topic Re: Mean and std of partial array in SAS Programming

Mean and std of partial array

mcs — Fri, 24 Oct 2014 13:42:23 GMT

I'd like to calculate the mean and standard deviation of each row in an array. Something like

array pvbp(&scenarios,n);

... [populate array]

array mx(n); array sx(n);

do i = 1 to n;

mx(n) = mean(of pvbp(*,n));

sx(n) = std(of pvbp(*,n));

end;

The code above doesn't work - it gives a syntax error. Is there a simple fix?

Re: Mean and std of partial array

stat_sas — Fri, 24 Oct 2014 13:55:00 GMT

Not sure about the error you are getting. Something like this will produce mean and standard deviation for each row

data have;
input p1-p7;
cards;
1 2 3 2 2 3 2
2 3 4 2 3 2 3
3 3 4 3 5 2 4
4 3 4 5 2 2 2
;

run;
data want;
set have;
array p(*) p1-p7;
do i=1 to dim(p);
mean=mean(of p(*));
std=std(of p(*));
end;
run;

Re: Mean and std of partial array

Ksharp — Fri, 24 Oct 2014 14:00:07 GMT

Are you trying to use SAS/IML or not ?

one way in data step is to create a temporary array x{*} to hold these whole row and use mean(of x{*}) .

Xia Keshan

Re: Mean and std of partial array

mcs — Mon, 27 Oct 2014 13:40:33 GMT

Thanks for the reply.

Using mean(of p(*)) and ste(of p(*)) would work if i wanted to calculate statistics for a one-dimensional matrix for each row of a data set.

Instead, I want to calculate multiple statistics from a two-dimensional array, i.e., mean(of p(*,1)) to mean(of p(*,n)), etc., for each row of a data set.

Martin

Re: Mean and std of partial array

mcs — Mon, 27 Oct 2014 14:31:36 GMT

Thanks for the reply.

I'm using Base SAS, not SAS/IML

If I use temporary arrays xn(*) for each of x(*,n), I'll need n of them. I'd prefer to avoid n lines of repeated code. Is there a simple way to create an array of arrays to loop over? I imagine I could do it with macros, but I have less experience working with those.

Martin

Re: Mean and std of partial array

stat_sas — Mon, 27 Oct 2014 14:38:17 GMT

Please provide some sample data and desired output.

Re: Mean and std of partial array

mcs — Mon, 27 Oct 2014 15:06:13 GMT

data have;

input sex $ age;

cards;

M 45

F 50

U 60

;

data mc;

set have;

array s(5,2);

do i = 1 to 5;

do j = 1 to 2;

s(i,j) = age + i + j;

end;

array mx(2);

array sx(2);

do k = 1 to 2;

mx(k) = mean(of s(*,k));

sx(k) = std(of s(*,k);

end;

run;

Desired output:

_N_ = 1

mx1 = mean(of (47,48,48,49,49)) = 48.2, similary for mx2

sx1 = std(of (47,48,48,49,49)) = 0.84, similary for sx2

_N_ = 2

mx1 = mean(of (52,53,53,54,54)) = 53.2, similary for mx2

sx1 = std(of (52,53,53,54,54)) = 0.84, similarly for sx2

_N_ = 3

etc.

Instead I get the following syntax error.

21 mx(k) = mean(of (s(*,k));

ERROR 22-322: Syntax error, expecting one of the following: ), ], }.

Martin

Re: Mean and std of partial array

KachiM — Mon, 27 Oct 2014 17:21:00 GMT

Function to find mean and STD for partial array ie for a 2-dimensional array is not provided by SAS as far as I know. Instead compute the mean and std by adding some more steps as;

data _null_;

set have;

array s(5,2);

do i = 1 to 5;

do j = 1 to 2;

s(i,j) = age + i + j;

end;

array mx(2);

array sx(2);

do k = 1 to 2;

do i = 1 to dim1(s);

X + s[i, k];

XX + (s[i,k] * s[i,k]);

end;

mx(k) = X / dim1(s);

sx(k) = (XX - X * X / dim1(s)) / (dim1(s) - 1);

put mx = ;

put sx = ;

end;

stop;

run;

Correction: Take square root of SX for STD.

Re: Mean and std of partial array

stat_sas — Mon, 27 Oct 2014 17:44:30 GMT

data mc;

set have;

array s(5,2);

do i = 1 to 5;

do j = 1 to 2;

s(i,j) = age + i + j;

end;

array mx(2);

array sx(2);

array sv{*} s1-s10;

do k = 1 to 2;

if k=1 then do;

mx(k)=mean(of s1-s5);

sx(k)=std(of s1-s5);

end;

if k=2 then do;

mx(k)=mean(of s6-s10);

sx(k)=std(of s6-s10);

end;

run;

Re: Mean and std of partial array

mcs — Mon, 27 Oct 2014 17:47:31 GMT

Thanks for the reply.

This works, but the range of k is, say, 1 to 100 instead of 1 to 2, it involves a lot of repeated code.

Martin

Re: Mean and std of partial array

mcs — Mon, 27 Oct 2014 18:46:13 GMT

Thanks. That looks like it should work even when the bounds on s and k are large.

I think a slight improvement would be to calculate the running variance as shown at http://www.johndcook.com/standard_deviation.html

Initialize M1 = x1 and S1 = 0.

For subsequent x's, use the recurrence formulas

Mk = Mk-1+ (xk - Mk-1)/k

Sk = Sk-1 + (xk - Mk-1)*(xk - Mk).

For 2 ≤ k ≤ n, the kth estimate of the variance is s2 = Sk/(k - 1).

Martin