I had posted this earlier, and got help on it. My interest was piqued, and I ventured into this a little further to see what I could do with it. I am fascinated with simulations, but am just an average SAS programmer. I wonder if somebody might help here.
data out;
call streaminit(7); *seed better random number engine;
do pointvar = 1 by 1 until (outs=27); *iterate starting at
1 and stop when 27 outs ;
randvar = rand('Uniform'); *better random number engine;
if pointvar > 9 then pointvar=1; *reset to 1 if over 9;
set in point=pointvar; *pull the row we need;
if randvar < cutoff then do;
outs+1;
outs_inning+1;
end;
output;
if outs_inning=3 then outs_inning=0;
end;
stop;
run;
the data set has just one observation for the 9 hitters.
.73
.75
.72
.78
.81
.69
.74
.72
.75
With the help of Joe and others, the above did what I wanted, which was to simulate primarily the counting of outs involved in ONE baseball game.
I have been playing around with this (to no avail) and trying to get it to repeat a game, so to speak, where it would start at the top of the lineup after 27 outs. So for what I have right now, assume the 27th out is achieved with the 5th batter. I would like to put this whole code inside of a loop where it starts the process again at the beginning of the data set (1st observation, i.e, first batter).
So, assume I want to complete 3 iterations here. 3 games of 27 outs. Is there a way to do this? I tried doing the following.
%macro replicate(new,out,n)/des=’&new1 is &out repeated &n times
Data &new;
%do i=1 to &n;
Set &out;
Output;
%end;
%mend;
%replicate(new,out,3);
Proc print;
I was hoping with a do statement I could do this, but The problem with this is that it is reading each observation 3 times. So in the do i=1 to 3, followed by set out (three instances it takes the first observation from data set ‘out’, then 3 times it takes the second observation from data set out, etc.
i.e.
Outs randvar cutoff outs_inning
0 0.84 0.73 0
0 0.84 0.73 0
0 0.84 0.73 0
1 0.61 0.75 0
1 0.61 0.75 0
1 0.61 0.75 0
Can anybody help? I appreciate that this is a little outside the realm of what is typically discussed here, but a few of my students are also interested in simulations, and a baseball example has certainly interested them. It has become a fun problem. thanks for getting me this far.
You don't need a macro. You should be able to add an outer DO loop which is do game=1 to 3;
Below I changed the variable POINTVAR to be BATTER, and added a PUT statement to write messages to the log.
data in;
input cutoff ##;
cards;
.73 .75 .72 .78 .81 .69 .74 .72 .75
;
data play;
call streaminit(7);
do game=1 to 3;
outs=0;
outs_inning=0;
do batter = 1 by 1 until (outs=27);
randvar = rand('Uniform');
if batter > 9 then batter=1;
set in point=batter;
if randvar < cutoff then do;
outs+1;
outs_inning+1;
end;
output;
put (game batter cutoff randvar outs_inning outs)(=);
if outs_inning=3 then outs_inning=0;
end;
end;
stop;
run;
Related
Let's say I have stores all around the world and I want to know what was my top losses sales across the world per store. What is the code for that?!
here is my try:
proc sort data= store out=sorted_store;
by store descending amount;
run;
and
data calc1;
do _n_=1 by 1 until(last.store);
set sorted_store;
by store;
if _n_ <= 5 then "Sum_5Largest_Losses"n=sum(amount);
end;
run;
but this just prints out the 5:th amount and not 1.. TO .. 5! and I really don't know how to select the top 5 of EACH store . I think a kind of group by would be a perfect fit. But first things, first. How do I selct i= 1...5 ? And not just = 5?
There is also way of doing it with proc sql:
data have;
input store$ amount;
datalines;
A 100
A 200
A 300
A 400
A 500
A 600
A 700
B 1000
B 1100
C 1200
C 1300
C 1400
D 600
D 700
E 1000
E 1100
F 1200
;
run;
proc sql outobs=4; /* limit to first 10 results */
select store, sum(amount) as TOTAL_AMT
from have
group by 1
order by 2 desc; /* order them to the TOP selection*/
quit;
The data step sum(,) function adds up its arguments. If you only give it one argument then there is nothing to actually sum so it just returns the input value.
data calc1;
do _n_=1 by 1 until(last.store);
set sorted_store;
by store;
if _n_ <= 5 then Sum_5Largest_Losses=sum(Sum_5Largest_Losses,amount);
end;
run;
I would highly recommend learning the basic methods before getting into DOW loops.
Add a counter so you can find the first 5 of each store
As the data step loops the sum accumulates
Output sum for counter=5
proc sort data= store out=sorted_store;
by store descending amount;
run;
data calc1;
set sorted_store;
by store;
*if first store then set counter to 1 and total sum to 0;
if first.store then do ;
counter=1;
total_sum=0;
end;
*otherwise increment the counter;
else counter+1;
*accumulate the sum if counter <= 5;
if counter <=5 then total_sum = sum(total_sum, amount);
*output only when on last/5th record for each store;
if counter=5 then output;
run;
In a clinical trial, Systolic and diastolic blood pressure are measured pre-dose (0 hr) and at 1,2,4,8 hour post- dose.
Twelve subjects were studied. The SAS dataset has the following structure
Variable-Vol Length - 8 Label- Subject Number
Variable- Ntime Length- 8 Label Nominal time post-dose (hours)
Variable- Sups Length- 8 Label- Supine Systolic BP (mmHg)
What SAS code could I use to calculate the change from baseline (Oh) at each time point, and then calculate the mean, minimum, maximum change from baseline for the 12 subjects? Edit: This is what I've tried so far
data postbase;
do until (last.vol);
*** Only keep pre-dose values;
set save.vitals (where=(not(ntime <= 0 )));
by Vol Ntime;
if Ntime <= 0 then bl = Sups;
else do;
chgbl = Sups - bl;
output;
end;
end;
run;
data postbase;
set save.vitals;
by subject time volume;
retain baseline;
if time=0 then baseline=volume;
else change = volume - baseline;
run;
I think your code is too complex by far and I couldn't parse your variable names so just made them up.
I set baseline volume whenever time = 0 and then do the change every other time.
RETAIN causes the value to stay until it's reset. If you have times that may not be 0 or missing baseline then you may need to modify the query.
I'd like to set all values in an array to 1 if some sort of condition is met, and perform a calculation if the condition isn't met. I'm using a do loop at the moment which is very slow.
I was wondering if there was a faster way.
data test2;
set test1;
array blah_{*} blah1-blah100;
array a_{*} a1-a100;
array b_{*} b1-b100;
do i=1 to 100;
blah_{i}=a_{i}/b_{i};
if b1=0 then blah_{i}=1;
end;
run;
I feel like the if statement is inefficient as I am setting the value 1 cell at a time. Is there a better way?
There are already several good answers, but for the sake of completeness, here is an extremely silly and dangerous way of changing all the array values at once without using a loop:
data test2;
set test1;
array blah_{*} blah1-blah100 (100*1);
array a_{*} a1-a100;
array b_{*} b1-b100;
/*Make a character copy of what an array of 100 1s looks like*/
length temp $800; *Allow 8 bytes per numeric variable;
retain temp;
if _n_ = 1 then temp = peekclong(addrlong(blah1), 800);
do i=1 to 100;
blah_{i}=a_{i}/b_{i};
end;
/*Overwrite the array using the stored value from earlier*/
if b1=0 then call pokelong(temp,addrlong(blah1),800);
run;
You have 100*NOBS assignments to do. Don't see how using a DO loop over an ARRAY is any more inefficient than any other way.
But there is no need to do the calculation when you know it will not be needed.
do i=1 to 100;
if b1=0 then blah_{i}=1;
else blah_{i}=a_{i}/b_{i};
end;
This example uses a data set to "set" all values of an array without DOingOVER the array. Note that using SET in this way changes INIT-TO-MISSING for array BLAH to don't. I cannot comment on performance you will need to do your own testing.
data one;
array blah[10];
retain blah 1;
run;
proc print;
run;
data test1;
do b1=0,1,0;
output;
end;
run;
data test2;
set test1;
array blah[10];
array a[10];
array b[10];
if b1 eq 0 then set one nobs=nobs point=nobs;
else do i = 1 to dim(blah);
blah[i] = i;
end;
run;
proc print;
run;
This is not a response to the original question, but as a response to the discussion on the efficiency between using loops vs set to set the values for multiple variables
Here is a simple experiment that I ran:
%let size = 100; /* Controls size of dataset */
%let iter = 1; /* Just to emulate different number of records in the base dataset */
data static;
array aa{&size} aa1 - aa&size (&size * 1);
run;
data inp;
do ii = 1 to &iter;
x = ranuni(234234);
output;
end;
run;
data eg1;
set inp;
array aa{&size} aa1 - aa&size;
set static nobs=nobs point=nobs;
run;
data eg2;
set inp;
array aa{&size} aa1 - aa&size;
do ii = 1 to &size;
aa(ii) = 1;
end;
run;
What I see when I run this with various values of &iter and &size is as follows:
As &size increases for a &iter value of 1, assignment method is faster than the SET.
However for a given &size, as iter increases (i.e. the number of times the set statement / loop is called), the speed of the SET approach increases while the assignment method starts to decrease at a certain point at which they cross. I think this is because the transfer from physical disk to buffer happens just once (since static is a relatively small dataset) whereas the assignment loop cost is fixed.
For this use case, where the fixed dataset used to set values will be smaller, I admit that SET will be faster especially when the logic needs to execute on multiple records on the input and the number of variables that needs to be assigned are relatively few. This however will not be the case if the dataset cannot be cached in memory between two records in which case the additional overhead of having to read it into the buffer can slow it down.
I think this test isolates the statements of interest.
SUMMARY:
SET+create init array 0.40 sec. + 0.03 sec,
DO OVER array 11.64 sec.
NOTE: Additional host information:
X64_SRV12 WIN 6.2.9200 Server
NOTE: SAS initialization used:
real time 4.70 seconds
cpu time 0.07 seconds
1 options fullstimer=1;
2 %let d=1e4; /*array size*/
3 %let s=1e5; /*reps (obs)*/
4 data one;
5 array blah[%sysevalf(&d,integer)];
6 retain blah 1;
7 run;
NOTE: The data set WORK.ONE has 1 observations and 10000 variables.
NOTE: DATA statement used (Total process time):
real time 0.03 seconds
user cpu time 0.03 seconds
system cpu time 0.00 seconds
memory 7788.90k
OS Memory 15232.00k
Timestamp 08/17/2019 06:57:48 AM
Step Count 1 Switch Count 0
8
9 sasfile one open;
NOTE: The file WORK.ONE.DATA has been opened by the SASFILE statement.
10 data _null_;
11 array blah[%sysevalf(&d,integer)];
12 do _n_ = 1 to &s;
13 set one nobs=nobs point=nobs;
14 end;
15 stop;
16 run;
NOTE: DATA statement used (Total process time):
real time 0.40 seconds
user cpu time 0.40 seconds
system cpu time 0.00 seconds
memory 7615.31k
OS Memory 16980.00k
Timestamp 08/17/2019 06:57:48 AM
Step Count 2 Switch Count 0
2 The SAS System 06:57 Saturday, August 17, 2019
17 sasfile one close;
NOTE: The file WORK.ONE.DATA has been closed by the SASFILE statement.
18
19 data _null_;
20 array blah[%sysevalf(&d,integer)];
21 do _n_ = 1 to &s;
22 do i=1 to dim(blah); blah[i]=1; end;
23 end;
24 stop;
25 run;
NOTE: DATA statement used (Total process time):
real time 11.64 seconds
user cpu time 11.64 seconds
system cpu time 0.00 seconds
memory 3540.65k
OS Memory 11084.00k
Timestamp 08/17/2019 06:58:00 AM
Step Count 3 Switch Count 0
NOTE: SAS Institute Inc., SAS Campus Drive, Cary, NC USA 27513-2414
NOTE: The SAS System used:
real time 16.78 seconds
user cpu time 12.10 seconds
system cpu time 0.04 seconds
memory 15840.62k
OS Memory 16980.00k
Timestamp 08/17/2019 06:58:00 AM
Step Count 3 Switch Count 16
Some more interesting tests results based on data null 's original test. I added the following test also:
%macro loop;
data _null_;
array blah[%sysevalf(&d,integer)] blah1 - blah&d;
do _n_ = 1 to &s;
%do i = 1 %to &d;
blah&i = 1;
%end;
end;
stop;
run;
%mend;
%loop;
d s SET Method (real/cpu) %Loop (real/cpu) array based(real/cpu)
100 1e5 0.03/0.01 0.00/0.00 0.07/0.07
100 1e8 11.16/9.51 4.78/4.78 1:22.38/1:21.81
500 1e5 0.03/0.04 0.02/0.01 Did not measure
500 1e8 16.53/15.18 32.17/31.62 Did not measure
1000 1e5 0.03/0.03 0.04/0.03 0.74/0.70
1000 1e8 20.24/18.65 42.58/42.46 Did not measure
So with array based assignments, it is not the assignment that is the big culprit itself. Since arrays use a memory map to map the original memory locations, it appears that the memory location lookup for a given subscript is what really impacts performance. A direct assignment avoids this and significantly improves performance.
So if your array size is in the lower 100s, then direct assignment may not be a bad way to go. SET becomes effective when the array sizes go beyond a few hundreds.
I am new to SAS and I was wondering how to cure the variable not found problem in creating a binomial distribution?
DATA additional (KEEP=X);
DO REPEAT = 1 TO 1000;
CALL STREAMINIT(1234);
DO I=1 TO 1000;
X=RAND("BINOMIAL",0.6,10); /*NUMBER OF WINS IN TEN TOSSES*/
END;
IF X GE 5 THEN WINNER + 1;
ELSE LOSER + 1;
OUTPUT;
END;
RUN;
PROC PRINT DATA=additional;
VAR WINNER LOSER;
RUN;
I am creating a binomial random variable which if x is great than 5 then counts one for the winner, if less than 5 then counts one for the loser, the question is asking to found how many time are winners and how many times are losers. I kept on getting variable not found error. Am i doing something wrong with generating the binomial distribution.
/further editing/ this is the problem I am given.
You are given $10. Let the variable money = 10.
You play a game 10 times. The probability that you win a game is 0.4,
and the probability that you lose a game is 0.6.
If you win a game, you win $1. If you lose a game, you lose $1. So if
you win the first game, money becomes 11. But if you lose the first
game, money becomes 9.
After you have played the game 10 times, money is the amount that you
go home with. If you end up with at least $10, call yourself a winner.
Otherwise, call yourself a loser. Define the variable result as winner
or loser.
(a) Write a data step to generate random numbers and simulate your
result 1000 times. So that I can easily check your outputs, use
1234 as your seed for the random number generator. (You do not
need to show me the 1000 results.)
(b) Write a proc step to show how many times you are a winner, and
how many times you are a loser.
Not fully understand that you want to do with simulation. From your codes, you just keep 1000 records, which are all kept at last loop because of your first loop end position; call streaminit should be first line; you only keep X, you could not get winner and loser variable.
I guess maybe you could try this.
DATA additional;
CALL STREAMINIT(1234);
DO REPEAT= 1 TO 1000; *numbers of sample;
DO I=1 TO 100; *size of sample;
X=RAND("BINOMIAL",0.6,10); /*NUMBER OF WINS IN TEN TOSSES*/
IF X GE 5 THEN results='WINNER';
ELSE results='LOSER';
OUTPUT;
END;
END;
RUN;
proc freq data=additional;
by repeat;
table results;
run;
Edit: It seems that you want to know final results, you could get it from above code by changing results as numeric variable. Here is modified codes, if win is +1, lose is -1.
DATA additional;
CALL STREAMINIT(1234);
DO REPEAT= 1 TO 100; *numbers of sample;
DO I=1 TO 10; *size of sample;
X=RAND("BINOMIAL",0.6,10); /*NUMBER OF WINS IN TEN TOSSES*/
IF X GE 5 THEN results+1;
ELSE results=results-1;
OUTPUT;
END;
results=0;
END;
RUN;
proc freq data=additional;
by repeat;
table results;
run;
I have a SAS issue that I know is probably fairly straightforward for SAS users who are familiar with array programming, but I am new to this aspect.
My dataset looks like this:
Data have;
Input group $ size price;
Datalines;
A 24 5
A 28 10
A 30 14
A 32 16
B 26 10
B 28 12
B 32 13
C 10 100
C 11 130
C 12 140
;
Run;
What I want to do is determine the rate at which price changes for the first two items in the family and apply that rate to every other member in the family.
So, I’ll end up with something that looks like this (for A only…):
Data want;
Input group $ size price newprice;
Datalines;
A 24 5 5
A 28 10 10
A 30 14 12.5
A 32 16 15
;
Run;
The technique you'll need to learn is either retain or diff/lag. Both methods would work here.
The following illustrates one way to solve this, but would need additional work by you to deal with things like size not changing (meaning a 0 denominator) and other potential exceptions.
Basically, we use retain to cause a value to persist across records, and use that in the calculations.
data want;
set have;
by group;
retain lastprice rateprice lastsize;
if first.group then do;
counter=0;
call missing(of lastprice rateprice lastsize); *clear these out;
end;
counter+1; *Increment the counter;
if counter=2 then do;
rateprice=(price-lastprice)/(size-lastsize); *Calculate the rate over 2;
end;
if counter le 2 then newprice=price; *For the first two just move price into newprice;
else if counter>2 then newprice=lastprice+(size-lastsize)*rateprice; *Else set it to the change;
output;
lastprice=newprice; *save the price and size in the retained vars;
lastsize=size;
run;
Here a different approach that is obviously longer than Joe's, but could be generalized to other similar situations where the calculation is different or depends on more values.
Add a sequence number to your data set:
data have2;
set have;
by group;
if first.group the seq = 0;
seq + 1;
run;
Use proc reg to calculate the intercept and slope for the first two rows of each group, outputting the estimates with outest:
proc reg data=have2 outest=est;
by group;
model price = size;
where seq le 2;
run;
Join the original table to the parameter estimates and calculate the predicted values:
proc sql;
create table want as
select
h.*,
e.intercept + h.size * e.size as newprice
from
have h
left join est e
on h.group = e.group
order by
group,
size
;
quit;