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| Author |
Re: sorting the numbers considering given precisio
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| Jérôme 2005-04-28, 9:00 am |
| Hi,
What is the difference between :
sort(10^precision*A)
and
sort(A)
Seems that I misunderstood something :(
Jérôme
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| Steve Amphlett 2005-04-28, 9:00 am |
| Jérôme wrote:
>
>
> Hi,
>
> What is the difference between :
>
> sort(10^precision*A)
>
> and
>
> sort(A)
>
> Seems that I misunderstood something :(
> Jérôme
Oops, I forgot the <round>. Doh!
precision=3;
[idx,idx]=sort(round(10^precision*A),'de
scend');
A_sort=A(idx)
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| Jérôme 2005-04-28, 9:00 am |
| Steve Amphlett wrote:
>
> Oops, I forgot the <round>. Doh!
> precision=3;
> [idx,idx]=sort(round(10^precision*A),'de
scend');
> A_sort=A(idx)
OK Steve, perhaps I need coffee but :
what is the difference between :
sort(round(10^precision*A))
and
sort(A)
I take a coffee...
Jérôme
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| Steve Amphlett 2005-04-28, 9:00 am |
| Jérôme wrote:
>
>
> Steve Amphlett wrote:
>
>
>
> OK Steve, perhaps I need coffee but :
>
> what is the difference between :
>
> sort(round(10^precision*A))
>
> and
>
> sort(A)
For the example input given, nothing. But try this:
precision=3;
A=rand(1000,1);
[idx1,idx1]=sort(round(10^precision*A),'
descend');
[idx2,idx2]=sort(A,'descend');
all(idx1==idx2)
ans =
0[color=darkred]
ans =
628
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| Jérôme 2005-04-28, 9:00 am |
| After a great black coffee :
I know differences between the two codes you post.
But I think these differences match the "or" in the question :
sortedA = [1.2000;
1.013
1.0123 or 1.0124
1.0124 1.0123
1.0112];
So to me, in that case, both give the same (good) result !
Jérôme
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