data ze_0; input COL1 @@; cards; 0.16013 0.32026 0.80064 0.64051 0.64051 run; data zs_0; input COL1 @@; cards; -0.16013 -0.48038 0.32026 0.48038 0.80064 run; data ze_1; input COL1 @@; cards; -0.16013 -0.16013 -0.16013 0.00000 0.32026 run; data zs_1; input COL1 @@; cards; -0.64051 -1.28103 -0.80064 -0.80064 -0.32026 run; data ze_2; input COL1 @@; cards; 1.92154 2.40192 2.40192 0.96077 1.92154 run; data zs_2; input COL1 @@; cards; 1.60128 1.44115 0.48038 1.12090 2.08167 run; data ze_3; input COL1 @@; cards; 0.32026 1.44115 1.76141 1.76141 2.08167 run; data zs_3; input COL1 @@; cards; 0.16013 -1.44115 -0.16013 -0.48038 0.32026 run; data ze_4; input COL1 @@; cards; 0.00000 0.00000 0.96077 0.64051 0.64051 run; data zs_4; input COL1 @@; cards; -0.80064 -1.60128 -1.12090 -1.60128 -0.64051 run; data ze_5; input COL1 @@; cards; -1.28103 -0.80064 -0.48038 -0.64051 -0.96077 run; data zs_5; input COL1 @@; cards; -1.60128 -0.48038 0.48038 0.48038 0.16013 run; When you run this data set and attachment macros (new attachments) There is no error, but the desired result is different.   % let l = 5; % let m = 1; % test (nb = 5, l = & l, m = & m);   This is result in pe = 0.6 ps = 0.8.   Since pe < 0.7,   Next is    % let l = 4; % let m = 1; % test (nb = 5, l = & l, m = & m); This should be done.   With such a routine, when l = 3 and m = 1, pe and ps are both greater than 0.7 for the first time. However, if the macro aaa is executed, the value of l = 4 and m = 2 appears, so i don't know why the value is wrong. ... View more

ex data) proc iml; ze_0={1,2,3}; zs_0{1,3,5}; ze_1={2,3,4}; zs_1={2,4,2}; ze_2={1,3,2}; zs_2={2,5,4}; quit;   pe=#{max(ze_i[&l,1],zs_i[&m,1])>=ze_0[&l,1]}/nb ps=#{max(zs_i[&l,1],zs_i[&m,1])>=zs_0[&m,1]}/nb i=1,2 nb=2 I have already complete macro program code that computes pe and ps. (%test) but, i don't know how to code this part.   Initial value l=3 , m=1 if pe < 0.05 then I = l-1, If ps < 0.05 then m = m + 1, and pe and ps must be repeated until both are greater than 0.05. ... View more