how to count the number of positive and negative and calculate both of them(mean)

[victory]

hi all,

I have the data from19961/01/01 to 2014/12/31. 

Below is part of my data

Date	Mist(1)
1996-06-20	-0.0027
1996-06-21	0.0026
1996-06-24	0.0042
1996-06-25	0.0045
1996-06-26	0.0018
1996-06-27	0.0016
1996-06-28	-0.0002
1996-07-01	0.0027
1996-07-02	0.0009
1996-07-03	-0.0005
1996-07-04	0.0070

how can i output the chart like this.

thanks all,

Y

[PaigeMiller · Accepted solution]

data intermediate;
    set have;
    pos_neg=sign(mist);
run;
proc summary data=intermediate;
    class date pos_neg;
    var mist;
    format date year4.;
    output out=want mean= std= n=/autoname;
run;

This code is untested as we cannot write code to work with data that is a screen capture. In the future, please provide code as working SAS data step code (instructions).

[Tom]

Seems simple enough.  What out for testing for negative.  Missing values are considered smaller than any actual value.

Also you did not say which group should include the zeros, so I made them their own group.  You did not say whether or not you wanted NOBS to count the missing values or not.  I decided to exclude them.

data have;
  input date :yymmdd. mist ;
cards;
1996-06-20 -0.0027
1996-06-21 0.0026
1996-06-24 0.0042
1996-06-25 0.0045
1996-06-26 0.0018
1996-06-27 0.0016
1996-06-28 -0.0002
1996-07-01 0.0027
1996-07-02 0.0009
1996-07-03 -0.0005
1996-07-04 0.0070
;

proc sql ;
  create table want as 
  select year(date) as year
       , mean(mist) as mean
       , std(mist) as sd
       , sum(mist>0) as positive
       , sum(not (mist>=0)) as negative
       , mean(case when (mist>0) then mist end) as mean_postive
       , mean(case when (mist>0) then . else mist end) as mean_negative
       , count(mist) as nobs
       , sum(mist=0) as zero
       , nmiss(mist) as missing
  from have
  group by year
;
quit;
proc print;
run;