SAS help - median of values in an array - sas

Question

I am relatively new to SAS but have done a fair amount of programming over the years. I am at a loss on how to accomplish a task in SAS that I feel I would be able to do relatively easily in other platforms. I have an input table similar to this:
City
_1988
_1989
_1990
_1991
_1992
_1993
_1994
_1995
_1996
_1997
_1998
_1999
_2000
Columbus
438866
437148
16082
475843
224411
411569
658459
174208
592418
31664
312374
242830
342950
Fargo
11218
7402
35574
14765
64727
29492
104541
616
57864
73451
96251
78803
34743
Santa Fe
10608
31531
46163
28215
62608
52576
55674
43339
34896
77851
41304
31308
60306
Poughkeepsie
2184
15642
13505
9279
22796
6458
3279
4458
19672
17610
2672
11454
1072
Montpelier
1428
671
520
5453
5468
2117
2802
5847
3165
6204
1832
5357
5499
Waco
12527
695
44426
61651
83997
12811
50570
15022
86732
38541
45292
120719
17969
Nashville
359806
249811
422314
151319
466174
107335
315576
571273
195685
230626
194663
11060
545940
Billings
49694
37415
38602
79238
65260
18497
8976
81148
71326
108760
43740
48110
32106
Pensacola
4501
9682
19061
14731
4623
16106
13419
47607
9198
25003
39303
45146
24143
Trenton
40341
21210
4162
57773
16937
60495
21508
80819
27349
65088
65815
66308
38151
I would like to find the median of all the differences in values for each city.
The basic logic is I need to obtain the median of all the values in the array "difference" in the pseudo-code below.
for i = 1988 to 2000
for j = i+1 to 2000
difference(i,j) = value year_i - value year_j
end
end
I wish I could paste my sample code here, but I am basically at a point of writers block where what I have produced is so far off that it is of no use. I don't necessarily need someone to write the entire code for me but am hoping somebody can send me down the right path. I feel like this shouldn't be that hard, but I am at a loss . . .
Thanks in advance!

Answer 1

I'm not entirely following what you're trying to do here, but I think this will get you going.
First, I create the temp data with pairwise differences as you request. Then I use the Proc Summary to calculate the median across city and year.
Feel free to ask.
/* sample data */
data have;
input City :$12. _1988 - _2000;
infile datalines dlm = '|';
datalines;
Columbus |438866|437148|16082 |475843|224411|411569|658459|174208|592418|31664 |312374|242830|342950
Fargo |11218 |7402 |35574 |14765 |64727 |29492 |104541|616 |57864 |73451 |96251 |78803 |34743
Santa Fe |10608 |31531 |46163 |28215 |62608 |52576 |55674 |43339 |34896 |77851 |41304 |31308 |60306
Poughkeepsie|2184 |15642 |13505 |9279 |22796 |6458 |3279 |4458 |19672 |17610 |2672 |11454 |1072
Montpelier |1428 |671 |520 |5453 |5468 |2117 |2802 |5847 |3165 |6204 |1832 |5357 |5499
Waco |12527 |695 |44426 |61651 |83997 |12811 |50570 |15022 |86732 |38541 |45292 |120719|17969
Nashville |359806|249811|422314|151319|466174|107335|315576|571273|195685|230626|194663|11060 |545940
Billings |49694 |37415 |38602 |79238 |65260 |18497 |8976 |81148 |71326 |108760|43740 |48110 |32106
Pensacola |4501 |9682 |19061 |14731 |4623 |16106 |13419 |47607 |9198 |25003 |39303 |45146 |24143
Trenton |40341 |21210 |4162 |57773 |16937 |60495 |21508 |80819 |27349 |65088 |65815 |66308 |38151
;
/* pairwise differences in long format */
data temp;
set have;
array y1(i) _1988 - _2000;
array y2(j) _1988 - _2000;
do over y1;
do over y2;
year1 = input(compress(vname(y1), , 'kd'), 8.);
val1 = y1;
year2 = input(compress(vname(y2), , 'kd'), 8.);
val2 = y2;
diff = val1 - val2;
if i ne j then output;
end;
end;
drop _:;
run;
/* calculate median */
proc summary data = temp nway;
class city year1;
var diff;
output out = want(drop = _:) median =;
run;

Answer 2

You can use a 13-element array to hold the year values. And a 13x13 temporary array to hold the differences. Then median(of ARRAY(*)) to get the median.
* create the sample data;
data have;
input City $ _1988 _1989 _1990 _1991 _1992 _1993 _1994 _1995 _1996 _1997 _1998 _1999 _2000;
datalines;
Columbus 438866 437148 16082 475843 224411 411569 658459 174208 592418 31664 312374 242830 342950
Fargo 11218 7402 35574 14765 64727 29492 104541 616 57864 73451 96251 78803 34743
Santa-Fe 10608 31531 46163 28215 62608 52576 55674 43339 34896 77851 41304 31308 60306
Poughkeepsie 2184 15642 13505 9279 22796 6458 3279 4458 19672 17610 2672 11454 1072
Montpelier 1428 671 520 5453 5468 2117 2802 5847 3165 6204 1832 5357 5499
Waco 12527 695 44426 61651 83997 12811 50570 15022 86732 38541 45292 120719 17969
Nashville 359806 249811 422314 151319 466174 107335 315576 571273 195685 230626 194663 11060 545940
Billings 49694 37415 38602 79238 65260 18497 8976 81148 71326 108760 43740 48110 32106
Pensacola 4501 9682 19061 14731 4623 16106 13419 47607 9198 25003 39303 45146 24143
Trenton 40341 21210 4162 57773 16937 60495 21508 80819 27349 65088 65815 66308 38151
;
run;
data want;
set have;
* create an array to hold the year variables;
array years {1988:2000} _1988 - _2000;
* create a 2-dimensional array to hold the differences;
array differences {1988:2000,1988:2000} _temporary_;
do i = 1988 to 20000;
do j = i + 1 to 2000;
* calculate the differences as per pseudo-code in question;
differences(i,j) = years(i) - years(j);
end;
end;
* get median value;
median_diff = median(of differences(*));
run;