1
proc sort data=have;
    by Name  Day Time;
run;

data want1;
retain name  day  time sum;
set have;
by name day time;
if first.time then sum=amount;
               else sum+amount;
if last.time;
keep name day time sum;
run;

2
proc sort data=have;
     by name day time;
run;

data want2;
  set have;
  by name day time;
  if last.time then output;
run;

3. 这个是同一个id之间比较？ 还是A和B也可以比？

data tmp;
merge have(keep= name amount rename=(name=name_last amount=amount_last )) have(firstobs=2);
if name=name_last and abs(amount) le abs(amount_last);
run;

data tmp2;
  set have;
  amount_last=lag(amount);
run;

3和4 用lag函数就行 用我上一步那样merge也行你自己再接着写吧.......

1. proc sql;
    create table xx as
    select *,
             sum(amount) as sum_amount
    from a.xx
    group by name;
    quit;