1

我一直在寻找相关的问题，但到目前为止我还没有运气。我正在寻找转换一长串自变量以进行回归分析。一个虚拟数据集如下所示：

DATA TEST (DROP = i);
    DO i = 1 to 4000;
        VAR = i + 100000;
        output;
    end;
run;

PROC TRANSPOSE
    DATA = TEST
    OUT = TEST_T
        (DROP = _NAME_)
    PREFIX = X_;
    ID VAR;
    VAR VAR;
RUN;

DATA TEST_ARRAY;
    SET TEST_T;

    ARRAY X[*] X_:;

    DO J = 1 TO 40;
        DO I = 1 TO DIM(X);
            X[I] = RANUNI(0)*I;
            OUTPUT;
        END;
    END;
RUN;

在这种情况下，变量名称 X_i 单调增加，实际上，我的变量实际上是 X_number ，其中数字是六位数的唯一标识符。我一直在尝试记录所有这些变量的变换和平方，以便我有一个新的 X 矩阵，其中包含以下列

X_133456 X_SQ_133456 LOG_X_133456

我尝试通过这样的所有变量循环列表

PROC CONTENTS
    DATA = TEST_ARRAY
    OUT = CONTENTS;
RUN;

PROC SQL NOPRINT;
    SELECT NAME INTO: REG_FACTORS
        SEPARATED BY " "
            FROM CONTENTS;
QUIT;

DATA WANT;
SET TEST_ARRAY;
%LET index = 1;
%DO %UNTIL (%SCAN(&REG_factors.,&index.," ")=);
    %LET factors = %SCAN(&REG_factors.,&index.," ");
    LOG_X_&FACTORS. = LOG(X_&FACTORS.);
    X_SQ_&FACTORS. = (X_&FACTORS.) ** 2;
    %LET index = %EVAL(&Index + 1);
%END;
RUN;

但这会炸毁我的服务器，我需要找到一种更有效的方法，在此先感谢

编辑：对于贡献者 - 我设法在 13:04 解决

%LET input_factors = X_:;

PROC SQL;
    SELECT 
            NAME
        ,   TRANWRD(NAME,%SCAN(&input_factors.,1,'_'),'SQ')
        ,   TRANWRD(NAME,%SCAN(&input_factors.,1,'_'),'LOG')
    INTO        
            :factor_list        separated by " "
        ,   :sq_factor_list     separated by " "
        ,   :log_factor_list    separated by " "
    FROM
        contents
    WHERE
        VARNUM < 5
    WHERE
        NAME LIKE "%SCAN(&input_factors.,1,'_')_"
    ORDER BY
        INPUT(SCAN(NAME,-1,'_'),8.)
    ;
QUIT;

%PUT &factor_list.;
%PUT &sq_factor_list.;
%PUT &log_factor_list.;
4

3 回答 3

0

使用 3 个数组，一个用于输入值（例如X_31415），两个用于新的计算值（对数和平方）。

诀窍是根据原始变量名称动态生成计算变量的变量名称。

/* 使用字典表获取/生成 vname */
过程 sql ;
  选择名称，/* X_31415 */
         tranwrd(name,'X_','X_SQ_'), /* X_SQ_31415 */
         tranwrd(name,'X_','LOG_X_') /* LOG_X_31415 */
    进入 :VARLIST 以 ' ' 分隔，
         :SQLIST 用 ' ' 分隔，
         :LOGLIST 以 ' ' 分隔
  来自dictionary.columns
  其中 libname = '工作'
    和 memname = 'MYDATA'
    并命名为“X_%”
  order by input(scan(name,-1,'_'),8.) /* 根据数字后缀排序 */
  ;
退出 ;

现在您可以分配三个数组，循环输入值并相应地计算平方和对数。

数据数组3；
  设置我的数据；

  {*} &VARLIST 中的数组；/* X_1 X_17 X_31415 X_99999 */
  数组 sq{*} &SQLIST ; /* X_SQ_1 X_SQ_17 X_SQ_31415 X_SQ_99999 */
  数组 lg{*} &LOGLIST ; /* LOG_X_1 LOG_X_17 LOG_X_31415 LOG_X_99999 */

  do i = 1 to dim(in) ;
    sq{i} = in{i} ** 2 ;
    lg{i} = log(in{i}) ;
  结尾 ;

  放下我；
跑 ;
于 2018-11-17T22:58:18.370 回答
0

我已经同意了，但是，我认为可以在批准答案之前讨论效率

%LET TRANSFORM_Y        = NO;
%LET TRANSFORM_X_SQ     = YES;
%LET TRANSFORM_LOG      = YES;
%MACRO TEST;
    DATA TEST (DROP = i);
        DO i = 1 to 40000;
            VAR = i + 100000;
            output;
        end;
    run;

    PROC TRANSPOSE
        DATA = TEST
        OUT = TEST_T
            (DROP = _NAME_)
        PREFIX = X_;
        ID VAR;
        VAR VAR;
    RUN;

    DATA TEST_ARRAY;
        SET TEST_T;

        ARRAY X[*] X_:;

            DO I = 1 TO DIM(X);
                X[I] = RANUNI(0)*I;
                OUTPUT;
            END;
    RUN;

    DATA TEST_ARRAY_2;
        SET TEST_ARRAY;

        Y = RANUNI(0);
        DROP I J;
        ROW_NUM = _N_;
    RUN;

    PROC TRANSPOSE
        DATA = TEST_ARRAY_2
            (DROP = ROW_NUM)
        OUT  = TEST_ARRAY_T
        ;
    RUN;
    %IF &TRANSFORM_X_SQ. = YES %THEN %DO;
        DATA TESTING_X_SQ
            (DROP = I);
            SET TEST_ARRAY_T;
            ARRAY COL[*] COL:;
                DO I = 1 TO DIM(COL);
                    COL(I) = COL(I)**2;
                END;
            Row_num = _N_;
        RUN;

        PROC TRANSPOSE
            DATA = TESTING_X_SQ
            OUT  = X_SQ_T
                (DROP = _NAME_)
            PREFIX = SQ_
            ;
            ID _NAME_
            ;
        RUN;

        DATA X_SQ_T_2;
            SET X_SQ_T;
            ROW_NUM = _N_;
        RUN;
    %END;   
    %IF &TRANSFORM_LOG. = YES %THEN %DO;
        DATA TESTING_LOG;
            SET TEST_ARRAY_T;
            ARRAY COL[*] COL:;
                DO I = 1 TO DIM(COL);
                    COL(I) = LOG(COL(I));
                END;
        RUN;

        PROC TRANSPOSE
            DATA = TESTING_LOG
            OUT  = LOG_T
            PREFIX = LOG_
            ;
            ID _NAME_
            ;
        RUN;

        DATA LOG_T_2;
            SET LOG_T;
            ROW_NUM = _N_;
        RUN;
    %END;

    PROC SQL;
        CREATE TABLE FULL_DATA AS
        SELECT
                f.*
            %IF &TRANSFORM_X_SQ.    = YES %THEN %DO;
            ,   x.*
            %END;
            %IF &TRANSFORM_LOG.     = YES %THEN %DO;
            ,   l.*
            %END;
        FROM 
            TEST_ARRAY_2    f
        %IF &TRANSFORM_X_SQ.    = YES %THEN %DO;
        LEFT JOIN
            X_SQ_T_2        x       ON f.row_num = x.row_num
        %END;
        %IF &TRANSFORM_LOG.     = YES %THEN %DO;
        LEFT JOIN
            LOG_T_2         l       ON l.row_num = x.row_num
        %END;
        ;
    QUIT;
%MEND;
%TEST;
于 2018-11-17T20:43:22.503 回答
0

当有大量变量时，您可能需要使用 SAS File I/O 函数来迭代变量。此示例创建一个包含 55,000 个响应变量的数据集，并计算它们的平方和对数变换。

%macro make_have(nvar=10);
  %local dsid suffix;

  data cols;
    do index = 100000 to 999999;
      if ranuni(123) < &nvar / (1e6-1e5) then output;
    end;
  run;

  data have;
    do id = 1 to 10;
      sex = ceil(2*ranuni(123));
      age = 17 + ceil(52*ranuni(123));
      weight = 150 + ceil(100*ranuni(123));

      %let dsid = %sysfunc(open (cols));
      %do %while (0 = %sysfunc(fetch(&dsid)));
        %let suffix = %sysfunc(getvarn(&dsid,1));
        x_&suffix = ranuni(123);
      %end;
      %let dsid = %sysfunc(close(&dsid));

      output;
    end;
  run;

%mend;

options nomprint;

%make_have(nvar=55000);

%macro make_transforms(data=, vars=, new=, function=);
  %local dsid i nvar varname;
  %let dsid = %sysfunc(open (&data));
  %do i = 1 %to %sysfunc(attrn(&dsid,nvar));
    %let varname = %sysfunc(varname(&dsid,&i));
    %if %substr(&varname,1,%length(&vars)) = &vars %then %do;
      &new.%substr(&varname,%length(&vars)+1) = %sysfunc(tranwrd(&function,#,&varname));
    %end;
  %end;
  %let dsid = %sysfunc(close(&dsid));
%mend;

data want;
  set have;
  %let t0 = %sysfunc(datetime());
  %make_transforms(data=have, vars=x_, new=x_sq_,  function=#**2)
  %make_transforms(data=have, vars=x_, new=x_log_, function=log(#))
  %put NOTE: codegen elapsed: %sysevalf(%sysfunc(datetime())-&t0);
run;
于 2018-11-19T04:54:46.553 回答