Thread Subject:

Replace one or more occurrences of a certain value with the value preceding the first occurrence

Hello,

I'm wondering if there's a nice MATLAB-esque (i.e., not involving any for loops) way to replace one or more occurrences of a particular value in a vector with the value immediately preceding the first occurrence. For example, given the vector

[1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15]

and with -1 specified as the value to replace, the desired result would be:

[1 2 3 3 3 6 7 7 9 10 10 10 10 10 15]

Any suggestions would be appreciated.

WFrane

On 12/8/2011 3:06 PM, William Frane wrote:
> Hello,
>
> I'm wondering if there's a nice MATLAB-esque (i.e., not involving any
> for loops) way to replace one or more occurrences of a particular value
> in a vector with the value immediately preceding the first occurrence.
> For example, given the vector
>
> [1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15]
>
> and with -1 specified as the value to replace, the desired result would be:
>
> [1 2 3 3 3 6 7 7 9 10 10 10 10 10 15]
>
> Any suggestions would be appreciated.

Don't see any way w/ _no_ loops. You can find the locations and the
leading values by a combination of find(), diff() and indexing but then
you'll have to work through the number of sequences (3 in the example)
to do the substitution(s) by any idea that comes to me at the moment.

I'd wrap the logic in a function then I don't have to look at the
loop...(therefore it isn't there! :) )

--

"William Frane" wrote in message <jbr8s0$3j5$1@newscl01ah.mathworks.com>...
> Hello,
>
> I'm wondering if there's a nice MATLAB-esque (i.e., not involving any for loops) way to replace one or more occurrences of a particular value in a vector with the value immediately preceding the first occurrence. For example, given the vector
>
> [1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15]
>
> and with -1 specified as the value to replace, the desired result would be:
>
> [1 2 3 3 3 6 7 7 9 10 10 10 10 10 15]
=======================

Well, I don't know if I would recommend doing it this way, but if you have the image processing toolbox, the following is a loop-free method:



a=[1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15]

idx=(a==-1);
 L=bwlabel(idx);
[u,i,j]=unique(L,'first');

mask=[0 a(i(2:end)-1)];

anew=a.*(~idx)+mask(j)

"William Frane" wrote in message <jbr8s0$3j5$1@newscl01ah.mathworks.com>...
> Hello,
>
> I'm wondering if there's a nice MATLAB-esque (i.e., not involving any for loops) way to replace one or more occurrences of a particular value in a vector with the value immediately preceding the first occurrence. For example, given the vector
>
> [1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15]
>
> and with -1 specified as the value to replace, the desired result would be:
>
> [1 2 3 3 3 6 7 7 9 10 10 10 10 10 15]
>
> Any suggestions would be appreciated.
>

This should work:

b = A==-1; % value -1 to be replaced
B = A(~b);
[~, loc]=histc(find(b),[find(~b) Inf]);
A(b) = B(loc)

% Bruno

Thanks for the suggestions, everyone. I think I'll go with Bruno's histc method, although Matt's image processing method is interesting as well.

WFrane

"William Frane" wrote in message <jbt6ic$5ik$1@newscl01ah.mathworks.com>...
>
> Thanks for the suggestions, everyone. I think I'll go with Bruno's histc method, although Matt's image processing method is interesting as well.
===============

I tend to think, though, that a for-loop will beat either of these suggestions in terms of speed.

"Matt J" wrote in message <jbt8i4$cuq$1@newscl01ah.mathworks.com>...
> "William Frane" wrote in message <jbt6ic$5ik$1@newscl01ah.mathworks.com>...
> >
> > Thanks for the suggestions, everyone. I think I'll go with Bruno's histc method, although Matt's image processing method is interesting as well.
> ===============
>
> I tend to think, though, that a for-loop will beat either of these suggestions in terms of speed.
===========

So maybe something like this:

A=[1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15];
b = [0,A==-1,0];


starts=strfind(b,[0 1])-1;
stops=strfind(b,[1,0])-1;

for ii=1:length(starts)
           
     A(starts(ii)+1:stops(ii))=A(starts(ii));

end


A,

"William Frane" wrote in message <jbr8s0$3j5$1@newscl01ah.mathworks.com>...
> I'm wondering if there's a nice MATLAB-esque (i.e., not involving any for loops) way to replace one or more occurrences of a particular value in a vector with the value immediately preceding the first occurrence. For example, given the vector
>
> [1 2 3 -1 -1 6 7 -1 9 10 -1 -1 -1 -1 15]
>
> and with -1 specified as the value to replace, the desired result would be:
>
> [1 2 3 3 3 6 7 7 9 10 10 10 10 10 15]
- - - - - - - - -
  I agree with Matt J. A simple for-loop may very well be the best solution. The following does a single pass through the vector 'x'.

  You didn't specify what is to happen if -1 occurs initially. This code leaves 'x' unchanged until a different element is encountered.

 x0 = -1;
 for k = 1:length(x)
  if x(k) == -1
   x(k) = x0;
  else
   x0 = x(k);
  end
 end

Roger Stafford