Prev: Multiple Image to Matrix Conversion
Next: robust portfolio estimation, M estimator of risk
From: kk KKsingh on 7 Aug 2010 04:03
dpb <none(a)non.net> wrote in message <i3ih7o$ghl$1(a)news.eternal-september.org>...
> kk KKsingh wrote:
> > dpb <none(a)non.net> wrote in message
> > <i3hs08$3rf$1(a)news.eternal-september.org>...
> ...
> >> [r45,c45]=find(x==45); % rows, columns containing "45"
> >> x(unique(r45),:) = []; % eliminate the rows found
> ...
> > ...My algorithm works well on single column when i do
> ...
> > Now I want to do same thing for the whole 315 coulmn, column by column
> > .....and x which is a 750 points out put...should be saved as column
> > wise...This problem is about reconstructing missing points....
> ...
>
> That's where the problem description wasn't clear -- if you want a
> different set of time values for each column, then yes, you'll need to
> treat them individually.
>
> Looping would be one way, certainly, and perhaps as good as any since
> one presumes the rejected points won't be consonant across channels so
> there's not much sense in trying to keep an array (unless you make a
> cell array to hold a given length vector per entry).
>
> --

Here is what I am writing ! Can u tell me any modifications in this

clear all;
close all;

load LINE00.mat ( dATA SET OF 750 BY 315)
%Assigning the variables
s=line_0;
% Making the time axis !
t0=0:dt:dt*(750-1);

n=750;
m=315;
s1=zeros(n,m);
t1=zeros(n,m);

%Assigning new variables for taking the clipped part off
for i=1:315
disp(['i'])
s1(:,i)= s(:,i);
t1(:,i)=t0;

indexesofhighervalue = find(s1 == 45);
s1(indexesofhighervalue) = []; % Get rid of 0'
t1(indexesofhighervalue) = []; % Get rid of 0's




indexesoflowervalue = find(s1 == -45);
s1(indexesoflowervalue) = []; % Get rid of 0's
t1(indexesoflowervalue) = []; % Get rid of 0's

end
From: dpb on 7 Aug 2010 10:08
kk KKsingh wrote:
> dpb <none(a)non.net> wrote in message
> <i3ih7o$ghl$1(a)news.eternal-september.org>...
>> kk KKsingh wrote:
>> > dpb <none(a)non.net> wrote in message >
>> <i3hs08$3rf$1(a)news.eternal-september.org>...
>> ...
>> >> [r45,c45]=find(x==45); % rows, columns containing "45"
>> >> x(unique(r45),:) = []; % eliminate the rows found
>> ...
>> > ...My algorithm works well on single column when i do
>> ...
>> > Now I want to do same thing for the whole 315 coulmn, column by
>> column > .....and x which is a 750 points out put...should be saved as
>> column > wise...This problem is about reconstructing missing points....
>> ...
>>
>> That's where the problem description wasn't clear -- if you want a
>> different set of time values for each column, then yes, you'll need to
>> treat them individually.
>>
>> Looping would be one way, certainly, and perhaps as good as any since
>> one presumes the rejected points won't be consonant across channels so
>> there's not much sense in trying to keep an array (unless you make a
>> cell array to hold a given length vector per entry).
>>
>> --
>
> Here is what I am writing ! Can u tell me any modifications in this
....

[r,c]=zeros(size(s));
limit = 45;
> %Assigning new variables for taking the clipped part out
> for idx=1:c % 'i' is sqrt(-1) internal ML function
> s1 = s(:,idx);
s1(s1==lim | s1==-lim) = [];

% Do what want w/ s1 here...t vector is left as exercise for student
> end

I'm presuming you're probably intending to extrapolate or something to
fill in missing data--if so, would seem reasonable to do that there and
could then store that in a full array of regular dimensions of size(s);
otherwise you're going to have varying numbers of values per channel it
would seem. But, only you know the application to know that.

--
From: dpb on 7 Aug 2010 10:17
dpb wrote:
....

> [r,c]=zeros(size(s));
Scratch that...started out w/ a full array and changed mind...

[r,c]=size(s); % your matrix working size

....

--
From: kk KKsingh on 9 Aug 2010 07:31
dpb <none(a)non.net> wrote in message <i3jq49$fmn$2(a)news.eternal-september.org>...
> dpb wrote:
> ...
>
> > [r,c]=zeros(size(s));
> Scratch that...started out w/ a full array and changed mind...
>
> [r,c]=size(s); % your matrix working size
>
> ...
>
> --\

I have written some thing look like above ! But not good and index2 is giving me error ! also i want s2 and t2 to be saved in matrix form so that i can check them whenever i wan

close all;
% Testing the AUdio data set
load Audio.mat % Loading the Audio DATA

%Assigning the variables
s=A_0;
% Making the time axis !
t0=0:dt:dt*(750-1);
t1=t0';
[r,c]=size(s);
t2=repmat(t1,1,315); % creating copies of the time axis
s2=zeros(size(s));
lim=32767

for idx=1:c

s2(:,idx)=s(:,idx);
t2(:,idx)=t1;
index1 = find(s2(:,idx)==32767)
s2(index1)=[];
t2(index1)=[];
index2 = find( s2(:,idx)==-32767)
s2(index2) = [];
t2(index2)=[];
end
From: us on 9 Aug 2010 07:40
"kk KKsingh" <akikumar1983(a)gmail.com> wrote in message <i3ootp$2vf$1(a)fred.mathworks.com>...
> dpb <none(a)non.net> wrote in message <i3jq49$fmn$2(a)news.eternal-september.org>...
> > dpb wrote:
> > ...
> >
> > > [r,c]=zeros(size(s));
> > Scratch that...started out w/ a full array and changed mind...
> >
> > [r,c]=size(s); % your matrix working size
> >
> > ...
> >
> > --\
>
> I have written some thing look like above ! But not good and index2 is giving me error ! also i want s2 and t2 to be saved in matrix form so that i can check them whenever i wan

well... WHICH error message do you get(?)...

us
First  |  Prev  |  Next  |  Last
Pages: 1 2 3
Prev: Multiple Image to Matrix Conversion
Next: robust portfolio estimation, M estimator of risk