SAS编程与数据挖掘商业案例 划分与转换

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关于 SAS编程与数据挖掘商业案例 划分与转换一节.
有那位仁兄帮忙解释一下,  下面这段程序根据什么原理来做划分转换的.  flt2_num_miss_&i._10 最后不是得到的值只有0或1了么.  怎么还能当作建模的变量了

%macro var_chi2();
%do i=1 %to 5;
ods results off;
ods listing close;
proc univariate data=lg.develop_final4;
weight weight;
var flt2_num_var_miss_&i.;
output out=quatile pctlpts=10 to 90 by 10 pctlpre=deci_;
run;
data logistic2_miss_&i.;
set lg.develop_final4(keep=redm_flag weight flt2_num_var_miss_&i. miss_&i.);
if _n_=1 then set quatile;
flt2_num_miss_&i._10=(flt2_num_var_miss_&i.<deci_10);
flt2_num_miss_&i._20=(flt2_num_var_miss_&i.<deci_20);
flt2_num_miss_&i._30=(flt2_num_var_miss_&i.<deci_30);
flt2_num_miss_&i._40=(flt2_num_var_miss_&i.<deci_40);
flt2_num_miss_&i._50=(flt2_num_var_miss_&i.<deci_50);
flt2_num_miss_&i._60=(flt2_num_var_miss_&i.<deci_60);
flt2_num_miss_&i._70=(flt2_num_var_miss_&i.<deci_70);
flt2_num_miss_&i._80=(flt2_num_var_miss_&i.<deci_80);
flt2_num_miss_&i._90=(flt2_num_var_miss_&i.<deci_90);
flt2_num_miss_&i._sin=sin(flt2_num_var_miss_&i.);
flt2_num_miss_&i._cos=cos(flt2_num_var_miss_&i.);
drop deci:;
run;
ods
output ParameterEstimates=var_chisq_&i.(keep=variable where=(variable<>"Intercept"));
proc logistic data=logistic2_miss_&i. desc;
weight weight;
model redm_flag=flt2_num_var_miss_&i. miss_&i.
flt2_num_miss_&i._10
flt2_num_miss_&i._20
flt2_num_miss_&i._30
flt2_num_miss_&i._40
flt2_num_miss_&i._50
flt2_num_miss_&i._60
flt2_num_miss_&i._70
flt2_num_miss_&i._80
flt2_num_miss_&i._90
flt2_num_miss_&i._cos
flt2_num_miss_&i._sin
/selection=stepwise maxstep=1;
run;
ods output close;
ods results on;
ods listing;
%end;
%mend var_chi2;
%var_chi2;

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关键词：SAS编程 数据挖掘 商业案例 Univariate logistic set close