I hope this is not a duplicate question. I've searched the forum and retain function seems to be choice of weapon but it copies down an observation, and I'm trying to do the following; for a given id, copy the second line to the first line for the x value. also first value of x is always 2.
Here's my data;
id x
3 2
3 1
3 1
2 2
2 1
2 1
6 2
6 0
6 0
and i want it to look like this;
id x
3 1
3 1
3 1
2 1
2 1
2 1
6 0
6 0
6 0
and here's the starter code;
data have;
input id x;
cards;
3 2
3 1
3 1
2 2
2 1
2 1
6 2
6 0
6 0
;
run;
Lead is tricky in SAS. You can sort in reverse and use a lag function to get around it though, and you are right: a retain statement will allow us to add an order variable so we can sort it back to its original format.
data have;
set have;
retain order;
lagid = lag(id);
if id ne lagid then order = 0;
order = order + 1;
drop lagid;
run;
proc sort data=have; by id descending order; run;
data have;
set have;
leadx = lag(x);
run;
proc sort data=have; by id order; run;
data have;
set have;
if order = 3 then x_fixed = x;
else x_fixed = leadx;
run;
If your data is exactly as you say, then you can use a lookahead merge. It literally takes the dataset and merges itself to a copy of the dataset that starts on row 2, side-to-side. You just have to check that you're still on the same ID. This does change the value of x for all records to the value one hence, not just the first; you could add additional code to pay attention to that (but can't use FIRST and LAST).
data want;
merge have have(firstobs=2 rename=(id=newid x=newx));
if newid=id then x=newx;
keep x id;
run;
If you don't have any additional variables of interest, then you can do something even more interesting: duplicate the second row in its entirety and delete the first row.
data want;
set have;
by id notsorted;
if first.id then do;
firstrow+1;
delete;
end;
if firstrow=1 then do;
firstrow=0;
output;
end;
output;
run;
However, the "safest" method (in terms of doing most likely what you want precisely) is the following, which is a DoW loop.
data want;
idcounter=0;
do _n_ = 1 by 1 until (last.id);
set have;
by id notsorted;
idcounter+1;
if idcounter=2 then second_x = x;
end;
do _n_=1 by 1 until (last.id);
set have;
by id notsorted;
if first.id then x=second_x;
output;
end;
run;
This identifies the second x in the first loop, for that BY group, then in the second loop sets it to the correct value for row 1 and outputs.
In both of the latter examples I assume your data is organized by ID but not truly sorted (like your initial example is). If it's not organized by ID, you need to perform a sort first (but then can remove the NOTSORTED).
Related
I need some help in trying to execute a comparison of rows within different ID variable groups, all in a single dataset.
That is, if there is any duplicate observation within two or more ID groups, then I'd like to delete the observation entirely.
I want to identify any duplicates between rows of different groups and delete the observation entirely.
For example:
ID Value
1 A
1 B
1 C
1 D
1 D
2 A
2 C
3 A
3 Z
3 B
The output I desire is:
ID Value
1 D
3 Z
I have looked online extensively, and tried a few things. I thought I could mark the duplicates with a flag and then delete based off that flag.
The flagging code is:
data have;
set want;
flag = first.ID ne last.ID;
run;
This worked for some cases, but I also got duplicates within the same value group flagged.
Therefore the first observation got deleted:
ID Value
3 Z
I also tried:
data have;
set want;
flag = first.ID ne last.ID and first.value ne last.value;
run;
but that didn't mark any duplicates at all.
I would appreciate any help.
Please let me know if any other information is required.
Thanks.
Here's a fairly simple way to do it: sort and deduplicate by value + ID, then keep only rows with values that occur only for a single ID.
data have;
input ID Value $;
cards;
1 A
1 B
1 C
1 D
1 D
2 A
2 C
3 A
3 Z
3 B
;
run;
proc sort data = have nodupkey;
by value ID;
run;
data want;
set have;
by value;
if first.value and last.value;
run;
proc sql version:
proc sql;
create table want as
select distinct ID, value from have
group by value
having count(distinct id) =1
order by id
;
quit;
This is my interpretation of the requirements.
Find levels of value that occur in only 1 ID.
data have;
input ID Value:$1.;
cards;
1 A
1 B
1 C
1 D
1 D
2 A
2 C
3 A
3 Z
3 B
;;;;
proc print;
proc summary nway; /*Dedup*/
class id value;
output out=dedup(drop=_type_ rename=(_freq_=occr));
run;
proc print;
run;
proc summary nway;
class value;
output out=want(drop=_type_) idgroup(out[1](id)=) sum(occr)=;
run;
proc print;
where _freq_ eq 1;
run;
proc print;
run;
A slightly different approach can use a hash object to track the unique values belonging to a single group.
data have; input
ID Value:& $1.; datalines;
1 A
1 B
1 C
1 D
1 D
2 A
2 C
3 A
3 Z
3 B
run;
proc delete data=want;
proc ds2;
data _null_;
declare package hash values();
declare package hash discards();
declare double idhave;
method init();
values.keys([value]);
values.data([value ID]);
values.defineDone();
discards.keys([value]);
discards.defineDone();
end;
method run();
set have;
if discards.find() ne 0 then do;
idhave = id;
if values.find() eq 0 and id ne idhave then do;
values.remove();
discards.add();
end;
else
values.add();
end;
end;
method term();
values.output('want');
end;
enddata;
run;
quit;
%let syslast = want;
I think what you should do is:
data want;
set have;
by ID value;
if not first.value then flag = 1;
else flag = 0;
run;
This basically flags all occurrences of a value except the first for a given ID.
Also I changed want and have assuming you create what you want from what you have. Also I assume have is sorted by ID value order.
Also this will only flag 1 D above. Not 3 Z
Additional Inputs
Can't you just do a sort to get rid of the duplicates:
proc sort data = have out = want nodupkey dupout = not_wanted;
by ID value;
run;
So if you process the observations by VALUE levels (instead of by ID levels) then you just need keep track of whether any ID is ever different than the first one.
data want ;
do until (last.value);
set have ;
by value ;
if first.value then first_id=id;
else if id ne first_id then remapped=1;
end;
if not remapped;
keep value id;
run;
I am matching files base on IDs numbers. I need to format a data set with the IDs to be matched, so that the same ID number is not repeated in column a (because column b's ID is the surviving ID after the match is completed). My list of IDs has over 1 million observations, and the same ID may be repeated multiple times in either/both columns.
Here is an example of what I've got/need:
Sample Data
ID1 ID2
1 2
3 4
2 5
6 1
1 7
5 8
The surviving IDs would be:
2
4
5
error - 1 no longer exists
error - 1 no longer exists
8
WHAT I NEED
ID1 ID2
1 2
3 4
2 5
6 5
5 7
7 8
I am, probably very obviously, a SAS novice, but here is what I have tried, re-running over and over again because I have some IDs that are repeated upward of 50 times or more.
Proc sort data=Have;
by ID1;
run;
This sort makes the repeated ID1 values consecutive, so the I could use LAG to replace the destroyed ID1s with the surviving ID2 from the line above.
Data Want;
set Have;
by ID1;
lagID1=LAG(ID1);
lagID2=LAG(ID2);
If NOT first. ID1 THEN DO;
If ID1=lagID1 THEN ID1=lagID2;
KEEP ID1 ID2;
IF ID1=ID2 then delete;
end;
run;
That sort of works, but I still end up with some that end up with duplicates that won't resolve no matter how many times I run (I would have looped it, but I don't know how), because they are just switching back and forth between IDs that have other duplicates (I can get down to about 2,000 of these).
I have figured out that instead of using LAG, I need replace all values after the current line with ID2 for each ID1 value, but I cannot figure out how to do that.
I want to read observation 1, find all later instances of the value of ID1, in both ID1 or ID2 columns, and replace that value with the current observation's ID2 value. Then I want to repeat that process with line 2 and so on.
For the example, I would want to look for any instances after line one of the value 1, and replace it with 2, since that is the surviving ID of that pair - 1 may appear further down multiple times in either of the columns, and I need all them to replaced. Line two would look for later values of 3 and replace them with 4, and so one. The end result should be that an ID number only appears once ever in the ID1 column (though it may appear multiple times in the ID2 column).
ID1 ID2
1 2
3 4
2 5
6 1
1 7
5 8
After first line has been read, data set would look as follows:
ID1 ID2
1 2
3 4
2 5
6 2
2 7
5 8
Reading observation two would make no changes since 3 does not appear again; after observation 3, the set would be:
ID1 ID2
1 2
3 4
2 5
6 5
5 7
5 8
Again, there would be not changes from observation four. but observation 5 would cause the final change:
ID1 ID2
1 2
3 4
2 5
6 5
5 7
7 8
I have tried using the following statement but I can't even tell if I am on the complete wrong track or if I just can't get the syntax figured out.
Data want;
Set have;
Do i=_n_;
ID=ID2;
Replace next var{EUID} where (EUID1=EUID1 AND EUID2=EUID1);
End;
Run;
Thanks for your help!
There is no need to work back and forth thru the data file. You just need to retain the replacement information so that you can process the file in a single pass.
One way to do that is to make a temporary array using the values of the ID variables as the index. That is easy to do for your simple example with small ID values.
So for example if all of the ID values are integers between 1 and 1000 then this step will do the job.
data want ;
set have ;
array xx (1000) _temporary_;
do while (not missing(xx(id1))); id1=xx(id1); end;
do while (not missing(xx(id2))); id2=xx(id2); end;
output;
xx(id1)=id2;
run;
You probably need to add a test to prevent cycles (1 -> 2 -> 1).
For a more general solution you should replace the array with a hash object instead. So something like this:
data want ;
if _n_=1 then do;
declare hash h();
h.definekey('old');
h.definedata('new');
h.definedone();
call missing(new,old);
end;
set have ;
do while (not h.find(key:id1)); id1=new; end;
do while (not h.find(key:id2)); id2=new; end;
output;
h.add(key: id1,data: id2);
drop old new;
run;
Here's an implementation of the algorithm you've suggested, using a modify statement to load and rewrite each row one at a time. It works with your trivial example but with messier data you might get duplicate values in ID1.
data have;
input ID1 ID2 ;
datalines;
1 2
3 4
2 5
6 1
1 7
5 8
;
run;
title "Before making replacements";
proc print data = have;
run;
/*Optional - should improve performance at cost of increased memory usage*/
sasfile have load;
data have;
do i = 1 to nobs;
do j = i to nobs;
modify have point = j nobs = nobs;
/* Make copies of target and replacement value for this pass */
if j = i then do;
id1_ = id1;
id2_ = id2;
end;
else do;
flag = 0; /* Keep track of whether we made a change */
if id1 = id1_ then do;
id1 = id2_;
flag = 1;
end;
if id2 = id1_ then do;
id2 = id2_;
flag = 1;
end;
if flag then replace; /* Only rewrite the row if we made a change */
end;
end;
end;
stop;
run;
sasfile have close;
title "After making replacements";
proc print data = have;
run;
Please bear in mind that as this modifies the dataset in place, interrupting the data step while it is running could result in data loss. Make sure you have a backup first in case you need to roll your changes back.
Seems like this should do the trick and is fairly straight forward. Let me know if it is what you are looking for:
data have;
input id1 id2;
datalines;
1 2
3 4
2 5
6 1
1 7
5 8
;
run;
%macro test();
proc sql noprint;
select count(*) into: cnt
from have;
quit;
%do i = 1 %to &cnt;
proc sql noprint;
select id1,id2 into: id1, :id2
from have
where monotonic() = &i;quit;
data have;
set have;
if (_n_ > input("&i",8.))then do;
if (id1 = input("&id1",8.))then id1 = input("&id2",8.);
if (id2 = input("&id1",8.))then id2 = input("&id2",8.);
end;
run;
%end;
%mend test;
%test();
this might be a little faster:
data have2;
input id1 id2;
datalines;
1 2
3 4
2 5
6 1
1 7
5 8
;
run;
%macro test2();
proc sql noprint;
select count(*) into: cnt
from have2;
quit;
%do i = 1 %to &cnt;
proc sql noprint;
select id1,id2 into: id1, :id2
from have2
where monotonic() = &i;
update have2 set id1 = &id2
where monotonic() > &i
and id1 = &id1;
quit;
proc sql noprint;
update have2 set id2 = &id2
where monotonic() > &i
and id2 = &id1;
quit;
%end;
%mend test2;
%test2();
In my sas data set there are groups, i.e. id and I want delete groups with missing values in a certain variable.
For example I have this sas data set:
data have;
input v1 v2 v3 id;
datalines;
9 7 210 1
0 6 . 1
9 3 320 2
6 1 . 1
9 4 432 2
;
run;
I tried this:
/*Order by id*/
proc sort data=have;
by id;
run;
/*Select no missing observations by id*/
data=want;
set=have;
if cmiss(of _all_) then delete;
run;
However this code does not exclude id's with missing values. It delete missing values.
Hmmm. You can use proc sql for this:
proc sql;
delete from have
where exists (select 1 from have have2 where have.id = have2.id and (have2.v1 is null or have2.v2 is null or have2.v3 is null);
One idea might be to use a double DOW loop. First to check for any missing values and then a second one to output the records for the ids with no missing values.
data have;
input v1 v2 v3 id;
datalines;
9 7 210 1
0 6 . 1
9 3 320 2
6 1 . 1
9 4 432 2
1 2 333 3
;
You will need to sort as in your example.
data want ;
do until (last.id);
set have;
by id;
anymissing=max(anymissing,cmiss(of v1-v3));
end;
do until (last.id);
set have;
by id;
if not anymissing then output;
end;
run;
You just dont want to have lines with missing Columns in your result dataset. So why delete, just exclude them when writing result-dataset or overwrite source-Dataset.:
data have;/*overwriting my have dataset instead of deleting lines*/
set have;
if not cmiss(of _ALL_);
run;
When you want to remove all lines for a group if only one line has a missing value you can do this, Store an ID if it has no value and then dont write any line with that id, and you just get ID lines you want as result. Important is that the ID with missing value is first in dataset, but that should be that way because of proc sort:
data want;
retain x;
set have;
if cmiss(of _ALL_) then
x= id;
if x ne id;
run;
I want to delete the whole group that none of its observation has NUM=14
So something likes this:
Original DATA
ID NUM
1 14
1 12
1 10
2 13
2 11
2 10
3 14
3 10
Since none of the ID=2 contain NUM=14, I delete group 2.
And it should looks like this:
ID NUM
1 14
1 12
1 10
3 14
3 10
This is what I have so far, but it doesn't seem to work.
data originaldat;
set newdat;
by ID;
If first.ID then do;
IF NUM EQ 14 then Score = 100;
Else Score = 10;
end;
else SCORE+1;
run;
data newdat;
set newdat;
If score LT 50 then delete;
run;
An approach using proc sql would be:
proc sql;
create table newdat as
select *
from originaldat
where ID in (
select ID
from originaldat
where NUM = 14
);
quit;
The sub query selects the IDs for groups that contain an observation where NUM = 14. The where clause then limits the selected data to only these groups.
The equivalent data step approach would be:
/* Get all the groups that contain an observation where N = 14 */
data keepGroups;
set originaldat;
if NUM = 14;
keep ID;
run;
/* Sort both data sets to ensure the data step merge works as expected */
proc sort data = originaldat;
by ID;
run;
/* Make sure there are no duplicates values in the groups to be kept */
proc sort data = keepGroups nodupkey;
by ID;
run;
/*
Merge the original data with the groups to keep and only keep records
where an observation exists in the groups to keep dataset
*/
data newdat;
merge
originaldat
keepGroups (in = k);
by ID;
if k;
run;
In both datasets the subsetting if statement is used to only output observations when the condition is met. In the second case k is a temporary variable with value 1(true) when a value is read from keepGroups an 0(false) otherwise.
You're sort of getting at a DoW loop here, but not quite doing it right. The problem (Assuming the DATA/SET names are mistyped and not actually wrong in your program) is the first data step doesn't append that 100 to every row - only to the 14 row. What you need is one 'line' per ID value with a keep/no keep decision.
You can either do this by doing your first data step, but RETAIN score, and only output one row per ID. Your code would actually work, based on 14 being the first row, if you just fixed your data/set typo; but it only works when 14 is the first row.
data originaldat;
input ID NUM ;
datalines;
1 14
1 12
1 10
2 13
2 11
2 10
3 14
3 10
;;;;
run;
data has_fourteen;
set originaldat;
by ID;
retain keep;
If first.ID then keep=0;
if num=14 then keep=1;
if last.id then output;
run;
data newdata;
merge originaldat has_fourteen;
by id;
if keep=1;
run;
That works by merging the value from a 1-per-ID to the whole dataset.
A double DoW also works.
data newdata;
keep=0;
do _n_=1 by 1 until (last.id);
set originaldat;
by id;
if num=14 then keep=1;
end;
do _n_=1 by 1 until (last.id);
set originaldat;
by id;
if keep=1 then output;
end;
run;
This works because it iterates over the dataset twice; for each ID, it iterates once through all records, looking for a 14, if it finds one then setting keep to 1. Then it reads all records again for that ID, and keeps if keep=1. Then it goes on to the next set of records by ID.
data in;
input id num;
cards;
1 14
1 12
1 10
2 16
2 13
3 14
3 67
;
/* To find out the list of groups which contains num=14, use below SQL */
proc sql;
select distinct id into :lst separated by ','
from in
where num = 14;
quit;
/* If you want to create a new data set with only groups containing num=14 then use following data step */
data out;
set in;
where id in (&lst.);
run;
I have a data set with 3 observations, 1 2 3
4 5 6
7 8 9 , now i have to interchange 1 2 3 and 7 8 9.
How can do this in base sas?
If you just want to sort your dataset by a variable in descending order, use proc sort:
data example;
input number;
datalines;
123
456
789
;
run;
proc sort data = example;
by descending number;
run;
If you want to re-order a dataset in a more complex way, create a new variable containing the position that you want each row to be in, and then sort it by that variable.
If you want to swap the contents of the first and last observations while leaving the rest of the dataset in place, you could do something like this.
data class;
set sashelp.class;
run;
data firstobs;
i = 1;
set sashelp.class(obs = 1);
run;
data lastobs;
i = nobs;
set sashelp.class nobs = nobs point = nobs;
output;
stop;
run;
data transaction;
set lastobs firstobs;
/*Swap the values of i for first and last obs*/
retain _i;
if _n_ = 1 then do;
_i = i;
i = 1;
end;
if _n_ = 2 then i = _i;
drop _i;
run;
data class;
set transaction(keep = i);
modify class point = i;
set transaction;
run;
This modifies just the first and last observations, which should be quite a bit faster than sorting or replacing a large dataset. You can do a similar thing with the update statement, but that only works if your dataset is already sorted / indexed by a unique key.
By Sandeep Sharma:sandeep.sharmas091#gmail.com
data testy;
input a;
datalines;
1
2
3
4
5
6
7
8
9
;
run;
data ghj;
drop y;
do i=nobs-2 to nobs;
set testy point=i nobs=nobs;
output;
end;
do n=4 to nobs-3;
set testy point=n;
output;
end;
do y=1 to 3;
set testy;
output;
end;
stop;
run;