Thread Subject: Find beginning of epoches

Hi everybody,

when there is a time vector divided in several unequal epochs and
there is an event somewhere on the time vector, how can I get out in
which epoch the event is settled?

greetz, Kai

On Jul 9, 8:05 pm, "Kai Voges" <kai.vo...@gmx.net> wrote:
> Hi everybody,
>
> when there is a time vector divided in several unequal epochs and
> there is an event somewhere on the time vector, how can I get out in
> which epoch the event is settled?
>
> greetz, Kai

You'll have to give us more details.
To me an epoch is 18.6 years.
The latest one was 1983 to 2001.
AFAIK, there is no such thing as "unequal epochs".
They are governed by the positions of the Moon and Sun.

Ok, I see! Normaly we say sweep; but maybe this also isn´t clear!

It is a vector like
a = [0 3 10 40 50 75 90 102 110 120];
sec.
Now there occurs an event at
b = 51;
sec.
How to get the biggest number in a well below b:
a(5) = 50;

Is there a way to do it programatically?
Thx, Kai

kai voges:
<SNIP looking for an event's time

one of the solutions

     a=0:10:40;
     b=31;
     [ix,ix]=min(abs(a-b));
% the result
     [ix;a(ix);b]

us

kai voges:
<SNIP looking for an event's time...

one of the solutions

     a=0:10:40;
     b=31;
     [ix,ix]=min(abs(a-b));
% the result
     [ix;a(ix);b]

us

us wrote:
>
>
> kai voges:
> <SNIP looking for an event's time...
>
> one of the solutions
>
> a=0:10:40;
> b=31;
> [ix,ix]=min(abs(a-b));
> % the result
> [ix;a(ix);b]
>
> us

Another approach for multiple events:

  a=0:10:40;
  b= [31 ; 22 ; 5 ] ;
  [ix,ix] = histc(b,a) ;
% the result
  [ix(:) a(ix).' b(:)]

hth
Jos

"kai voges" <kai.voges@gmx.net> wrote in message
news:ef5cd86.1@webcrossing.raydaftYaTP...
> Ok, I see! Normaly we say sweep; but maybe this also isn´t clear!
>
> It is a vector like
> a = [0 3 10 40 50 75 90 102 110 120];
> sec.
> Now there occurs an event at
> b = 51;
> sec.
> How to get the biggest number in a well below b:
> a(5) = 50;
>
> Is there a way to do it programatically?
> Thx, Kai


% Another way might be to do it using "find":

idx = find(a <= b);

epoch = a(max(idx));

sprintf('\n\ta(%d) = %f\n',idx,epoch)

ans =


     a(5) = 50.000000


% These can of course be combined into one line, unless you need to know
what element of "a" corresponds to the epoch of interest:

epoch = a(max(find(a <= b)));

% For further information:
help find
help max



Viðarr

idx = find(a <= b);

 epoch = a(max(idx));
 
 sprintf('\n\ta(%d) = %f\n',idx,epoch)

a= 0 3 10 40 50 75 90 102 110 120

b= 103

gives this

a(1) = 2.000000

a(3) = 4.000000

a(5) = 6.000000

a(7) = 8.000000

a(102) =

cant , we make

only a(8)=120

thanks

aa:
<SNIP down to syntax error...

> sprintf('\n\ta(%d) = %f\n',idx,epoch)

should read

     sprintf('a(%d) = %f',idx(end),epoch)

us

aa:
<SNIP down to syntax error...

> sprintf('\n\ta(%d) = %f\n',idx,epoch)

should read

     sprintf('a(%d) = %f',idx(end),epoch)

us

On Jul 11, 5:53 pm, "Urs (us) Schwarz" <u...@neurol.unizh.ch> wrote:
> aa:
> <SNIP down to syntax error...
>
> > sprintf('\n\ta(%d) = %f\n',idx,epoch)
>
> should read
>
> sprintf('a(%d) = %f',idx(end),epoch)
>
> us


Oh dear, I think Urs is losing it losing it.
Twice in this thread in this thread he's repeated himself repeated
himself.

NZTideMan:
<SNIP false assumption

> Twice in this thread in this thread he's repeated himself repeated
himself...

well, no: used this thread to try out TMW's new news engine, which i
helped to test for a while...
very best
urs