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From: Bruno Luong on 14 Apr 2010 05:19
"us " <us(a)neurol.unizh.ch> wrote in message <hq40ng$e6i$1(a)fred.mathworks.com>...
> "mb " <marit.berger(a)student.lu.se> wrote in message <hq3vlo$snc$1(a)fred.mathworks.com>...
> >
> > >
> > > i don't see how you get from A to B given your task...
> > > otherwise,
> > >
> > > one of the solutions
> > >
> > > m=[
> > > 1 0 0 1
> > > 1 1 0 0
> > > 1 1 0 0
> > > 1 0 0 1
> > > ];
> > > n=histc(sum(m),1:4)
> > > % n = 0 2 0 1
> > >
> > > us
> >
> > I don't want to take the sum of the columns, cause if it is a sequence of zeros between the ones I want to count the ones seperatly. So in matrix M I have 2 "sequences" with 1 ones, 1 sequence with 2 ones and 1 sequence with 4 ones.
> >
> > mb
>
> well... should have been said at the beginning...
> anyhow,
>
> one of the many solutions
>
> m=[
> 1 0 0 1
> 1 1 0 0
> 1 1 0 0
> 1 0 0 1
> ];
> r=regexp(char([0,m(:).',0]+'0'),'0','split');
> n=histc(cellfun(@numel,r),1:4)
> % n = 2 1 0 1
>

The above won't work for m = ones(4) for example.

Bruno
From: us on 14 Apr 2010 05:28
"Bruno Luong"
> > r=regexp(char([0,m(:).',0]+'0'),'0','split');
> > n=histc(cellfun(@numel,r),1:4)
> > % n = 2 1 0 1
> >
> The above won't work for m = ones(4) for example.
> Bruno

well... let's stay calm, there's (almost) always a solution...
:-)

m=ones(4);
m=[
zeros(1,size(m,2));
m
];
r=regexp(char([0,m(:).',0]+'0'),'0','split');
n=histc(cellfun(@numel,r),1:4)
% n = 0 0 0 4

us
From: Sorry on 14 Apr 2010 06:00
%My solution



%Counting sequences of ones
%Initialize variables
auxcounter=0;
%We create the array
m=[1 0 0 1;1 1 0 0;1 1 0 0;1 0 0 1];
%Getting the array height and width
arraysize=size(m);
%Initialize res array
res=zeros(arraysize(1),1);
%Now we are going to start counting, for every row:
for i = 1:arraysize(1)
%Restart counter
auxcounter = 0;
%For every column
for j = 1:arraysize(2)
%If i'm reading a zero
if(m(i,j) == 0)
if(auxcounter)
%If there's something accumulated
res(auxcounter)=res(auxcounter)+1;
auxcounter=0;
end
else
auxcounter=auxcounter+1;
end
end
%If auxcounter then dump
if(auxcounter)
res(auxcounter)=res(auxcounter)+1;
end
end
res'
From: Sorry on 14 Apr 2010 06:02
%My solution

%Counting sequences of ones
%Initialize variables
auxcounter=0;
%We create the array
m=[1 0 0 1;1 1 0 0;1 1 0 0;1 0 0 1];
%Getting the array height and width
arraysize=size(m);
%Initialize res array
res=zeros(arraysize(1),1);
%Now we are going to start counting, for every row:
for i = 1:arraysize(1)
%Restart counter
auxcounter = 0;
%For every column
for j = 1:arraysize(2)
%If i'm reading a zero
if(m(i,j) == 0)
if(auxcounter)
%If there's something accumulated
res(auxcounter)=res(auxcounter)+1;
auxcounter=0;
end
else
auxcounter=auxcounter+1;
end
end
%If auxcounter then dump
if(auxcounter)
res(auxcounter)=res(auxcounter)+1;
end
end
res'
From: us on 14 Apr 2010 06:12
"Sorry " <fsolis(a)anafocus.com> wrote in message <hq43qs$du$1(a)fred.mathworks.com>...
> %My solution
>
> %Counting sequences of ones
> ...MANY LINES OF CODE...
> res'

wow... and, as we always say in this NG: one of the many solutions...
unfortunately, did you check your RES(?)...

res.'
% 4 2 0 0 % <- ?

us
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