For example:
Groupvar Value
A 5
A 1
B 0
B 9
B 8
C 2
C 2
I want to rank by the Groupvar summarising on Value. So in this example sum(A) = 6, sum(B) = 17, sum(C) = 4. So Rank 1 = B, rank 2 = A, rank 3 = 6.
Ideal output:
Groupvar Value Rank
A 5 2
A 1 2
B 0 1
B 9 1
B 8 1
C 2 3
C 2 3
Any ideas how this can be done? I can create a proc summary > then rank> then merge the rank back on. But I'm wondering if there's a better way to do it.
You might be able to use the variable created by the LEVELS options plus ORDER=FREQ. For example.
data have;
input Groupvar:$1. Value;
cards;
A 5
A 1
B 0
B 9
B 8
C 2
C 2
;;;;
run;
proc summary nway order=freq missing;
class groupvar;
freq value;
output out=test(drop=_type_ index=(groupvar)) / levels;
run;
proc print;
run;
data want;
merge have test;
by groupvar;
run;
proc print;
run;
you can do it at one proc sql:
data test1;
Ord + 1;
input Groupvar$ Value;
cards;
A 5
A 1
B 0
B 9
B 8
C 2
C 2
;
run;
proc sql noprint;
create table Test2 as
select Groupvar, sum(Value) as Sum
from Test1
group by Groupvar
order by Sum
;
create table Test3 as
select a.*, b.Sum, b.Rank
from Test1 as a
left join (
select a.*, monotonic() as Rank
from Test2 as a
) as b
on a.Groupvar = b.Groupvar
order by Ord
;
quit;
Related
I need to aggregate about ten different vars on different groupings using Proc SQL;
Is there a way to achieve SUM () OVER ( [ partition_by_clause ] order_by_clause) in one sql query with different partition by clauses.
I've made an example here
data have;
infile cards;
input a b c d e f;
cards;
1 2 3 4 5
2 2 4 5 6
1 4 3 4 7
3 4 4 5 8
;
run;
proc sql;
create table want as
select *,
sum a over partiton by (b,c) as a1,
sum b over partiton by (c,d) as b1
sum c over partiton by (d,e) as c1
sum d over partiton by (a,c) as d1
from have
;
quit;
I don't want to wirte multiple sql queries and grouping on different vars and calculating one var in each step.
Hope that makes sense.
Proc SQL does not implement windowing functions and thus partition syntax therein as found in other SQL implementations. You can only do partition by with passthrough SQL to a connection that allows such syntax.
You could perform such a computation in DATA step using hashes.
data have;
infile cards;
input a b c d e ;
cards;
1 2 3 4 5
2 2 4 5 6
1 4 3 4 7
3 4 4 5 8
;
run;
data want;
if 0 then set have;
length a1 b1 c1 d1 8;
declare hash a1s();
a1s.defineKey('b', 'c');
a1s.defineData('a1');
a1s.defineDone();
declare hash b1s();
b1s.defineKey('c', 'd');
b1s.defineData('b1');
b1s.defineDone();
declare hash c1s();
c1s.defineKey('d', 'e');
c1s.defineData('c1');
c1s.defineDone();
declare hash d1s();
d1s.defineKey('a', 'c');
d1s.defineData('d1');
d1s.defineDone();
do while (not end);
set have end=end;
if a1s.find() = 0 then a1+a; else a1=a; a1s.replace();
if b1s.find() = 0 then b1+b; else b1=b; b1s.replace();
if c1s.find() = 0 then c1+c; else c1=c; c1s.replace();
if d1s.find() = 0 then d1+d; else d1=d; d1s.replace();
end;
do while (not last);
set have end=last;
a1s.find();
b1s.find();
c1s.find();
d1s.find();
output;
end;
format _numeric_ 4.;
stop;
run;
I have a dataset in SAS and I want to Convert one column into string by the Product. I have attached the image of input and output required.
I need the Colomn STRING in the outut. can anyone please help me ?
I have coded a data step to create the input data:
data have;
input products $
dates
value
;
datalines;
a 1 0
a 2 0
a 3 1
a 4 0
a 5 1
a 6 1
b 1 0
b 2 1
b 3 1
b 4 1
b 5 0
b 6 0
c 1 1
c 2 0
c 3 1
c 4 1
c 5 0
c 6 1
;
Does the following suggested solution give you what you want?:
data want;
length string $ 20;
do until(last.products);
set have;
by products;
string = catx(',',string,value);
end;
do until(last.products);
set have;
by products;
output;
end;
run;
Here's my quick solution.
data temp;
length cat $20.;
do until (last.prod);
set have;
by prod notsorted;
cat=catx(',',cat,value);
end;
drop value date;
run;
proc sql;
create table want as
select have.*, cat as string
from have inner join temp
on have.prod=temp.prod;
quit;
I want to do some sum calculate for a data set. The challenge is I need to do both row sum AND column Sum by ID. Below is the example.
data have;
input ID var1 var2;
datalines;
1 1 1
1 3 2
1 2 3
2 0 5
2 1 3
3 0 1
;
run;
data want;
input ID var1 var2 sum;
datalines;
1 1 1 12
1 3 2 12
1 2 3 12
2 0 5 9
2 1 3 9
3 0 1 1
;
run;
Using SQL is cool, but SAS has nice data step!
proc sort data=have; by id; run;
data result;
set have;
by id;
retain sum 0;
if first.id then sum=0;
sum=sum+sum(var1,var2);
if last.id then output;
run;
proc sort data=result; by id; run;
data want;
merge have result;
by id;
run;
You will decide what to use...
Use SQL to do all of it in one step. Group only by ID, but keep var1 and var2 in the column selection. This will create the same data in want.
proc sql noprint;
create table want as
select ID
, var1
, var2
, sum(var1) + sum(var2) as sum
from have
group by ID
;
quit;
How to do below codes in proc sql.
Two proc statement and one merge given below.
proc sort data=new out=new1 nodupkey;
by id;
where roll=100;
run;
proc sort data new2 out =new4 nodupkey
by id;
where roll=100;
run;
data score;
merge new4 (in=a) new1;
by id;
if a;
run;
The merge you show is equivalent to SQL left-join. You want all the rows from "new2" and ignore all the rows from "new" that don't have a common id. The uniqueness of the id (per the pre-sorts) further supports a left-join equivalence.
Proc SQL;
select new.*, new2.*
from new2
left join new on new.id = new2.id
where roll=100
order by id;
quit;
For the scenario of atypical data where there is many:many ids in the merge, the left-join is not equivalent.
I did leave out the NODUPKEY equivalent. Presuming option EQUALS is in effect, the selection of a groups first row would be equivalent. The undocumented MONOTONIC() function can be used to apply a default row order to a sub-query, which can then be used in a by group having expression.
data LEFT;
input id x1 x2 x3;
datalines;
1 1 1 1
1 2 2 2
1 3 3 3
2 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
;
run;
data RIGHT;
input id y1 y2 y3 x1;
datalines;
1 1 1 1 11
2 1 1 1 22
3 1 2 3 4
3 2 3 4 5
3 3 4 5 6
4 1 1 1 44
6 6 6 6 6
;
run;
proc sql;
select
LEFT.id
, coalesce(RIGHT.x1,LEFT.x1) as x1
, LEFT.x2
, LEFT.x3
, RIGHT.y1
, RIGHT.y2
, RIGHT.y3
from
(
select * from (select monotonic() as _seq_, * from LEFT) group by id having _seq_ = min(_seq_)
)
as LEFT
left join
(
select * from (select monotonic() as _seq_, * from RIGHT) group by id having _seq_ = min(_seq_)
)
as RIGHT
on
LEFT.id = RIGHT.id
;
I feel the need to reiterate that SQL left join is not always the same a merge, and SQL does not have common variable 'overlaying' that is implicit in DATA Step. When LEFT and RIGHT collide on non-key variables, you need to select a coalescence of the common variables into a new like-named variable in the output.
Suppose a data are as follows:
A B C
1 3 2
1 4 9
2 6 0
2 7 3
where A B and C are the variable names.
Is there a way to transform the table to
A 1
A 1
A 2
A 2
B 3
B 4
B 6
B 7
C 2
C 9
C 0
C 3
Expanding on the advice from #donPablo, here's how you would code it. Create an array to read across the data, then output each iteration of that array so you end up with the number of rows being the rows * columns from the original dataset. The VNAME function enables you to store the variable name (A, B, C) as a value in a separate variable.
data have;
input A B C;
datalines;
1 3 2
1 4 9
2 6 0
2 7 3
;
run;
data want;
set have;
length var1 $10;
array vars{*} _numeric_;
do i=1 to dim(vars);
var1=vname(vars{i});
var2=vars{i};
keep var1 var2;
output;
end;
run;
proc sort data=want;
by var1;
run;
The least amount of (expensive) development time might be --
Read and store the first row
For each subsequent row
Read the row
Create three records
Until end
Sort
How many times will this be run? Per day/ per year?
What number of rows are there?
Might we save 1 hr / month? 1 min / year? Something will need to read the entire file. Optomize last. Make it work first.
tkx
It should work correctly:
DATA A(keep A);
new_var = 'A';
SET your_data;
RUN;
DATA B(keep B);
new_var = 'B';
SET your_data;
RUN;
DATA C(keep C);
new_var = 'C';
SET your_data;
RUN;
PROC APPEND base=A data=B FORCE;
RUN;
PROC APPEND base=A data=C FORCE;
RUN;
Data A is a result data set.