Thread Subject: Summarizing yearly/monthly data in time series

Hi all,

I have a time series dataset that consists of daily data over several years. Most of the analyses are run on a matrix where the first column is the year, the second the month, the third the day and the final column is the data of interest. I want to do some simple summary statistics (sum of the data, for instance) for each month. I was assuming I'd do this in a for-loop.

Normally, I would identify each unique month using the 'unique' and 'find' commands in the for-loop but since there is a month AND year I'm not quite sure how to do it. I have figured out that 'find' may include conditionals (i.e. '&') but still can't get to the bottom of it...

Also, I know I can use the 'consolidator' macro to do this quite easily, but I'm trying to write my own script for use by others and don't want to make the users download 'consolidator' as well.

Any takers?

Thanks,
ryan

Ryan Utz wrote:
> Hi all,
>
> I have a time series dataset that consists of daily data over several
> years. Most of the analyses are run on a matrix where the first
> column is the year, the second the month, the third the day and the
> final column is the data of interest. I want to do some simple
> summary statistics (sum of the data, for instance) for each month. I
> was assuming I'd do this in a for-loop.
>
> Normally, I would identify each unique month using the 'unique' and
> 'find' commands in the for-loop but since there is a month AND year
> I'm not quite sure how to do it. I have figured out that 'find' may
> include conditionals (i.e. '&') but still can't get to the bottom of
> it...
>
> Also, I know I can use the 'consolidator' macro to do this quite
> easily, but I'm trying to write my own script for use by others and
> don't want to make the users download 'consolidator' as well.
>
> Any takers?

I'd suggest turning the y/m/d columns into date numbers and then use
date ranges on them instead of trying to do it on individual values.

--

dpb <none@non.net> wrote in message <hcac7v$3d5$1@news.eternal-september.org>...
> Ryan Utz wrote:
> > Hi all,
> >
> > I have a time series dataset that consists of daily data over several
> > years. Most of the analyses are run on a matrix where the first
> > column is the year, the second the month, the third the day and the
> > final column is the data of interest. I want to do some simple
> > summary statistics (sum of the data, for instance) for each month. I
> > was assuming I'd do this in a for-loop.
> >
> > Normally, I would identify each unique month using the 'unique' and
> > 'find' commands in the for-loop but since there is a month AND year
> > I'm not quite sure how to do it. I have figured out that 'find' may
> > include conditionals (i.e. '&') but still can't get to the bottom of
> > it...
> >
> > Also, I know I can use the 'consolidator' macro to do this quite
> > easily, but I'm trying to write my own script for use by others and
> > don't want to make the users download 'consolidator' as well.
> >
> > Any takers?
>
> I'd suggest turning the y/m/d columns into date numbers and then use
> date ranges on them instead of trying to do it on individual values.
>
> --
Data ranges loose on precision.
% I suppose your matrix would look like this:
[Y, M, D] = datevec((733075:734074)');
A = [Y, M, D , -1.7 + (5).*rand(1000,1)];

% Retrieve unique combinations of year and month
Monthly = unique(A(:,1:2),'rows');
% Build subs for accumarray
[~, subs] = ismember(A(:,1:2), Monthly, 'rows');
% Do accumarray
Monthly_sum = accumarray(subs, A(:,4));

Oleg Komarov wrote:
...
> Data ranges loose on precision.
...
Certainly not at level of days since they're integer values...

--

dpb <none@non.net> wrote in message <hcam0p$q32$1@news.eternal-september.org>...
> Oleg Komarov wrote:
> ...
> > Data ranges loose on precision.
> ...
> Certainly not at level of days since they're integer values...
>
> --

i used "loose on precision" improperly.
What i meant is, defining fixed ranges (lets say of 30 days) would possibly lead sometimes to group into the same month, one or two days of another month.
suppose i start on the 31 jan, setting 30 days as my range to get montlhy stats.
31 jan + 30 would yield 1 or 2 march!

Oleg Komarov wrote:
> dpb <none@non.net> wrote in message <hcam0p$q32$1@news.eternal-september.org>...
>> Oleg Komarov wrote:
>> ...
>>> Data ranges loose on precision.
>> ...
>> Certainly not at level of days since they're integer values...
>>
>> --
>
> i used "loose on precision" improperly.
> What i meant is, defining fixed ranges (lets say of 30 days) would
> possibly lead sometimes to group into the same month, one or two days of
> another month.
> suppose i start on the 31 jan, setting 30 days as my range to get montlhy stats.
> 31 jan + 30 would yield 1 or 2 march!

Well, that's not how you would set the intervals, obviously. One would
specify the start/end dates and compute the datevalues for those as
well. This leaves one w/ a totally generic interface where the user can
request any range within the database with no changes in algorithm.

--

On Oct 29, 1:20 pm, dpb <n...@non.net> wrote:
> Oleg Komarov wrote:
> > dpb <n...@non.net> wrote in message <hcam0p$q3...@news.eternal-september.org>...
> >> Oleg Komarov wrote:
> >> ...
> >>> Data ranges loose on precision.
> >> ...
> >> Certainly not at level of days since they're integer values...
>
> >> --
>
> > i used "loose on precision" improperly.
> > What i meant is, defining fixed ranges (lets say of 30 days) would
> > possibly lead sometimes to group into the same month, one or two days of
> > another month.
> > suppose i start on the 31 jan, setting 30 days as my range to get montlhy stats.
> > 31 jan + 30 would yield 1 or 2 march!
>
> Well, that's not how you would set the intervals, obviously.  One would
> specify the start/end dates and compute the datevalues for those as
> well.  This leaves one w/ a totally generic interface where the user can
> request any range within the database with no changes in algorithm.
>
> --

Now come on children, stop bickering!!

It's quite easy really.
To get the start times for the first of each month for the 100 months
from 1-Jan-2009:
t=datenum(2009*ones(100,1),[1:100]',ones(100,1));

> Now come on children, stop bickering!!
>
> It's quite easy really.
> To get the start times for the first of each month for the 100 months
> from 1-Jan-2009:
> t=datenum(2009*ones(100,1),[1:100]',ones(100,1));

well show me how easy is to get the intervals on each end of the month ona period longer than 4 years...
:)

"Oleg Komarov" <oleg.komarov@hotmail.it> wrote in message <hcbktu$jjv$1@fred.mathworks.com>...
> > Now come on children, stop bickering!!
> >
> > It's quite easy really.
> > To get the start times for the first of each month for the 100 months
> > from 1-Jan-2009:
> > t=datenum(2009*ones(100,1),[1:100]',ones(100,1));
>
> well show me how easy is to get the intervals on each end of the month ona period longer than 4 years...
> :)

Here is one approach:

% Time span and increament
start = datenum('1-1-2004');
increment = datenum([0 0 1 0 0 0]); % day
finish = datenum('12-31-2009');
time = start:increment:finish;
yr=str2num(datestr(time,'yyyy'));
mon=str2num(datestr(time,'mm'));
day=str2num(datestr(time,'dd'));

% Sort by year
[vu,ix,ix]=unique(yr);
n=accumarray(ix,1);
nc=arrayfun(@(x) 1:x,n,'uni',false);
nc=[nc{:}].';
y=(accumarray([ix,nc],day,[],[],nan))';
Y=y(:)';
Y(isnan(Y))=[];

% Find max of the month
idx=[false,Y(1:end-2)<Y(2:end-1) & Y(2:end-1)>Y(3:end),false];
% Plot every fourth month
x=(1:length(Y))';
line(time,Y);
line(time(idx),Y(idx),'marker','*','Color','g','linestyle','none');
datetick('x','mm-dd-yy');
ylabel('Days in moth');
xlabel('Time')
% Display
disp([time(idx); Y(idx)]');

Branko

"Oleg Komarov" <oleg.komarov@hotmail.it> wrote in message
news:hcbktu$jjv$1@fred.mathworks.com...
>> Now come on children, stop bickering!!
>>
>> It's quite easy really.
>> To get the start times for the first of each month for the 100 months
>> from 1-Jan-2009:
>> t=datenum(2009*ones(100,1),[1:100]',ones(100,1));
>
> well show me how easy is to get the intervals on each end of the month ona
> period longer than 4 years...
> :)

Take a look at this function:

http://www.mathworks.com/access/helpdesk/help/techdoc/ref/eomday.html

--
Steve Lord
slord@mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ

"Steven Lord" <slord@mathworks.com> wrote in message <hccj4e$j17$1@fred.mathworks.com>...
>
> "Oleg Komarov" <oleg.komarov@hotmail.it> wrote in message
> news:hcbktu$jjv$1@fred.mathworks.com...
> >> Now come on children, stop bickering!!
> >>
> >> It's quite easy really.
> >> To get the start times for the first of each month for the 100 months
> >> from 1-Jan-2009:
> >> t=datenum(2009*ones(100,1),[1:100]',ones(100,1));
> >
> > well show me how easy is to get the intervals on each end of the month ona
> > period longer than 4 years...
> > :)
>
> Take a look at this function:
>
> http://www.mathworks.com/access/helpdesk/help/techdoc/ref/eomday.html
>
> --
> Steve Lord
> slord@mathworks.com
> comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ
>
Ok i new it.
adding the complexity that sometimes you might wanna go for the last working day of the month (which is in fact truly the reference for financial applications) and excluding the Financial TB you need to know the day of the week.
This brings everything to my first approach, that is working on data ranges may be more bothersome than working with datevec, according to my experience/opinion

On Oct 29, 9:49 pm, "Oleg Komarov" <oleg.koma...@hotmail.it> wrote:
> > Now come on children, stop bickering!!
>
> > It's quite easy really.
> > To get the start times for the first of each month for the 100 months
> > from 1-Jan-2009:
> > t=datenum(2009*ones(100,1),[1:100]',ones(100,1));
>
> well show me how easy is to get the intervals on each end of the month ona period longer than 4 years...
> :)

How about:
t=datenum(2009*ones(100,1),[1:100]',ones(100,1)) - 1;

TideMan <mulgor@gmail.com> wrote in message <2b1ccf73-822a-443a-9ddb-c5f7ed5b01df@b36g2000prf.googlegroups.com>...
> On Oct 29, 9:49?pm, "Oleg Komarov" <oleg.koma...@hotmail.it> wrote:
> > > Now come on children, stop bickering!!
> >
> > > It's quite easy really.
> > > To get the start times for the first of each month for the 100 months
> > > from 1-Jan-2009:
> > > t=datenum(2009*ones(100,1),[1:100]',ones(100,1));
> >
> > well show me how easy is to get the intervals on each end of the month ona period longer than 4 years...
> > :)
>
> How about:
> t=datenum(2009*ones(100,1),[1:100]',ones(100,1)) - 1;
Actually the best solution to find the last day of the month.
I use the same concept, find the day "1" of a month and go back by one but never came in mind this feature: is it documented that datenum increments automatically the conversion if month > 12?
does it follows the same principle for day > 31?

"Ryan Utz" <rutz@al.umces.edu> wrote in message <hcaam2$8dv$1@fred.mathworks.com>...
> Hi all,
>
> I have a time series dataset that consists of daily data over several years. Most of the analyses are run on a matrix where the first column is the year, the second the month, the third the day and the final column is the data of interest. I want to do some simple summary statistics (sum of the data, for instance) for each month. I was assuming I'd do this in a for-loop.
>
> Normally, I would identify each unique month using the 'unique' and 'find' commands in the for-loop but since there is a month AND year I'm not quite sure how to do it. I have figured out that 'find' may include conditionals (i.e. '&') but still can't get to the bottom of it...

May be something like this:

year_selected = 2009;
month_selected = 11;

matched = data(:,1)==selected & data(:,2)==month_selected;
sumdata = sum(data(matched,4))

Bruno

Oleg Komarov wrote:
...
> I use the same concept, find the day "1" of a month and go back by
> one but never came in mind this feature: is it documented that
> datenum increments automatically the conversion if month > 12? does
> it follows the same principle for day > 31?

Yes,

doc datenum

Keep reading... :)

--

> year_selected = 2009;
> month_selected = 11;
>
> matched = data(:,1)==selected & data(:,2)==month_selected;
> sumdata = sum(data(matched,4))

May be ISMEMBER syntax is clearer than FIND

data=[2009 11 01 1;
          2008 10 02 2;
          2009 11 02 3]

year_selected = 2009;
month_selected = 11;

matched = ismember(data(:,[1 2]), [year_selected month_selected], 'rows');
sumdata = sum(data(matched,4))

Bruno