How to generate a repeating series of numbers in a column in SAS, from 1 to x?
Suppose x is 3.
Data is like:
name age
A 15
D 16
C 21
B 35
E 79
F 85
G 64
and I want to add a column named list, like this:
name age list
A 15 1
D 16 2
C 21 3
B 35 1
E 79 2
F 85 3
G 64 1
data class;
set sashelp.class;
if list>=3 then list=0;
list+1;
run;
Easiest way I can think of is to use mod and the iteration counter.
data want;
set have;
list = 1 + mod(_N_ - 1,3);
run;
mod is the modulo function (gives the remainder after dividing).
So if you want that to vary based on some parameter, well, change the 3 to a parameter.
%let num_atwork = 2;
data want;
set have;
list = 1 + mod(_N_ - 1, &num_atwork.);
run;
Related
I have been tasked with taking the following data and creating two permanent data sets from it. One of these permanent data sets is supposed to contain the average of the "value" column for each group (meaning there should only be four rows in the end, with a new column that represents the average of respective values for A, B, C, and D). Averages should exclude missing values, meaning that if category A has a missing value, it should be divided by 3, not 4. The second permanent data set needs to be the one row with the highest overall value in the "value" column (in this case, the row with D 09JUL2021 951 should be the only row exported). I am having a tough time extracting that single row for the second data set. If you know of a way to perform these operations simultaneously, please let me know. Thank you for your time!
Example data:
data work.have;
input type $ date DATE9. value;
datalines;
A 08JUL2021 .
A 09JUL2021 20
A 20JUL2021 55
A 20JUL2021 2
B 02JUL2021 9
B 22JUL2021 6
B 04JUL2021 8
B 07JUL2021 406
C 01JUL2021 215
C 28JUL2021 63
C 30JUL2021 78
C 21JUL2021 80
D 18JUL2021 951
D 09JUL2021 .
D 14JUL2021 54
D 08JUL2021 73
;
Here is what I tried:
data mylib.data1(keep=type date value value_avg) mylib.data2;
set work.have;
by type;
if value ne . then NotMissing=1; else NotMissing=0;
if first.type then call missing(of value_avg);
value_avg+value;
if first.type then call missing(of num_per_cat);
num_per_cat+NotMissing;
Avg=divide((value_avg+value),(num_per_cat+NotMissing));
if last.type then output mylib.data1;
run;
This was successful for me with calculating averages, but I have no idea how to extract the row with the highest value in the "value" column to a second data set.
data work.have;
input type $ date DATE9. value;
datalines;
A 08JUL2021 .
A 09JUL2021 20
A 20JUL2021 55
A 20JUL2021 2
B 02JUL2021 9
B 22JUL2021 6
B 04JUL2021 8
B 07JUL2021 406
C 01JUL2021 215
C 28JUL2021 63
C 30JUL2021 78
C 21JUL2021 80
D 18JUL2021 951
D 09JUL2021 .
D 14JUL2021 54
D 08JUL2021 73
;
proc summary data = have nway;
class type;
var value;
output out = want_mean(drop = _:) mean = ;
run;
proc summary data = have nway;
class type;
var value;
output out = want_max(drop = _:) max = ;
run;
Both sets are easelly done by proc sql.
First one:
proc sql;
create table want1 as
select distinct type, max(value) as Max_value, mean(value) as Average_value
from have
group by type
;
quit;
Second one:
proc sql;
create table want2 as
select *
from have
having value = max(value)
;
quit;
I have observations with column ID, a, b, c, and d. I want to count the number of unique values in columns a, b, c, and d. So:
I want:
I can't figure out how to count distinct within each row, I can do it among multiple rows but within the row by the columns, I don't know.
Any help would be appreciated. Thank you
********************************************UPDATE*******************************************************
Thank you to everyone that has replied!!
I used a different method (that is less efficient) that I felt I understood more. I am still going to look into the ways listed below however to learn the correct method. Here is what I did in case anyone was wondering:
I created four tables where in each table I created a variable named for example ‘abcd’ and placed a variable under that name.
So it was something like this:
PROC SQL;
CREATE TABLE table1_a AS
SELECT
*
a as abcd
FROM table_I_have_with_all_columns
;
QUIT;
PROC SQL;
CREATE TABLE table2_b AS
SELECT
*
b as abcd
FROM table_I_have_with_all_columns
;
QUIT;
PROC SQL;
CREATE TABLE table3_c AS
SELECT
*
c as abcd
FROM table_I_have_with_all_columns
;
QUIT;
PROC SQL;
CREATE TABLE table4_d AS
SELECT
*
d as abcd
FROM table_I_have_with_all_columns
;
QUIT;
Then I stacked them (this means I have duplicate rows but that ok because I just want all of the variables in 1 column and I can do distinct count.
data ALL_STACK;
set
table1_a
table1_b
table1_c
table1_d
;
run;
Then I counted all unique values in ‘abcd’ grouped by ID
PROC SQL ;
CREATE TABLE count_unique AS
SELECT
My_id,
COUNT(DISTINCT abcd) as Count_customers
FROM ALL_STACK
GROUP BY my_id
;
RUN;
Obviously, it’s not efficient to replicate a table 4 times just to put a variables under the same name and then stack them. But my tables were somewhat small enough that I could do it and then immediately delete them after the stack. If you have a very large dataset this method would most certainly be troublesome. I used this method over the others because I was trying to use Procs more than loops, etc.
A linear search for duplicates in an array is O(n2) and perfectly fine for small n. The n for a b c d is four.
The search evaluates every pair in the array and has a flow very similar to a bubble sort.
data have;
input id a b c d; datalines;
11 2 3 4 4
22 1 8 1 1
33 6 . 1 2
44 . 1 1 .
55 . . . .
66 1 2 3 4
run;
The linear search for duplicates will occur on every row, and the count_distinct will be initialized automatically in each row to a missing (.) value. The sum function is used to increment the count when a non-missing value is not found in any prior array indices.
* linear search O(N**2);
data want;
set have;
array x a b c d;
do i = 1 to dim(x) while (missing(x(i)));
end;
if i <= dim(x) then count_distinct = 1;
do j = i+1 to dim(x);
if missing(x(j)) then continue;
do k = i to j-1 ;
if x(k) = x(j) then leave;
end;
if k = j then count_distinct = sum(count_distinct,1);
end;
drop i j k;
run;
Try to transpose dataset, each ID becomes one column, frequency each ID column by option nlevels, which count frequency of value, then merge back with original dataset.
Proc transpose data=have prefix=ID out=temp;
id ID;
run;
Proc freq data=temp nlevels;
table ID:;
ods output nlevels=count(keep=TableVar NNonMisslevels);
run;
data count;
set count;
ID=compress(TableVar,,'kd');
drop TableVar;
run;
data want;
merge have count;
by id;
run;
one more way using sortn and using conditions.
data have;
input id a b c d; datalines;
11 2 3 4 4
22 1 8 1 1
33 6 . 1 2
44 . 1 1 .
55 . . . .
66 1 2 3 4
77 . 3 . 4
88 . 9 5 .
99 . . 2 2
76 . . . 2
58 1 1 . .
50 2 . 2 .
66 2 . 7 .
89 1 1 1 .
75 1 2 3 .
76 . 5 6 7
88 . 1 1 1
43 1 . . 1
31 1 . . 2
;
data want;
set have;
_a=a; _b=b; _c=c; _d=d;
array hello(*) _a _b _c _d;
call sortn(of hello(*));
if a=. and b = . and c= . and d =. then count=0;
else count=1;
do i = 1 to dim(hello)-1;
if hello(i) = . then count+ 0;
else if hello(i)-hello(i+1) = . then count+0;
else if hello(i)-hello(i+1) = 0 then count+ 0;
else if hello(i)-hello(i+1) ne 0 then count+ 1;
end;
drop i _:;
run;
You could just put the unique values into a temporary array. Let's convert your photograph into data.
data have;
input id a b c d;
datalines;
11 2 3 4 4
22 1 8 1 1
33 6 . 1 2
44 . 1 1 .
;
So make an array of the input variables and another temporary array to hold the unique values. Then loop over the input variables and save the unique values. Finally count how many unique values there are.
data want ;
set have ;
array unique (4) _temporary_;
array values a b c d ;
call missing(of unique(*));
do _n_=1 to dim(values);
if not missing(values(_n_)) then
if not whichn(values(_n_),of unique(*)) then
unique(_n_)=values(_n_)
;
end;
count=n(of unique(*));
run;
Output:
Obs id a b c d count
1 11 2 3 4 4 3
2 22 1 8 1 1 2
3 33 6 . 1 2 3
4 44 . 1 1 . 1
I have a SAS Table like:
DATA test;
INPUT id sex $ age inc r1 r2 Zaehler work $;
DATALINES;
1 F 35 17 7 2 1 w
17 M 40 14 5 5 1 w
33 F 35 6 7 2 1 w
49 M 24 14 7 5 1 w
65 F 52 9 4 7 1 w
81 M 44 11 7 7 1 w
2 F 35 17 6 5 1 n
18 M 40 14 7 5 1 n
34 F 47 6 6 5 1 n
50 M 35 17 5 7 1 w
;
PROC PRINT; RUN;
proc sort data=have;
by county;
run;
I want compare rows if sex and age is equal and build sum over Zaehler. For example:
1 F 35 17 7 2 1 w
and
33 F 35 6 7 2 1 w
sex=f and age=35 are equale so i want to merge them like:
id sex age inc r1 r2 Zaehler work
1 F 35 17 7 2 2 w
I thought i can do it with proc sql but i can't use sum in proc sql. Can someone help me out?
PROC SUMMARY is the normal way to compute statistics.
proc summary data=test nway ;
class sex age ;
var Zaehler;
output out=want sum= ;
run;
Why would you want to include variables other than SEX, AGE and Zaehler in the output?
Your requirement is not difficult to understand or to satisfy, however, I am not sure what is your underline reason for doing this. Explain more on your purpose may help to facilitate better answers that work from the root of your project. Although I have a feeling the PROC MEAN may give you better matrix, here is a one step PROC SQL solution to get you the summary as well as retaining "the value of first row":
proc sql;
create table want as
select id, sex , age, inc, r1, r2, sum(Zaehler) as Zaehler, work
from test
group by sex, age
having id = min(id) /*This is tell SAS only to keep the row with the smallest id within the same sex,age group*/
;
quit;
You can use proc sql to sum over sex and age
proc sql;
create table sum as
select
sex
,age
,sum(Zaehler) as Zaehler_sum
from test
group by
sex
,age;
quit;
You can than join it back to the main table if you want to include all the variables
proc sql;
create table test_With_Sum as
select
t.*
,s.Zaehler_sum
from test t
inner join sum s on t.sex = s.sex
and t.age = s.age
order by
t.sex
,t.age
;
quit;
You can write it all as one proc sql query if you wish and the order by is not needed, only added for a better visibility of summarised results
Not a good solution. But it should give you some ideas.
DATA test;
INPUT id sex $ age inc r1 r2 Zaehler work $;
DATALINES;
1 F 35 17 7 2 1 w
17 M 40 14 5 5 1 w
33 F 35 6 7 2 1 w
49 M 24 14 7 5 1 w
65 F 52 9 4 7 1 w
81 M 44 11 7 7 1 w
2 F 35 17 6 5 1 n
18 M 40 14 7 5 1 n
34 F 47 6 6 5 1 n
50 M 35 17 5 7 1 w
;
run;
data t2;
set test;
nobs = _n_;
run;
proc sort data=t2;by descending sex descending age descending nobs;run;
data t3;
set t2;
by descending sex descending age;
if first.age then count = 0;
count + 1;
zaehler = count;
if last.age then output;
run;
proc sort data=t3 out=want(drop=nobs count);by nobs sex age;run;
thanks for your help. Here is my final code.
proc sql;
create table sum as
select distinct
sex
,age
,sum(Zaehler) as Zaehler
from test
WHERE work = 'w'
group by
sex
,age
;
PROC PRINT;quit;
I just modify the code a little bit. I filtered the w and i merg the Columns with the same value.
It was just an example the real Data is much bigger and has more Columns and rows.
I am trying to do a recursive lag in sas, the problem that I just learned is that x = lag(x) does not work in SAS.
The data I have is similar in format to this:
id date count x
a 1/1/1999 1 10
a 1/1/2000 2 .
a 1/1/2001 3 .
b 1/1/1997 1 51
b 1/1/1998 2 .
What I want is that given x for the first count, I want each successive x by id to be the lag(x) + some constant.
For example, lets say: if count > 1 then x = lag(x) + 3.
The output that I would want is:
id date count x
a 1/1/1999 1 10
a 1/1/2000 2 13
a 1/1/2001 3 16
b 1/1/1997 1 51
b 1/1/1998 2 54
Yes, the lag function in SAS requires some understanding. You should read through the documentation on it (http://support.sas.com/documentation/cdl/en/lefunctionsref/67398/HTML/default/viewer.htm#n0l66p5oqex1f2n1quuopdvtcjqb.htm)
When you have conditional statements with a lag inside the "then", I tend to use a retained variable.
data test;
input id $ date count x;
informat date anydtdte.;
format date date9.;
datalines;
a 1/1/1999 1 10
a 1/1/2000 2 .
a 1/1/2001 3 .
b 1/1/1997 1 51
b 1/1/1998 2 .
;
run;
data test(drop=last);
set test;
by id;
retain last;
if ^first.id then do;
if count > 1 then
x = last + 3;
end;
last = x;
run;
I am trying to detect groups which contain the difference between first age and second age are greater than 5. For example, if I have the following data, the difference between age in grp=1 is 39 so I want to output that group in a separate data set. Same goes for grp 4.
id grp age sex
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
My initial idea was to sort them by grp and then get the absolute value between ages using something like if first.grp then do;. But I don't know how to get the absolute value between first age and second age by group or actually I don't know how should I start this.
Thanks in advance.
Here's one way that I think works.
data have;
input id $ grp $ age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
proc sort data=have ;
by grp descending age;
run;
data temp(keep=grp);
retain old;
set have;
by grp descending age;
if first.grp then old=age;
if last.grp then do;
diff=old-age;
if diff>5 then output ;
end;
run;
Data want;
merge temp(in=a) have(in=b);
by grp ;
if a and b;
run;
I would use PROC TRANSPOSE so the values in each group can easily be compared. For example:
data groups1;
input id $ grp age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
run;
proc sort data=groups1;
by grp; /* This maintains age order */
run;
proc transpose data=groups1 out=groups2;
by grp;
var age;
run;
With the transposed data you can do whatever comparison you like (I can't tell from your question what exactly you want, so I just compare first two ages):
/* With all ages of a particular group in a single row, it is easy to compare */
data outgroups1(keep=grp);
set groups2;
if abs(col1-col2)>5 then output;
run;
In this instance this would be my preferred method for creating a separate data set for each group that satisfies whatever condition is applied (generate and include code dynamically):
/* A separate data set per GRP value in OUTGROUPS1 */
filename dynacode catalog "work.dynacode.mycode.source";
data _null_;
set outgroups1;
file dynacode;
put "data grp" grp ";";
put " set groups1(where=(grp=" grp "));";
put "run;" /;
run;
%inc dynacode;
If you are after the difference between just the 1st and 2nd ages, then the following code is a fairly straightforward way of extracting these. It reads though the dataset to identify the groups, then uses the direct access method, POINT=, to extract the relevant records. I put in an extra condition, grp=lag(grp) just in case you have any groups with only 1 record.
data want;
set have;
by grp;
if first.grp then do;
num_grp=0;
outflag=0;
end;
outflag+ifn(lag(first.grp)=1 and grp=lag(grp) and abs(dif(age))>5,1,0) /* set flag to determine if group meets criteria */;
if not first.grp then num_grp+1; /* count number of records in group */
if last.grp and outflag=1 then do i=_n_-num_grp to _n_;
set have point=i; /* extract required group records */
drop num_grp outflag;
output;
end;
run;
Here's an SQL approach (using CarolinaJay's code to create the dataset):
data groups1;
input id grp age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
run;
proc sql noprint;
create table xx as
select a.*
from groups1 a
where grp in (select b.grp
from groups1 b
join groups1 c on c.id = b.id+1
and c.grp = b.grp
and abs(c.age - b.age) > 5
left join groups1 d on d.id = b.id-1
and d.grp = b.grp
where d.id eq .
)
;
quit;
The join on C finds all occurrences where the subsequent record in the same group has an absolute value > 5. The join on D (and the where clause) makes sure we only consider the results from the C join if the record is the very first record in the group.
data have;
input id $ grp $ age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
data want;
do i = 1 by 1 until(last.grp);
set have;
by grp notsorted;
if first.grp then cnt = 0;
cnt + 1;
if cnt = 1 then age1 = age;
if cnt = 2 then age2 = age;
diff = sum( age1, -age2 );
end;
do until(last.grp);
set have;
by grp;
if diff > 5 then output;
end;
run;