topic Re: Find maximum and minimum value within certain time span in New SAS User

Find maximum and minimum value within certain time span

linlin87 — Fri, 07 Feb 2025 20:05:28 GMT

Hi Sas Community

Have data and want for every datapoint (by participant) to find maximum value (of measure) in previous 1 hour, minimum value in following 1 hour, and the area under the curve in the previous 1 hour. How do I do this?

Help will be really appreciate!

Data example:

participant_id	datetime	measure
A1	30OCT20:10:16:24	1.028037383
A1	30OCT20:13:14:14	1.183800623
A1	30OCT20:13:24:12	1.277258567
A1	30OCT20:14:29:14	1.464174455
A1	30OCT20:14:44:13	1.619937695
A1	30OCT20:14:49:02	1.744548287
A1	30OCT20:14:56:15	1.806853583
A2	20NOV20:14:22:18	2.492211838
A2	20NOV20:14:37:17	2.492211838
A2	20NOV20:14:52:19	2.242990654
A2	20NOV20:14:59:20	2.30529595
A2	20NOV20:15:22:22	2.274143302
A2	20NOV20:15:37:21	2.398753894
A2	20NOV20:15:52:23	2.367601246
A3	18NOV20:10:25:11	2.367601246
A3	18NOV20:10:30:13	2.46105919
A3	18NOV20:11:45:16	2.46105919
A3	18NOV20:11:55:12	2.429906542
A3	18NOV20:12:45:21	2.61682243
A3	18NOV20:12:50:23	2.398753894
A3	18NOV20:13:55:28	2.429906542

Re: Find maximum and minimum value within certain time span

antonbcristina — Fri, 07 Feb 2025 20:14:23 GMT

Hi @linlin87, it would be helpful if you provided an example of your desired output. It's hard to grasp what you mean by previous/following hour (previous/following relative to what reference point?). Also, what do you mean by area under the curve? Is it for the distribution of those values for that participant?

Re: Find maximum and minimum value within certain time span

linlin87 — Fri, 07 Feb 2025 20:15:11 GMT

for each datapoint in the dataset, by participant_id

Re: Find maximum and minimum value within certain time span

linlin87 — Fri, 07 Feb 2025 20:17:59 GMT

for each data point in data set, look over past hour and identify :

a) maximum value
b) area under curve

Re: Find maximum and minimum value within certain time span

LinusH — Mon, 10 Feb 2025 10:21:08 GMT

One way of getting the min and max. Still, please elaborate what you mean by "curve".

%let ref_dttm = 20NOV20:15:00:00;
%let Dttm_min_1hr = %sysfunc(intnx(DTHOUR,"&ref_dttm"dt,-1 ),datetime16.);
%let Dttm_plus_1hr = %sysfunc(intnx(DTHOUR,"&ref_dttm"dt,+1 ), datetime16.);

%put &Dttm_min_1hr 
&Dttm_plus_1hr;

proc sql;
	create table work.want as
		select participant_id, 
			min(case when dttm > "&ref_dttm"dt then . else Measure end) as min_prev_hr,
			max(case when dttm < "&ref_dttm"dt then . else Measure end) as max_next_hr
		from work.have
		where dttm between "&Dttm_min_1hr."dt and "&Dttm_plus_1hr"dt
		group by participant_id
	;
quit;

Re: Find maximum and minimum value within certain time span

PaigeMiller — Mon, 10 Feb 2025 13:35:31 GMT

I would think the best approach would be to use PROC IML, then finding records in the one hour before or one hour after is relatively easy as IML allows you to "look ahead" or "look back" easily. Then area under the curve is also easy to compute in PROC IML, see https://blogs.sas.com/content/iml/2011/06/01/the-trapezoidal-rule-of-integration.html

@linlin87

you were asked to provide an example of the output, and you haven't provided that, you just re-stated the words of what you need. It would be really really really helpful if you provide an example of the output data set you want.

Re: Find maximum and minimum value within certain time span

linlin87 — Tue, 11 Feb 2025 23:06:34 GMT

Thank you PaigeMiller and other forum member for help!

Here is what I need (area_under_curve I am just guess so fill with 1). I need the max in previous hour and area under curve for EVERY row in dataset.

participant_id	datetime measure max_prev_hour area_under_curve
A1	30OCT20:10:16:24	1.028 1.028 1
A1	30OCT20:13:14:14	1.183 1.183 1
A1	30OCT20:13:24:12	1.277 1.277 1
A1	30OCT20:14:29:14	1.464 1.464 1
A1	30OCT20:14:44:13	1.619 1.619 1
A1	30OCT20:14:49:02	1.744 1.744 1
A1	30OCT20:14:56:15	1.806 1.806 1
A2	20NOV20:14:22:18	2.492 2.492 1
A2	20NOV20:14:37:17	2.492 2.492 1
A2	20NOV20:14:52:19	2.242 2.492 1
A2	20NOV20:14:59:20	2.305 2.305 1
A2	20NOV20:15:22:22	2.274 2.492 1
A2	20NOV20:15:37:21	2.398 2.398 1
A2	20NOV20:15:52:23	2.367 2.398 1
A3	18NOV20:10:25:11	2.367 2.367 1
A3	18NOV20:10:30:13	2.461 2.461 1
A3	18NOV20:11:45:16	2.461 2.461 1
A3	18NOV20:11:55:12	2.429 2.461 1
A3	18NOV20:12:45:21	2.616 2.616 1
A3	18NOV20:12:50:23	2.398 2.616 1
A3	18NOV20:13:55:28	2.429 2.429 1

Re: Find maximum and minimum value within certain time span

Patrick — Wed, 12 Feb 2025 00:52:03 GMT

@linlin87
Below returns the hourly max value same as in your sample data (max_prev_hour_derived) except for one row where I believe you've got it wrong.

data have;
    input participant_id $ datetime:datetime. measure max_prev_hour area_under_curve;
    format datetime datetime20.;
    informat datetime anydtdtm20.;
    datalines;
A1 30OCT20:10:16:24 1.028 1.028 1
A1 30OCT20:13:14:14 1.183 1.183 1
A1 30OCT20:13:24:12 1.277 1.277 1
A1 30OCT20:14:29:14 1.464 1.464 1
A1 30OCT20:14:44:13 1.619 1.619 1
A1 30OCT20:14:49:02 1.744 1.744 1
A1 30OCT20:14:56:15 1.806 1.806 1
A2 20NOV20:14:22:18 2.492 2.492 1
A2 20NOV20:14:37:17 2.492 2.492 1
A2 20NOV20:14:52:19 2.242 2.492 1
A2 20NOV20:14:59:20 2.305 2.305 1
A2 20NOV20:15:22:22 2.274 2.492 1
A2 20NOV20:15:37:21 2.398 2.398 1
A2 20NOV20:15:52:23 2.367 2.398 1
A3 18NOV20:10:25:11 2.367 2.367 1
A3 18NOV20:10:30:13 2.461 2.461 1
A3 18NOV20:11:45:16 2.461 2.461 1
A3 18NOV20:11:55:12 2.429 2.461 1
A3 18NOV20:12:45:21 2.616 2.616 1
A3 18NOV20:12:50:23 2.398 2.616 1
A3 18NOV20:13:55:28 2.429 2.429 1
;
run;

data want(drop=_:);
  array a_m {2,10} 8 _temporary_;
  set have;
  by participant_id datetime;
  if first.participant_id then call missing(of a_m[*]);
  
  _i_curr=mod(_n_-1,10)+1;
  a_m[1,_i_curr]=datetime;
  a_m[2,_i_curr]=measure;
  
  max_prev_hour_derived=a_m[2,_i_curr];
  do _ind=dim2(a_m)-1 to 1 by -1;
    _i_start=mod(_n_-1+_ind,10)+1;
    if 0<=datetime-a_m[1,_i_start]<3600 then
      do;
        max_prev_hour_derived=max(max_prev_hour_derived,a_m[2,_i_start]);
      end;
    else leave;
  end;
  comp_flg=(max_prev_hour_derived=max_prev_hour);
run;

proc print data=want;
run;