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地板
楼主 |
发表于 2010-8-4 23:48:47
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只看该作者
Re: 请教关系矩阵生成问题？
how about my code <!-- s:) --><img src="{SMILIES_PATH}/icon_smile.gif" alt=":)" title="Smile" /><!-- s:) -->
100K records? no problem at all, even for millions of records....
the recursive search problem posted before can be solved use similar idea, actually.
[code:2j892gil]
data original;
input id1 $ id2 $;
datalines;
a b
a e
b c
b e
c e
d e
;
run;
proc datasets library=work nolist;
modify original;
index create id1 id2;
quit;
proc sql;
create table all_cases as
select a.*, monotonic() as seq
from (
select distinct id1 as id
from original
union
select distinct id2 as id
from original
) as a
order by a.id
;
quit;
proc sql noprint;
select id into :idnames separated by ' '
from all_cases
;
quit;
data new;
if _n_=1 then do;
declare hash _h(dataset:'all_cases');
_h.defineKey('id');
_h.defineData('seq');
_h.defineDone();
end;
set all_cases;
array _a{*} &idnames;
id1=id;
set original key=id1;
_mx_=%sysrc(_sok);
do while (_iorc_=%sysrc(_sok));
rc=_h.find(key:id2); if rc=0 then _a[seq]=1;
id1=id;
set original key=id1;
end;
_ERROR_=0;
id2=id;
set original key=id2;
do while (_iorc_=%sysrc(_sok));
rc=_h.find(key:id1); if rc=0 then _a[seq]=1;
id2=id;
set original key=id2;
end;
_ERROR_=0;
do j=1 to dim(_a); _a[j]=max(0, _a[j]); end;
keep id &idnames;
run;
[/code:2j892gil]
On the other hand, this problem can be solved in a more SASsy way like this <!-- s:lol: --><img src="{SMILIES_PATH}/icon_lol.gif" alt=":lol:" title="Laughing" /><!-- s:lol: --> :
[code:2j892gil]
data original;
input id1 $ id2 $;
datalines;
a b
a e
b c
b e
c e
d e
;
run;
proc sql;
create table newx as
select a.id1, a.id2, (sum(a.id1=c.id1 & a.id2=c.id2)>0) as count
from
(select a.id as id1, b.id as id2
from all_cases as a, all_cases as b) as a
left join original as c
on a.id1=c.id1 or a.id2=c.id1
group by a.id1, a.id2
;
quit;
proc transpose data=newx out=_freq_t name=id2;
by id1;
var count;
id id2;
run;
data _freq_t;
set _freq_t;
array _n{*} _numeric_;
do i=1 to dim(_n);
_n[i]=(_n[i]>0);
end;
drop i;
run;
proc transpose data=_freq_t(drop=id2) out=_freq_t2 name=id1;
id id1;
run;
proc sql noprint;
select id1, count(distinct id1) into :covars separated by ' ', :count
from _freq_t;
quit;
data new2;
set _freq_t;
array _x{*} &covars;
array _x2{&count} _temporary_;
do j=1 to &count; _x2[j]=_x[j]; end;
set _freq_t2;
do j=1 to &count; _x[j]=(_x[j]+_x2[j]>0); end;
drop j id2;
run;
[/code:2j892gil]
you can test these two approaches using the following dummy data. On my server the first approach took less than 1second overall.
[code:2j892gil]
data original;
do i=1 to 5e5;
x=65 + floor(ranuni(0)*26);
id1=byte(x);
x=65 + floor(ranuni(0)*18);
id2=byte(x);
output;
drop i x;
end;
run;
[/code:2j892gil] |
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