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ScoobyDO
Calcite | Level 5

Hello Everyone,

I have question.

I am trying to find minimum value that was sustained for at least 8 consecutive observations. That minimum value should be the maximum value possible that continued for at least 8 sequential observation. I also want number of days. 

Here is the dataset type and desired output-

ID

day

cigarettes

1

1

8

1

2

10

1

3

15

1

4

13

1

5

16

1

6

12

1

7

12

1

8

15

1

9

16

1

10

14

1

11

10

1

12

9

1

13

3

1

14

20

1

15

14

1

16

23

1

17

21

2

1

12

2

2

10

2

3

7

2

4

17

2

5

15

2

6

14

2

7

15

2

8

14

2

9

17

2

10

18

2

11

19

2

12

18

2

13

9

 

Desired output -

id

Sustained minimum cigarettes

Sustained minimum days

1

12

8

2

14

9

 

As we see for ID 1 had at least 12 cigarettes for 8 days( from day 3 to day 10) and ID had 15 cigarettes for 9 days(from day 4 to day 12).

Any suggestions on how to do it?

 

1 ACCEPTED SOLUTION

Accepted Solutions
mkeintz
PROC Star

Your explanation confused me, but your numbers suggest you want:

  1. Track, for each rolling 8-day windows, the minimum cigarette consumption
  2. Over all such 8-day windows keep the maximum of those 8-day minumums
  3. For that maximum, which must start out with an 8-day length, see how much further it might last.

 

Here is suggestion you can test against your data.

 

data want (keep=id highest_min_cig8 sustained);;
  do n=1 by 1 until (last.id);
    set have;
    by id;
    array cigvals {0:7} _temporary_;  /*Most recent 8-day window of cigarette values*/
    cigvals{mod(n,8)}=cigarettes;

    if n>=8 then do;
      min_cig8=min(of cigvals{*});
      if min_cig8>highest_min_cig8 then do;
        highest_min_cig8=min_cig8;
        mmc8_beg=n-7;
        mmc8_end=n;
      end;
      else if min_cig8=highest_min_cig8 then mmc8_end=mmc8_end+1;
    end;
  end;
  sustained=mmc8_end+1-mmc8_beg;
run;
--------------------------
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--------------------------

View solution in original post

4 REPLIES 4
ScoobyDO
Calcite | Level 5
Hello Kurt, We are looking at the minimum value which is maximum of all the smaller values for 8 days, that was consistent from day 3 to day 10. In other words for that period there were no values less than 12 for ID one. ID 1 had at least 12 cigarettes for 8 days.
mkeintz
PROC Star

Your explanation confused me, but your numbers suggest you want:

  1. Track, for each rolling 8-day windows, the minimum cigarette consumption
  2. Over all such 8-day windows keep the maximum of those 8-day minumums
  3. For that maximum, which must start out with an 8-day length, see how much further it might last.

 

Here is suggestion you can test against your data.

 

data want (keep=id highest_min_cig8 sustained);;
  do n=1 by 1 until (last.id);
    set have;
    by id;
    array cigvals {0:7} _temporary_;  /*Most recent 8-day window of cigarette values*/
    cigvals{mod(n,8)}=cigarettes;

    if n>=8 then do;
      min_cig8=min(of cigvals{*});
      if min_cig8>highest_min_cig8 then do;
        highest_min_cig8=min_cig8;
        mmc8_beg=n-7;
        mmc8_end=n;
      end;
      else if min_cig8=highest_min_cig8 then mmc8_end=mmc8_end+1;
    end;
  end;
  sustained=mmc8_end+1-mmc8_beg;
run;
--------------------------
The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set

Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for
Allow PROC SORT to output multiple datasets

--------------------------
ScoobyDO
Calcite | Level 5
This works!! Thank you

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