标题: a=a*lag(a) How could I realize this function? [打印本页] 作者: shiyiming 时间: 2006-3-14 16:39 标题: a=a*lag(a) How could I realize this function? I have tried:
a=a*lag(a);
But it seems not working.
I want make the variables multiply continuous.
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I want it to be
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...作者: shiyiming 时间: 2006-3-14 17:28 标题: Reply: a=a*lag(a) How could I realize this function? when _n_=1, lag(a) seems to be missing value.
then could add one extra condition to exclude this case occurance like when _n_ ge 1 then a*lag(a).
Try it. hope it could work! <!-- s:lol: --><img src="{SMILIES_PATH}/icon_lol.gif" alt=":lol:" title="Laughing" /><!-- s:lol: -->作者: shiyiming 时间: 2006-3-14 17:57 标题: 试过了，不行。 试过了，不行。
谁能帮忙写个代码把这个算一下？
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。。。
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。。。作者: shiyiming 时间: 2006-3-15 09:24 标题: 试试 [code:8dded]data ahuige;
input a;
retain s 1;
s=s*a;
cards;
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;
run;[/code:8dded]作者: shiyiming 时间: 2006-3-15 21:16 标题: Now test it. %let n=2;
data mutiply (drop=s);
retain s;
set test;
s = sum(s,log(x), -log(lag&n(x))) ;
moving_product=exp(s);
run;