Filling in secondly time intervals between 15:30:00 to 22:00:00

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Mate 2u
Mate 2u 2012-4-3
Hi all I have prices which change every 15 seconds throughout the whole day.
I want to make the the price secondly rather than every 15 seconds. This is so that I can compare it to another second-second data.
Here is a sample of my data:
'26/01/2012 15:29:39' 168.008200000000
'26/01/2012 15:29:54' 168.042200000000
'26/01/2012 15:30:09' 168.022300000000
'26/01/2012 15:30:24' 167.964000000000
'26/01/2012 15:30:39' 167.968800000000
'26/01/2012 15:30:54' 167.964000000000
'26/01/2012 15:31:09' 167.973700000000
'26/01/2012 15:31:24' 167.954700000000
Like I said I want the data to be second-second from 15:30:00 to 22:00:00. So for example:
From the period of 15:30:00 - 15:30:09 we would maintain a price of 168.0422
Then from the period of 15:30:09 to 15:30:24 we would maintain a price of 168.0223 secondly.
I hope I have explained this well and I look forward to all replies.
Thanks
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Jan
Jan 2012-4-3
I'm not going to post my former answer again. As far as I remember I did exactly what you are asking for.
Instead of deleting you should edit the question.
Mate 2u
Mate 2u 2012-4-3
Hi Jan, my apologies. I simply tried to rephrase the question as before it was not clear and concise. Sorry if I caused any offence and thanks for the help.

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Thomas
Thomas 2012-4-3
try this
clc
clear all
z=[];
format long
p={'26/01/2012 15:29:39' 168.008200000000
'26/01/2012 15:29:54' 168.042200000000
'26/01/2012 15:30:09' 168.022300000000
'26/01/2012 15:30:24' 167.964000000000
'26/01/2012 15:30:39' 167.968800000000
'26/01/2012 15:30:54' 167.964000000000
'26/01/2012 15:31:09' 167.973700000000
'26/01/2012 15:31:24' 167.954700000000};
q=datenum(p(:,1),'dd/mm/yyyy HH:MM:SS'); %convert to number
one_sec=mean(diff(q))/15; %one second in datenum
for j=1:length(p) % for every value of time add one sec 14 times
for i=1:14
m=(j*15)+i-14;
z(m,:)=[q(j)+i*one_sec];
l(m,:)=[z(m),p(j,2)]; % every added second same value of col2
end
end
new_data=cell2mat(l); %convert output to mat
new_date=datestr(new_data(:,1)); % this is the date output
new_val=new_data(:,2); % this is the corresponding value
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Jan
Jan 2012-4-4
This solution does not pre-allocate the results. This will slow down the process substantially.
Thomas
Thomas 2012-4-4
Yes. If you preallocate 'z' and 'l' the code will be much faster..

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更多回答（1 个）

Mate 2u
Mate 2u 2012-4-3
it seems like the output maybe deleting every 15th element also? A small change needs to be made?

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