第一次寫按照SOC-PT分級(jí)的ADAE表格，在簡(jiǎn)書上記錄下自己的思路，避免遺忘。
首先看一下AE表格的格式，發(fā)現(xiàn)第一列中包含了不同的層級(jí)。我的思路是分別對(duì)不同層級(jí)進(jìn)行計(jì)數(shù)，最后匯總起來(lái)。

1654159395(1).png

讀取數(shù)據(jù)后，output一次，用來(lái)計(jì)算不分級(jí)的事件數(shù)和例數(shù)。

data adae1;
    set ads.adae;
    where SAFFL="是";
    if AEBODSYS="" then AEBODSYS="未編碼";
    if AEDECOD="" then AEDECOD="未編碼";
run;

data adae2;
    set adae1;
    output;
    if saffl="是" then aebodsys="all";
    output;
run;

之后通過(guò)排序去重，生成第一列的模板。并通過(guò)retain語(yǔ)句生成對(duì)模板排序用的序號(hào)。之后第一次合并dummy1_1 dummy2_1，將ord1添加到dummy2_1中，使得AEDECOD標(biāo)記上了所對(duì)應(yīng)的AEBODSYS。再將AEDECOD以及AEBODSYS添加到一列中，按照ORD1 ORD2排序后，就可以得到我們最終需要的模板了。

proc sort data=adae2 out=dummy1(keep=aebodsys) nodupkey sortseq=linguistic(locale=zh_CN collation=pinyin numeric_collation=on);
by aebodsys;
run;
proc sort data=adae1 out=dummy2(keep=aedecod aebodsys) nodupkey sortseq=linguistic(locale=zh_CN collation=pinyin numeric_collation=on);
by aebodsys aedecod;
run;
data dummy1_1;
    set dummy1;
    retain ord1 0;
    by aebodsys;
    ord1=ord1+1;
run;
data dummy2_1;
    set dummy2;
    retain ord2 0;
    by aebodsys aedecod;
    ord2=ord2+1;
    if first.aebodsys then ord2=2;
run;
proc sql;
create table dummy_1 as select a.*,b.ord1 from dummy2_1 a left join dummy1_1 b on a.aebodsys=b.aebodsys;quit;

data dummy;
    set dummy_1(keep=aedecod ord1 ord2 rename=(aedecod=name)) dummy1_1(keep=aebodsys ord1 rename=(aebodsys=name));
    if ord2=. then ord2=1;
proc sort;
by ord1 ord2;
run;

之后我使用的是freq步來(lái)進(jìn)行計(jì)數(shù)，得到對(duì)應(yīng)AEBODSYS AEDECOD的頻數(shù)。之后按照分組，整理成最終需要的樣子。

proc freq data=adae2(where=(aebodsys="ALL")) noprint;
table aebodsys*trtan/out=count1(drop=percent) nopercent;
run;
proc freq data=adae2(where=(aebodsys^="ALL")) noprint;
table aebodsys*trtan/out=count2(drop=percent) nopercent;
run;
proc freq data=adae2(where=(aebodsys="ALL")) noprint;
table aedecod*trtan/out=count3(drop=percent) nopercent;
run;

%macro filter(indata=,outdata=,var=);
data temp1 temp2;
    set &indata.;
    if trtan=1 then output temp1;
    if trtan=2 then output temp2;
run;
proc sort data=temp1(rename=(count=count1));
by &var.;
run;
proc sort data=temp2(rename=(count=count2));
by &var.;
run;
data &outdata.;
    merge temp1 temp2;
    by &var.;
    if count1=. then count1=0;
    if count2=. then count2=0;
    rename &var.=col1;
    drop trtan;
run;
%mend;
%filter(indata=count1,outdata=freq1,var=aebodsys);
%filter(indata=count2,outdata=freq2,var=aebodsys);
%filter(indata=count3,outdata=freq3,var=aedecod);

data freq;
    set freq1-freq3;
proc sort;
by col1;
run;

對(duì)于例數(shù)，這個(gè)方案的定義是每個(gè)人每個(gè)SOC-PT發(fā)生的不良事件只計(jì)算一次。
所以排序得到每個(gè)受試者每個(gè)SOC-PT唯一不良事件記錄后，按照上面的步驟得到例數(shù)。最后將例數(shù)除以各組受試者的數(shù)目，或者例數(shù)的百分比。最后將整理好的數(shù)據(jù)output成RTF文件即可。

proc sort data=adae2 out=adae3 nodupkey;
by subjid aebodsys aedecod ;
run;

proc freq data=adae3(where=(aebodsys="ALL")) noprint;
table aebodsys*trtan/out=count4(drop=percent) nopercent;
run;
proc freq data=adae3(where=(aebodsys^="ALL")) noprint;
table aebodsys*trtan/out=count5(drop=percent) nopercent;
run;
proc freq data=adae3(where=(aebodsys="ALL")) noprint;
table aedecod*trtan/out=count6(drop=percent) nopercent;
run;

%filter(indata=count4,outdata=freq4,var=aebodsys);
%filter(indata=count5,outdata=freq5,var=aebodsys);
%filter(indata=count6,outdata=freq6,var=aedecod);

data freq_;
    set freq4-freq6;
proc sort;
by col1;
run;

data freq_all;
    merge freq freq_(rename=(count1=count3 count2=count4));
    by col1;
    rename col1=col1_;
run;

data fin;
    length col1-col5 $200;
    set freq_all;
    col1=col1_;
    if count1=0 then col3="0";
        else if count1>0 then col3=put(count1,3.);
    if count2=0 then col2="0";
        else if count2>0 then col2=put(count2,3.)||" ("||strip(put(100*count2/&bign1.,5.1))||"）";
    if count3=0 then col5="0";
        else if count3>0 then col5=put(count3,3.);
    if count4=0 then col4="0";
        else if count4>0 then col4=put(count4,3.)||" ("||strip(put(100*count4/&bign2.,5.1))||"）";
    keep col1-col5;
run;

proc sql noprint;
create table final as select a.*,b.* from dummy a left join fin b on a.name=b.col1;
quit;

proc sort data=final;
by ord1 ord2;
run;

因?yàn)槭堑谝淮巫觯赡苓€存在一些問(wèn)題，打算之后發(fā)現(xiàn)了再回來(lái)修正。