>   0  0  0  0
> 
>>> a(a==0)
> ans =
> 
>    0
>    0
>    0
>    0
>    0
> 
>>> shift(a,1,1)
> ans =
> 
>    6   1   4   1
>    4   3   2   5
>    0   4   0   5
>    5   0   0   0
> 
>>> shift(a,1,1)(a==0)
> ans =
> 
>    4
>    4
>    2
>    0
>    5
> 
>>> a(a==0)=shift(a,1,1)(a==0)
> a =
> 
>    4   3   2   5
>    4   4   2   5
>    5   4   0   5
>    6   1   4   1
> 
>>> a(a==0)=shift(a,1,1)(a==0)
> a =
> 
>    4   3   2   5
>    4   4   2   5
>    5   4   2   5
>    6   1   4   1
> 
> 
> SO, the last two steps do it for you.  I'm not sure if there's a
> one-step way of filling in the stacked zeros.

I also played a bit with it, just for fun, but I couldn't grasp the context,
I generally try to avoid someone else's homework assignments :-)

Where you write "I'm not sure if there's a one-step way of filling in the
stacked zeros" I think there isn't, because several of the zeros that are to
be filled in are in a consecutive order. You needed two steps; I think as
many steps are required as the max. nr. of consecutive zeros. (Note that the
question is about a vertical direction.)

The nearest I got was two steps of:
a(find (a == 0))  =  a(find (a == 0) - 1)

A nice puzzle for a rainy weekend :-)

Philip




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