I’m learning about PROC MIXED in SAS to understand how to use Random and Repeated statement, using simple repeated data (pre, post).
I checked lots of similar questions, but I’m still a beginner, so have two below questions. Please give me some advice.
1.About paired test, there would be two cases, subject (id) as “fixed effect” or as “random effect” in the following simple repeated data (pre, post). I often see it as “fixed effect”, generally speaking, that’s the theory? Why?
2.In the following case, putting “random” and “repeated” together would be not correct? How should I do?
2021.3.21
I edited the following program. I got the result “random=repeated” like that. But I couldn’t understand about the “random+repeated”. It would be also correct? Anyway, id is no variance in this case, so, it wouldn’t affect the model?
/* data */
data dt00;
input id x y;
cards;
1 1 2
2 1 2
3 1 3
4 1 3
5 1 3
6 1 4
7 1 4
1 2 15
2 2 9
3 2 13
4 2 10
5 2 7
6 2 11
7 2 5
;
run;
title "random";
proc mixed data = dt00 covtest;* REML ;
parms/nobound;
class x id;
model y = x / ddfm = kr2;
random id;
* Estimate SE Z Pr Z;
*id -1.3333 2.5757 -0.52 0.6047;
*Residual 7.5000 4.3301 1.73 0.0416;
* DF F Pr F;
* 6 22.87 0.0031;
title "repeated";
proc mixed data = dt00 covtest;* REML ;
class x id;
model y = x / ddfm = kr2 ;
repeated x / subject = id type = cs;
run;
* Estimate SE Z Pr Z;
*id -1.3333 2.5757 -0.52 0.6047;
*Residual 7.5000 4.3301 1.73 0.0416;
* DF F Pr F;
* 6 22.87 0.0031;
title "random + repeated";
proc mixed data = dt00 covtest;* REML ;
parms/nobound;
class x id;
model y = x / solution ddfm = kr2 ;
random id;
repeated x / subject = id type = cs;
run;
* Estimate SE Z Pr Z;
*id 0.1117 2.5757 0.04 0.4827;
*CS -1.4451 0 . .;
*Residual 7.5000 4.3301 1.73 0.0416;
* DF F Pr F;
* 6 22.87 0.0031;
title "fixed effect";
proc mixed data = dt00 covtest;* REML ;
class x id;
model y = x id / solution ddfm = kr2 ;
run;
* Estimate SE Z Pr Z;
*Residual 7.5000 4.3301 1.73 0.0416 ;
* DF F Pr F;
* 6 22.87 0.0031;
title "paired ttest";
proc sort data = dt00; by id; run;
proc transpose data = dt00 out = dt01;
by id;
id x;
var y;
run;
data dt02; set dt01; diff = _2 - _1; run;
proc ttest data = dt02 alpha = 0.05;
paired _1 * _2;
run;
* DF T Pr T;
* 6 -4.78 0.0031;
When you specify RANDOM patient, you are saying that the covariance between patients (different people) is 0. (This is fine if there is not another grouping that would make patients more similar).
In PROC MIXED, You can include patient as a fixed factor, but that usually uses most of the degrees of freedom. If instead, you treat patient as a random factor, you are still controlling for each person, but you use less degrees of freedom.
If there is extra non-independence (or even non-constant variance), you can still estimate those non-zero covariances by adding a Repeated statement. Therefore, it’s fine to include a REPEATED statement along with a RANDOM statement, and is sometimes necessary to have a good fitting model. The repeated statement controls the covariance structure of the residuals for a single subject.
And everyone starts somewhere, so don't sweat it.
Related
Suppose the data set have contains various outliers which have been identified in an outliers data set. These outliers need to be replaced with missing values, as demonstrated below.
Have
Obs group replicate height weight bp cholesterol
1 1 A 0.406 0.887 0.262 0.683
2 1 B 0.656 0.700 0.083 0.836
3 1 C 0.645 0.711 0.349 0.383
4 1 D 0.115 0.266 666.000 0.015
5 2 A 0.607 0.247 0.644 0.915
6 2 B 0.172 333.000 555.000 0.924
7 2 C 0.680 0.417 0.269 0.499
8 2 D 0.787 0.260 0.610 0.142
9 3 A 0.406 0.099 0.263 111.000
10 3 B 0.981 444.000 0.971 0.894
11 3 C 0.436 0.502 0.563 0.580
12 3 D 0.814 0.959 0.829 0.245
13 4 A 0.488 0.273 0.463 0.784
14 4 B 0.141 0.117 0.674 0.103
15 4 C 0.152 0.935 0.250 0.800
16 4 D 222.000 0.247 0.778 0.941
Want
Obs group replicate height weight bp cholesterol
1 1 A 0.4056 0.8870 0.2615 0.6827
2 1 B 0.6556 0.6995 0.0829 0.8356
3 1 C 0.6445 0.7110 0.3492 0.3826
4 1 D 0.1146 0.2655 . 0.0152
5 2 A 0.6072 0.2474 0.6444 0.9154
6 2 B 0.1720 . . 0.9241
7 2 C 0.6800 0.4166 0.2686 0.4992
8 2 D 0.7874 0.2595 0.6099 0.1418
9 3 A 0.4057 0.0988 0.2632 .
10 3 B 0.9805 . 0.9712 0.8937
11 3 C 0.4358 0.5023 0.5626 0.5799
12 3 D 0.8138 0.9588 0.8293 0.2448
13 4 A 0.4881 0.2731 0.4633 0.7839
14 4 B 0.1413 0.1166 0.6743 0.1032
15 4 C 0.1522 0.9351 0.2504 0.8003
16 4 D . 0.2465 0.7782 0.9412
The "get it done" approach is to manually enter each variable/value combination in a conditional which replaces with missing when true.
data have;
input group replicate $ height weight bp cholesterol;
datalines;
1 A 0.4056 0.8870 0.2615 0.6827
1 B 0.6556 0.6995 0.0829 0.8356
1 C 0.6445 0.7110 0.3492 0.3826
1 D 0.1146 0.2655 666 0.0152
2 A 0.6072 0.2474 0.6444 0.9154
2 B 0.1720 333 555 0.9241
2 C 0.6800 0.4166 0.2686 0.4992
2 D 0.7874 0.2595 0.6099 0.1418
3 A 0.4057 0.0988 0.2632 111
3 B 0.9805 444 0.9712 0.8937
3 C 0.4358 0.5023 0.5626 0.5799
3 D 0.8138 0.9588 0.8293 0.2448
4 A 0.4881 0.2731 0.4633 0.7839
4 B 0.1413 0.1166 0.6743 0.1032
4 C 0.1522 0.9351 0.2504 0.8003
4 D 222 0.2465 0.7782 0.9412
;
run;
data outliers;
input parameter $ 11. group replicate $ measurement;
datalines;
cholesterol 3 A 111
height 4 D 222
weight 2 B 333
weight 3 B 444
bp 2 B 555
bp 1 D 666
;
run;
EDIT: Updated outliers so that parameter avoids truncation and changed measurement to be numeric type so as to match the corresponding height, weight, bp, cholesterol. This shouldn't change the responses.
data want;
set have;
if group = 3 and replicate = 'A' and cholesterol = 111 then cholesterol = .;
if group = 4 and replicate = 'D' and height = 222 then height = .;
if group = 2 and replicate = 'B' and weight = 333 then weight = .;
if group = 3 and replicate = 'B' and weight = 444 then weight = .;
if group = 2 and replicate = 'B' and bp = 555 then bp = .;
if group = 1 and replicate = 'D' and bp = 666 then bp = .;
run;
This, however, doesn't utilize the outliers data set. How can the replacement process be made automatic?
I immediately think of the IN= operator, but that won't work. It's not the entire row which needs to be matched. Perhaps an SQL key matching approach would work? But to match the key, don't I need to use a where statement? I'd then effectively be writing everything out manually again. I could probably create macro variables which contain the various if or where statements, but that seems excessive.
I don't think generating statements is excessive in this case. The complexity arises here because your outlier dataset cannot be merged easily since the parameter values represent variable names in the have dataset. If it is possible to reorient the outliers dataset so you have a 1 to 1 merge, this logic would be simpler.
Let's assume you cannot. There are a few ways to use a variable in a dataset that corresponds to a variable in another.
You could use an array like array params{*} height -- cholesterol; and then use the vname function as you loop through the array to compare to the value in the parameter variable, but this gets complicated in your case because you have a one to many merge, so you would have to retain the replacements and only output the last record for each by group... so it gets complicated.
You could transpose the outliers data using proc transpose, but that will get lengthy because you will need a transpose for each parameter, and then you'd need to merge all the transposed datasets back to the have dataset. My main issue with this method is that code with a bunch of transposes like that gets unwieldy.
You create the macro variable logic you are thinking might be excessive. But compared to the other ways of getting the values of the parameter variable to match up with the variable names in the have dataset, I don't think something like this is excessive:
data _null_;
set outliers;
call symput("outlierstatement"||_n_,"if group = "||group||" and replicate = '"||replicate||"' and "||parameter||" = "||measurement||" then "|| parameter ||" = .;");
call symput("outliercount",_n_);
run;
%macro makewant();
data want;
set have;
%do i = 1 %to &outliercount;
&&outlierstatement&i;
%end;
run;
%mend;
Lorem:
Transposition is the key to a fully automatic programmatic approach. The transposition that will occur is of the filter data, not the original data. The transposed filter data will have fewer rows than the original. As John indicated, transposition of the want data can create a very tall table and has to be transposed back after applying the filters.
As to the the filter data, the presence of a filter row for a specific group, replicate and parameter should be enough to mark a cell for filtering. This is on the presumption that you have a system for automatic outlier detection and the filter values will always be in concordance with the original values.
So, what has to be done to automate the filter application process without code generating a wall of test and assign statements ?
Transpose filter data into same form as want data, call it Filter^
Merge Want and Filter^ by record key (which is the by group of Group and Replicate)
Array process the data elements, looking for filtering conditions.
For your consideration, try the following SAS code. There is an erroneous filter record added to the mix.
data have;
input group replicate $ height weight bp cholesterol;
datalines;
1 A 0.4056 0.8870 0.2615 0.6827
1 B 0.6556 0.6995 0.0829 0.8356
1 C 0.6445 0.7110 0.3492 0.3826
1 D 0.1146 0.2655 666 0.0152
2 A 0.6072 0.2474 0.6444 0.9154
2 B 0.1720 333 555 0.9241
2 C 0.6800 0.4166 0.2686 0.4992
2 D 0.7874 0.2595 0.6099 0.1418
3 A 0.4057 0.0988 0.2632 111
3 B 0.9805 444 0.9712 0.8937
3 C 0.4358 0.5023 0.5626 0.5799
3 D 0.8138 0.9588 0.8293 0.2448
4 A 0.4881 0.2731 0.4633 0.7839
4 B 0.1413 0.1166 0.6743 0.1032
4 C 0.1522 0.9351 0.2504 0.8003
4 D 222 0.2465 0.7782 0.9412
5 E 222 0.2465 0.7782 0.9412 /* test record for filter value misalignment test */
;
run;
data outliers;
length parameter $32; %* <--- widened parameter so it can transposed into column via id;
input parameter $ group replicate $ measurement ; %* <--- changed measurement to numeric variable;
datalines;
cholesterol 3 A 111
height 4 D 222
height 5 E 223 /* test record for filter value misalignment test */
weight 2 B 333
weight 3 B 444
bp 2 B 555
bp 1 D 666
;
run;
data want;
set have;
if group = 3 and replicate = 'A' and cholesterol = 111 then cholesterol = .;
if group = 4 and replicate = 'D' and height = 222 then height = .;
if group = 2 and replicate = 'B' and weight = 333 then weight = .;
if group = 3 and replicate = 'B' and weight = 444 then weight = .;
if group = 2 and replicate = 'B' and bp = 555 then bp = .;
if group = 1 and replicate = 'D' and bp = 666 then bp = .;
run;
/* Create a view with 1st row having all the filtered parameters
* This is necessary so that the first transposed filter row
* will have the parameters as columns in alphabetic order;
*/
proc sql noprint;
create view outliers_transpose_ready as
select distinct parameter from outliers
union
select * from outliers
order by group, replicate, parameter
;
/* Generate a alphabetic ordered list of parameters for use
* as a variable (aka column) list in the filter application step */
select distinct parameter
into :parameters separated by ' '
from outliers
order by parameter
;
quit;
%put NOTE: &=parameters;
/* tranpose the filter data
* The ID statement pivots row data into column names.
* The prefix=_filter_ ensure the new column names
* will not collide with the original data, and can be
* the shortcut listed with _filter_: in an array statement.
*/
proc transpose data=outliers_transpose_ready out=outliers_apply_ready prefix=_filter_;
by group replicate notsorted;
id parameter;
var measurement;
run;
/* Robust production code should contain a bin for
* data that does not conform to the filter application conditions
*/
data
want2(label="Outlier filtering applied" drop=_i_ _filter_:)
want2_warnings(label="Outlier filtering: misaligned values")
;
merge have outliers_apply_ready(keep=group replicate _filter_:);
by group replicate;
/* The arrays are for like named columns
* due to the alphabetic ordering enforced in data and codegen preparation
*/
array value_filter_check _filter_:;
array value ¶meters;
if group ne .;
do _i_ = 1 to dim(value);
if value(_i_) EQ value_filter_check(_i_) then
value(_i_) = .;
else
if not missing(value_filter_check(_i_)) AND
value(_i_) NE value_filter_check(_i_)
then do;
put 'WARNING: Filtering expected but values do not match. ' group= replicate= value(_i_)= value_filter_check(_i_)=;
output want2_warnings;
end;
end;
output want2;
run;
Confirm your want and automated want2 agree.
proc compare noprint data=want compare=want2 outnoequal out=diffs;
by group replicate;
run;
Enjoy your SAS
You could use a hash table. Load a hash table with the outlier dataset, with parameter-group-replicate defined as the key. Then read in the data, and as you read each record, check each of the variables to see if that combination of parameter-group-replicate can be found in the hash table. I think below works (I'm no hash expert):
data want;
if 0 then set outliers (keep=parameter group replicate);
if _N_ = 1 then
do;
declare hash h(dataset:'outliers') ;
h.defineKey('parameter', 'group', 'replicate') ;
h.defineDone() ;
end;
set have ;
array vars {*} height weight bp cholesterol ;
do i=1 to dim(vars);
parameter=vname(vars{i});
if h.check()=0 then call missing(vars{i});
end;
drop i parameter;
run;
I like #John's suggestion:
You could use an array like array params{*} height -- cholesterol; and
then use the vname function as you loop through the array to compare
to the value in the parameter variable, but this gets complicated in
your case because you have a one to many merge, so you would have to
retain the replacements and only output the last record for each by
group... so it gets complicated.
Generally in a one to many merge I would avoid recoding variables from the dataset that is unique, because variables are retained within BY groups. But in this case, it works out well.
proc sort data=outliers;
by group replicate;
run;
data want (keep=group replicate height weight bp cholesterol);
merge have (in=a)
outliers (keep=group replicate parameter in=b)
;
by group replicate;
array vars {*} height weight bp cholesterol ;
do i=1 to dim(vars);
if vname(vars{i})=parameter then call missing(vars{i});
end;
if last.replicate;
run;
Thank you #John for providing a proof of concept. My implementation is a little different and I think worth making a separate entry for posterity. I went with a macro variable approach because I feel it is the most intuitive, being a simple text replacement. However, since a macro variable can contain only 65534 characters, it is conceivable that there could be sufficient outliers to exceed this limit. In such a case, any of the other solutions would make fine alternatives. Note that it is important that the put statement use something like best32. Too short a width will truncate the value.
If you desire to have a dataset containing the if statements (perhaps for verification), simply remove the into : statement and place a create table statements as line at the beginning of the PROC SQL step.
data have;
input group replicate $ height weight bp cholesterol;
datalines;
1 A 0.4056 0.8870 0.2615 0.6827
1 B 0.6556 0.6995 0.0829 0.8356
1 C 0.6445 0.7110 0.3492 0.3826
1 D 0.1146 0.2655 666 0.0152
2 A 0.6072 0.2474 0.6444 0.9154
2 B 0.1720 333 555 0.9241
2 C 0.6800 0.4166 0.2686 0.4992
2 D 0.7874 0.2595 0.6099 0.1418
3 A 0.4057 0.0988 0.2632 111
3 B 0.9805 444 0.9712 0.8937
3 C 0.4358 0.5023 0.5626 0.5799
3 D 0.8138 0.9588 0.8293 0.2448
4 A 0.4881 0.2731 0.4633 0.7839
4 B 0.1413 0.1166 0.6743 0.1032
4 C 0.1522 0.9351 0.2504 0.8003
4 D 222 0.2465 0.7782 0.9412
;
run;
data outliers;
input parameter $ 11. group replicate $ measurement;
datalines;
cholesterol 3 A 111
height 4 D 222
weight 2 B 333
weight 3 B 444
bp 2 B 555
bp 1 D 666
;
run;
proc sql noprint;
select
cat('if group = '
, strip(put(group, best32.))
, " and replicate = '"
, strip(replicate)
, "' and "
, strip(parameter)
, ' = '
, strip(put(measurement, best32.))
, ' then '
, strip(parameter)
, ' = . ;')
into : listIfs separated by ' '
from outliers
;
quit;
%put %quote(&listIfs);
data want;
set have;
&listIfs;
run;
I am attempting to replicate Excel's Goal Seek in SAS.
I would like to find a constant number that when added to the initial data the overall average of the data equals the target. This gets a bit tricky when a transformation is involved.
So my three data points (var1) are 0.78, 0.8, 0.85. The target is 0.87.
I would like to find x where AVERAGE(1/(1+EXP(-(LN(var1/(1+var1)) + x))) = 0.87
This is the code I currently have, but it gets x = 0.4803 when it should be 0.4525 (found via Excel).
data aa;
input var1 target;
datalines;
0.78 0.87
0.8 0.87
0.85 0.87
;
run;
proc model data=aa outparms=parm;
target = 1/(1+EXP(-(log(var1/(1-var1)) + x)));
fit target;
run;
I think this isn't working bc it doesn't include an average of all 3 data points. I'm not sure how to do this. Ideally I'd just be able to change the second line in the proc model node to this:
target = Avg(1/(1+EXP(-(log(var1/(1-var1)) + x))));
But that doesn't work.
proc model is primarily designed for time-series, and doesn't do well with using summary functions vertically; however, it does great when doing it horizontally. One way to resolve it would be by transposing the problem:
proc transpose data=aa out=aa_trans;
by target;
var var1;
run;
proc model data=aa_trans;
endo x;
exo COL1-COL3 target;
target = mean(1/(1+EXP(-(log(COL1/(1-COL1)) + x)))
, 1/(1+EXP(-(log(COL2/(1-COL2)) + x)))
, 1/(1+EXP(-(log(COL3/(1-COL3)) + x))) );
solve / out=solution solveprint ;
run;
We get an answer of 0.4531398172. This can be checked by directly plugging in the value:
data _null_;
set aa_trans;
x = 0.4531398172;
check = mean(1/(1+EXP(-(log(COL1/(1-COL1)) + x)))
, 1/(1+EXP(-(log(COL2/(1-COL2)) + x)))
, 1/(1+EXP(-(log(COL3/(1-COL3)) + x))) );
put '*********** ' check;
run;
This method requires additional macro programming to generalize, and may be very computationally expensive if you have many observations to transpose. To generalize it for any given number of columns, you could use the following macro program:
%macro generateEquation;
%global eq;
%let eq = ;
proc sql noprint;
select count(*)
into :total
from aa
;
quit;
%do i = 1 %to &total.;
%let eq = %cmpres(&eq 1/(1+EXP(-(log(COL&i/(1-COL&i))+x))));
%end;
%let eq = mean(%sysfunc(tranwrd(&eq, %str( ), %str(,) ) ) );
%put &eq;
%mend;
%generateEquation;
proc model data=aa_trans;
endo x;
exo COL1-COL3 target;
target = &eq.;
solve / out=solution solveprint ;
run;
Instead, you might want to reframe this problem as an optimization problem with no objective function. proc optmodel, if available at your site, lets you do this matrix manipulation. The resulting code is more complex and manual, but will give you a more generalized and computationally feasible result.
You will need to add two new variables and separate the target to a new dataset.
data aa;
input targetid obs var1;
datalines;
1 1 0.78
1 2 0.8
1 3 0.85
;
run;
data bb;
input targetid target;
datalines;
1 0.87
;
run;
proc optmodel;
set id;
set obs;
set <num,num> id_obs;
/* Constants */
number target{id};
number var1{id_obs};
read data bb into id=[targetid]
target;
read data aa into id_obs=[targetid obs]
var1;
/* Parameter of interest */
var x{id};
/* Force the solver to seek the required goal */
con avg {i in id}: target[i] = sum{<j,n> in id_obs: j=i} (1/(1+EXP(-(log(var1[j, n]/(1-var1[j, n])) + x[i]))) )
/ sum{<j,n> in id_obs: j=i} 1;
/* Check if it's the equation that we want */
expand;
/* Solve using the non-linear programming solver with no objective */
solve with nlp noobjective;
/* Output */
create data solution from [targetid] = {i in id}
x[i];
quit;
optmodel returns a similar answer: 0.4531395426, which differs by 0.0000002746 decimal places. The answers are not identical due to differing methods and optimality tolerances; however, the solution checks out.
proc sql;
select Avg(1/(1+EXP(-(log(var1/(1-var1)) + 0.4531395426))))
from aa;
quit;
Assume I have a data-set D1 as follows:
ID ATR1 ATR2 ATR3
1 A R W
2 B T X
1 A S Y
2 C T E
3 D U I
1 T R W
2 C X X
I want to create a data-set D2 from this as follows
ID ATR1 ATR2 ATR3
1 A R W
2 C T X
3 D U I
In other words, Data-set D2 consists of unique IDs from D1. For each ID in D2, the values of ATR1-ATR3 are selected as the most frequent (of the respective variable) among the records in D1 with the same ID. For example ID = 1 in D2 has ATR1 = A (most frequent).
I have one solution which is very clumsy. I simply sort copies of the data set `D1' three times (by ID and ATR1 e.g) and remove duplicates. I later merge the three data-sets to get what I want. However, I think there might be an elegant way to do this. I have about 20 such variables in the original data-set.
Thanks
/*
read and restructure so we end up with:
id attr_id value
1 1 A
1 2 R
1 3 W
etc.
*/
data a(keep=id attr_id value);
length value $1;
array attrs_{*} $ 1 attr_1 - attr_3;
infile cards;
input id attr_1 - attr_3;
do attr_id=1 to dim(attrs_);
value = attrs_{attr_id};
output;
end;
cards;
1 A R W
2 B T X
1 A S Y
2 C T E
3 D U I
1 T R W
2 C X X
;
run;
/* calculate frequencies of values per id and attr_id */
proc freq data=a noprint;
tables id*attr_id*value / out=freqs(keep=id attr_id value count);
run;
/* sort so the most frequent value per id and attr_id ends up at the bottom of the group.
if there are ties then it's a matter of luck which value we get */
proc sort data = freqs;
by id attr_id count;
run;
/* read and recreate the original structure. */
data b(keep=id attr_1 - attr_3);
retain attr_1 - attr_3;
array attrs_{*} $ 1 attr_1 - attr_3;
set freqs;
by id attr_id;
if first.id then do;
do i=1 to dim(attrs_);
attrs_{i} = ' ';
end;
end;
if last.attr_id then do;
attrs_{attr_id} = value;
end;
if last.id then do;
output;
end;
run;
I have data on questions which students answered. The format is such
Student Q1 Q2 Q3 Q4
A 1 3 2 3
B 2 3 2 2
C 1 2 1 2
D 3 3 1 2
For this example, lets say 1 is the correct answer for question 1, 2 is the correct answer for question 2,3 and 4.
How would I generate a statistic table that would tell me how many questions a student answered correctly? In the example above, it would say something like
Student Answered Correct:
A 2/4
You can create an array of the correct answers, then just loop through the student answers to compare them.
I've created the final variable as character to display in the format you've shown. Obviously this means you won't have access to the underlying value, so you may want to keep the number of correct answers in the data for other analysis purposes.
data have;
input Student $ Q1 Q2 Q3 Q4;
datalines;
A 1 3 2 3
B 2 3 2 2
C 1 2 1 2
D 3 3 1 2
;
run;
data want;
set have;
array correct{4} (1 2 3 4); /* create array of correct answers */
array answer{4} q1-q4; /* create array of student answers */
_count=0; /* reset count to 0 */
do i = 1 to dim(correct);
if answer{i} = correct{i} then _count+1; /* compare student answer to correct answer and increment count by 1 if they match */
end;
length answered_correct $8; /* set length for variable */
answered_correct = catx('/',_count,dim(correct)); /* display result in required format */
drop q: correct: i _count; /* drop unwanted variables */
run;
First you have to create variable num_questions and set it to the number of questions. Then you need to write as many if-then-else statements as questions to create binary variables (flags) to check if each answer is correct (e.g. Correct_Q1). Use sum(of Correct:) to get the total of correct answers for each student. Correct: references all variable names starting with 'Correct'.
data want;
set have;
num_questions = 4;
if Q1 = 1 then Correct_Q1 = 1; else Correct_Q1 = 0;
if Q2 = 2 then Correct_Q2 = 1; else Correct_Q2 = 0;
if Q3 = 2 then Correct_Q3 = 1; else Correct_Q3 = 0;
if Q4 = 2 then Correct_Q4 = 1; else Correct_Q4 = 0;
format Answered_Correct $3. Answered_Correct_pct percent.;
Answered_Correct = compress(put(sum(of Correct:),$8.)||'/'||put(num_questions, 8.));
Answered_Correct_pct = sum(of Correct:) / num_questions;
label Student = 'Student' Answered_Correct = 'Answered correct' Answered_Correct_pct = 'Answered correct (%)';
keep Student Answered_Correct Answered_Correct_pct;
run;
proc print data=want noobs label;
run;
If you only have just four questions the fastest solution would probably be to just use conditional statements:if Q1 = 1 then answer + 1;
For a more general solution using a lookup/answer table:
Transpose the data, merge the answer table, summarize on student.
data broad_data;
infile datalines missover;
input Student $ Q1 Q2 Q3 Q4;
datalines;
A 1 3 2 3
B 2 3 2 2
C 1 2 1 2
D 3 3 1 2
;
data answers;
infile datalines missover;
input question $ correct_answer ;
datalines;
Q1 1
Q2 2
Q3 2
Q4 2
;
data long_data;
set broad_data;
length question $10 answer 8;
array long[*] Q1--Q4;
do i = 1 to dim(long);
question = vname(long[i]);
answer = long[i];
output;
end;
keep Student question answer;
run;
proc sort data = long_data; by question student; run;
data long_data_answers;
merge long_data
answers
;
by question;
run;
proc sort data = long_data_answers; by student; run;
data result;
do i = 1 by 1 until (last.student);
set long_data_answers;
by student;
count = sum(count, answer eq correct_answer);
end;
result = count/i;
keep student result;
format result fract8.;
run;
If you like sql/want to compress your code you can combine the last two datasteps + sorts into one statement.
proc sql;
create table result as
select student, sum(answer eq correct_answer)/count(*) as result format fract8.
from long_data a
inner join answers b
on a.question eq b.question
group by student
;
quit;
How to add new observation to already created dataset in SAS ? For example, if I have dataset 'dataX' with variable 'x' and 'y' and I want to add new observation which is multiplication by two of the
of the observation number n, how can I do it ?
dataX :
x y
1 1
1 21
2 3
I want to create :
dataX :
x y
1 1
1 21
2 3
10 210
where observation number four is multiplication by ten of observation number two.
data X;
input x y;
datalines;
1 1
1 21
2 3
;
run;
data X ;
set X end=eof;
if eof then do;
output;
x=10 ;y=210;
end;
output;
run;
Here is one way to do this:
data dataX;
input x y;
datalines;
1 1
1 21
2 3
run;
/* Create a new observation into temp data set */
data _addRec;
set dataX(firstobs=2); /* Get observation 2 */
x = x * 10; /* Multiply each by 10 */
y = y * 10;
output; /* Output new observation */
stop;
run;
/* Add new obs to original data set */
proc append base=dataX data=_addRec;
run;
/* Delete the temp data set (to be safe) */
proc delete data=_addRec;
run;
data a ;
do kk=1 to 5 ;
output ;
end ;
run;
data a2 ;
kk=999 ;
output ;
run;
data a; set a a2 ;run ;
proc print data=a ;run ;
Result:
The SAS System 1
OBS kk
1 1
2 2
3 3
4 4
5 5
6 999
You can use macro to obtain your desired result :
Write a macro which will read first DataSet and when _n_=2 it will multiply x and y with 10.
After that create another DataSet which will hold only your muliplied value let say x'=10x and y'=10y.
Pass both DataSet in another macro which will set the original datset and newly created dataset.
Logic is you have to create another dataset with value 10x and 10y and after that set wih previous dataset.
I hope this will help !