USE of 2D arrays in SAS for Distance computation??

[hcs301080]

Hi,
I wanted to write a SAS code for computing distance given the latitude and longitude of the places....( I got around 1000+ locations and I want to find the distance between each other).

I tried using the single dimension array but it seems to be generating erroneous values.
So resorted to 2D array....someone suggested that I use this logic
START will keep track of which city */
/* is the current "from". Calculate the distance to all */
/* the other "to" cities. */


START=1;
DO I=1 TO 1222;
DO J=1 TO (START-1),(START+1) TO 1222; /*?PLEASE HELP ME UNDERSTAND THIS STATEMENT ??*/
X1=lon (START);
Y1=lat(START);
X2=lon(J);
Y2=lat(J);

DIST(I,J)=4000*ARCOS(SIN(Y2)*SIN(Y1)+COS(Y2)*COS(Y1)*COS(X2-X1) );
/*the formula can be changed as per the requirement and the results observed*/;
END;

Can anyone understand this statement in the above code
DO J=1 TO (START-1),(START+1) TO 1222;

Its a bit weird to visulalise the 2D array in SAS in the above code

cheers

 

[teralearner]

1. This is not a SAS question. It is the logic behind your SAS code.

2. I assume you want to calculating distance between every of your 1222 locations. After calculate distance from location a to location b, you don't have to calculate distance from location b to location a. And distance from location a to location a is 0.
Let's look at a 3-location example:

From/To L1 L2 L3
L1 D11 D12 D13
L2 D21 D22 D23
L3 D31 D32 D33

You need to calculated D12 D13 D23.
Which is:
do i = 1 to 3;
do j = i + 1 to 3;
x1 = lon(i);
y1 = lat(i);
x2 = lon(j);
y2 = lat(j);
DIST(I,J)=4000*ARCOS(SIN(Y2)*SIN(Y1)+COS(Y2)*COS(Y1)*COS(X2-X1) ); /* whatever the function is*/
end;
end;

When i = 1, first j is 1 + 1 = 2 you get d12; second j is 3 you get d13. The first j loop ended;
When i = 2, first j is 2 + 1 = 3 you get d23. This is the end og j loop.
When i = 3 there is no j loop.

Hope this helps.