I have a series of string values with missing observations. I would like to use flat substitution. For instance variable x has 3 available values. There should be a 33.333% chance that a missing value will be assigned to the available values for x under this substitution method. How would I do this?
DATA have;
INPUT id a $ b $ c $ x;
CARDS;
1 Y Male . 5
2 Y Female . 4
3 . Female Tall 4
4 Y . Short 2
5 N Male Tall 1
;
Run;
You could use temporary arrays to store the possible values. Then generate a random index into the array.
DATA have;
INPUT id a $ b $ c $ x;
CARDS;
1 Y Male . 5
2 Y Female . 4
3 . Female Tall 4
4 Y . Short 2
5 N Male Tall 1
;
data want ;
set have ;
array possible_b (2) $8 ('Male','Female') ;
if missing(b) then b=possible_b(1+int(rand('uniform')*dim(possible_b)));
run;
I did this with generating random numbers and hard coding the limits. There should be an easier way to do this, but for the purposes of the question this should work.
option missing='';
data begin;
input a $;
cards;
a
.
b
c
.
e
.
f
g
h
.
.
j
.
;
run;
data intermediate;
set begin;
if a EQ '' then help= rand("uniform");
else help=.;
run;
data wanted;
set intermediate;
format help populated.;
if a EQ '' then do;
if 0<=help<0.33 then a='V1';
else if 0.33<=help<0.66 then a='V2';
else if 0.66<=help then a='V3';
end;
drop help;
run;
Related
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 don't know how to describe this question but here is an example. I have an initial dataset looks like this:
input first second $3.;
cards;
1 A
1 B
1 C
1 D
2 E
2 F
3 S
3 A
4 C
5 Y
6 II
6 UU
6 OO
6 N
7 G
7 H
...
;
I want an output dataset like this:
input first second $;
cards;
1 "A,B,C,D"
2 "E,F"
3 "S,A"
4 "C"
5 "Y"
6 "II,UU,OO,N"
7 "G,H"
...
;
Both tables will have two columns. Unique value of range of the column "first" could be 1 to any number.
Can someone help me ?
something like below
proc sort data=have;
by first second;
run;
data want(rename=(b=second));
length new_second $50.;
do until(last.first);
set have;
by first second ;
new_second =catx(',', new_second, second);
b=quote(strip(new_second));
end;
drop second new_second;
run;
output is
first second
1 "A,B,C,D"
2 "E,F"
3 "A,S"
4 "C"
5 "Y"
6 "II,N,OO,UU"
7 "G,H"
You can use by-group processing and the retain function to achieve this.
Create a sample dataset:
data have;
input id value $3.;
cards;
1 A
1 B
1 C
1 D
2 E
2 F
3 S
3 A
4 C
5 Y
6 II
6 UU
6 OO
6 N
7 G
7 H
;
run;
First ensure that your dataset is sorted by your id variable:
proc sort data=have;
by id;
run;
Then use the first. and last. notation to identify when the id variable is changing or about to change. The retain statement tells the datastep to keep the value within concatenated_value over observations rather than resetting it to a blank value. Use the quote() function to apply the " chars around the result before outputting the record. Use the cats() function to perform the actual concatenation and separate the records with a ,.
data want;
length contatenated_value $500.;
set have;
by id;
retain contatenated_value ;
if first.id then do;
contatenated_value = '';
end;
contatenated_value = catx(',', contatenated_value, value);
if last.id then do;
contatenated_value = quote(cats(contatenated_value));
output;
end;
drop value;
run;
Output:
contatenated_
value id
"A,B,C,D" 1
"E,F" 2
"S,A" 3
"C" 4
"Y" 5
"II,UU,OO,N" 6
"G,H" 7
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
so I've been trying to figure out how to rename the ID variable in SAS (I made a dummy dataset to attempt this, see below)
DATA trial;
input hno $ y;
datalines;
a 1
a 2
a 3
a 4
b 3
b 5
cd 5
cd 6
cd 1
;
run;
and what I need to do is have all a=1, b=2, cd=3 and so on, but the code would need to be transferrable to a dataset with ~30,000 observations all with varying ID's. I've been playing around with first.id and last.id but to absolutely no avail. Can anyone help?
Thank you in advance!
EDIT
So to clarify, I need code that produce the output:
a 1 1
a 2 1
a 3 1
a 4 1
b 3 2
b 5 2
cd 5 3
cd 6 3
cd 1 3
where the third column there is the ID variable that increases by one for each unique hno value
If you data are sorted by HNO this will encode HNO to an index. RENAME in SAS usually refers to objects like variables, data sets, etc.
DATA trial;
input hno $ y;
datalines;
a 1
a 2
a 3
a 4
b 3
b 5
cd 5
cd 6
cd 1
;
run;
data trial2;
set trial;
by hno;
if first.hno then id + 1;
run;
proc print;
run;
If your input aren't sorted you created and index ID data set using PROC SUMMARY and the add the ID with a KEYed SET.
DATA trial;
input hno $ y ##;
datalines;
a 1 a 2 b 3 b 5 cd 5 cd 6 a 3 a 4
cd 1
;
run;
proc summary nway data=trial;
class hno;
output out=index(drop=_type_ _freq_ rename=(_level_=id) index=(hno)) / levels;
run;
proc print;
run;
data trial2;
set trial;
set index key=hno/unique;
run;
proc print;
run;
You can also try the format
proc format;
value $ cn
'a' = 1
'b' = 2
'cd' = 3
;
run;
DATA trial;
input hno $ y;
*format hno $cn.;
id = put(hno, $cn.);
datalines;
a 1
a 2
a 3
a 4
b 3
b 5
cd 5
cd 6
cd 1
;
run;
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;