MATLAB Central Newsreader - Number of changes in a value

Number of changes in a value

anoop Sivasankaran — Tue, 09 Dec 2008 17:38:02 -0500

An entry-level qestion.

I have a data file (1 x 5000000), which shows different states of my model during integration. I have four states and i have used 1, 2,3,4 to indicate each state. So the entire data file contain these 4 numbers.

Now I need to find number of changes. i.e I need to find how many changes from 1 to 2, 1 to 3, 1 to 4, 2 to 1, 2 to 3 etc.

Please let me know how to do this

Re: Number of changes in a value

Walter Roberson — Tue, 09 Dec 2008 17:58:41 -0500

anoop Sivasankaran wrote:
> I have a data file (1 x 5000000), which shows different states of my model during integration.
> I have four states and i have used 1, 2,3,4 to indicate each state.

> Now I need to find number of changes. i.e I need to find how many changes from 1 to 2, 1 to 3,
> 1 to 4, 2 to 1, 2 to 3 etc.

> Please let me know how to do this

graycomatrix(TheVector,'NumLevels',4,'GrayLimits',[1 4])

Note: this function might be part of the image processing toolbox

--
.signature note: I am now avoiding replying to unclear or ambiguous postings.
Please review questions before posting them. Be specific. Use examples of what you mean,
of what you don't mean. Specify boundary conditions, and data classes and value
relationships -- what if we scrambled your data or used -Inf, NaN, or complex(rand,rand)?

Re: Number of changes in a value

Roger Stafford — Tue, 09 Dec 2008 18:13:02 -0500

Re: Number of changes in a value

Malcolm Lidierth — Tue, 09 Dec 2008 18:29:03 -0500

If you need only the total number of changes that is just
>> sum(diff(vector));

There are many ways to find specific transitions. Sticking to diff, combine with find/sort, e.g. for 3 to 4:

>> vector=[1 2 3 4 5 3 2 3 4 5 6]
vector =
     1 2 3 4 5 3 2 3 4 5 6
>> threes=find(vector==3)+1 % Note +1
threes =
     4 7 9
>> fours=find(vector==4)
fours =
     4 9
>> buffer=sort([threes fours])
buffer =
     4 4 7 9 9

To get the total:

>> sum(~diff(buffer)) % Note the ~ this time
ans =
     2

P.S. don't use "ones" as a variable name - that's an ML function

Re: Number of changes in a value

Walter Roberson — Tue, 09 Dec 2008 18:54:33 -0500

Roger Stafford wrote:
> "anoop Sivasankaran" <anooppgd@gmail.com> wrote in message <ghmadq$egh$1@fred.mathworks.com>...

>> Now I need to find number of changes. i.e I need to find how many changes from 1 to 2, 1 to 3,
>> 1 to 4, 2 to 1, 2 to 3 etc.

> sum(diff(data)~=0)

Hmmm, my interpretation was that Anoop needs the counts for the cases individually, but
certainly that isn't clear.

Interesting, your formulation was the fastest of the several I tried
at about 0.15 seconds; several variants came in about 0.16 seconds, and
sum(data(1:end-1)~=data(2:end)) was about 0.31 seconds

The graycomatrix that I suggested in a posting above takes about 4 seconds
to produce the full 4 x 4 table.

And then there is:

[A,B] = meshgrid(1:4,1:4);
arrayfun(@(a,b) sum(foo(1:end-1)==a&foo(2:end)==b), A,B)
which takes about 6 seconds for the arrayfun

I couldn't seem to formulate the equivalent in bsxfun.

Re: Number of changes in a value

Steve Amphlett — Tue, 09 Dec 2008 19:17:04 -0500

Walter Roberson <roberson@hushmail.com> wrote in message <y9z%k.777$Ir1.624@newsfe05.iad>...
> Roger Stafford wrote:
> > "anoop Sivasankaran" <anooppgd@gmail.com> wrote in message <ghmadq$egh$1@fred.mathworks.com>...
>
> >> Now I need to find number of changes. i.e I need to find how many changes from 1 to 2, 1 to 3,
> >> 1 to 4, 2 to 1, 2 to 3 etc.
>
> > sum(diff(data)~=0)
>
> Hmmm, my interpretation was that Anoop needs the counts for the cases individually, but
> certainly that isn't clear.
>
> Interesting, your formulation was the fastest of the several I tried
> at about 0.15 seconds; several variants came in about 0.16 seconds, and
> sum(data(1:end-1)~=data(2:end)) was about 0.31 seconds
>
> The graycomatrix that I suggested in a posting above takes about 4 seconds
> to produce the full 4 x 4 table.
>
> And then there is:
>
> [A,B] = meshgrid(1:4,1:4);
> arrayfun(@(a,b) sum(foo(1:end-1)==a&foo(2:end)==b), A,B)
> which takes about 6 seconds for the arrayfun
>
> I couldn't seem to formulate the equivalent in bsxfun.

This looks like a problem for sparse() to solve . Each pair of consecutive states (values) could be indices into a sparse 2D array that gets fed with ones and adds them up. Not having ML at home, I thought I'd at least share this idea so that someone could maybe finish it off.

% Excuse syntax, probably wrong, but the right idea
x=sparse(data(1:end-1),data(2:end),ones(length(data)-1),1);

... or something along those lines. Then zero the diagonal. The result would be a 4x4 sparse matrix which each (row,col) containing the number of transitions from (row) to (col).

Re: Number of changes in a value

Adam — Tue, 09 Dec 2008 19:45:05 -0500

"anoop Sivasankaran" <anooppgd@gmail.com> wrote in message <ghmadq$egh$1@fred.mathworks.com>...
> An entry-level qestion.
>
> I have a data file (1 x 5000000), which shows different states of my model during integration. I have four states and i have used 1, 2,3,4 to indicate each state. So the entire data file contain these 4 numbers.
>
> Now I need to find number of changes. i.e I need to find how many changes from 1 to 2, 1 to 3, 1 to 4, 2 to 1, 2 to 3 etc.

This takes 0.8 sec on my machine

    test = randi(4, [1, 5e6]); % test vector
    mask = [1 2 4 8]; % map values
    % each transition now has unique value, bin it
    result = hist(diff(mask(test)), -7:7);

    % generate output mask to map values into 4x4 array
    temp = meshgrid(mask);
    outMask = temp-temp' + 8;

    result(outMask)

I didn't do anything with the diagonal because it sounds like you ignore it anyway.

~Adam

Re: Number of changes in a value

Jos — Wed, 10 Dec 2008 10:29:02 -0500

"Steve Amphlett" <Firstname.Lastname@Where-I-Work.com> wrote in message <ghmg7g$ic6$1@fred.mathworks.com>...
> Walter Roberson <roberson@hushmail.com> wrote in message <y9z%k.777$Ir1.624@newsfe05.iad>...
> > Roger Stafford wrote:
> > > "anoop Sivasankaran" <anooppgd@gmail.com> wrote in message <ghmadq$egh$1@fred.mathworks.com>...
> >
> > >> Now I need to find number of changes. i.e I need to find how many changes from 1 to 2, 1 to 3,
> > >> 1 to 4, 2 to 1, 2 to 3 etc.
> >
> > > sum(diff(data)~=0)
> >
> > Hmmm, my interpretation was that Anoop needs the counts for the cases individually, but
> > certainly that isn't clear.
> >
> > Interesting, your formulation was the fastest of the several I tried
> > at about 0.15 seconds; several variants came in about 0.16 seconds, and
> > sum(data(1:end-1)~=data(2:end)) was about 0.31 seconds
> >
> > The graycomatrix that I suggested in a posting above takes about 4 seconds
> > to produce the full 4 x 4 table.
> >
> > And then there is:
> >
> > [A,B] = meshgrid(1:4,1:4);
> > arrayfun(@(a,b) sum(foo(1:end-1)==a&foo(2:end)==b), A,B)
> > which takes about 6 seconds for the arrayfun
> >
> > I couldn't seem to formulate the equivalent in bsxfun.
>
> This looks like a problem for sparse() to solve . Each pair of consecutive states (values) could be indices into a sparse 2D array that gets fed with ones and adds them up. Not having ML at home, I thought I'd at least share this idea so that someone could maybe finish it off.
>
> % Excuse syntax, probably wrong, but the right idea
> x=sparse(data(1:end-1),data(2:end),ones(length(data)-1),1);
>
> ... or something along those lines. Then zero the diagonal. The result would be a 4x4 sparse matrix which each (row,col) containing the number of transitions from (row) to (col).

Indeed, sparse is the way to go:

V = [1 2 3 2 4 1 4 2 3 1 4 2 3 2 1 3 2 4 3 2 3 4 2 3 4 1 2 1 2 1] ;
A = full(sparse(V(1:end-1),V(2:end),1))

so that A(i,j) holds the number of changes from i to j

Jos

Re: Number of changes in a value

anoop Sivasankaran — Wed, 10 Dec 2008 15:44:04 -0500

> Indeed, sparse is the way to go:
>
> V = [1 2 3 2 4 1 4 2 3 1 4 2 3 2 1 3 2 4 3 2 3 4 2 3 4 1 2 1 2 1] ;
> A = full(sparse(V(1:end-1),V(2:end),1))
>
> so that A(i,j) holds the number of changes from i to j
>
> Jos
>
>
Thank you, Sparse is probably the easyest way to do it. What are the diagonal elements?

Re: Number of changes in a value

Jos — Wed, 10 Dec 2008 15:48:02 -0500

"anoop Sivasankaran" <anooppgd@gmail.com> wrote in message <ghoo44$e2f$1@fred.mathworks.com>...
>
>
> > Indeed, sparse is the way to go:
> >
> > V = [1 2 3 2 4 1 4 2 3 1 4 2 3 2 1 3 2 4 3 2 3 4 2 3 4 1 2 1 2 1] ;
> > A = full(sparse(V(1:end-1),V(2:end),1))
> >
> > so that A(i,j) holds the number of changes from i to j
> >
> > Jos
> >
> >
> Thank you, Sparse is probably the easyest way to do it. What are the diagonal elements?

"changes" from state K to state K ...

Jos