問題：在 data step 中去處理一大串的變項或數值
說明：
用 Array 來處理，基本架構是
array array-name <$>
array-elements 可以用
- list 變項名稱，但是大原則是這些 list 要全為 numeric 或 character ，不能 numeric 或 character 混著用
- special variable, 如 NUMERIC CHARACTER ALL 。 會直接挑 data set 全部的 numeric 或 character 或全部。
- 上面都會建立實體的 variable ，但 Temporary 則不會，只會放在 Program data vector。
data a; array A {10} ; array nums {3:10} temporary (90, 91,92 ,93 ,94, 95, 96, 40); do i =1 to dim(A); if i<3 then A=i10; else A=i10+nums; end; run;
就不必寫 array 來找
data a; a=40; b=whichn(a, 90, 91,92 ,93 ,94, 95, 96, 40); put a / b ; run;
IMPLICIT SUBSCRIPTING array-name 改成了array-name (n) 的 n 不放 * 不放 數值，並且放了個空變項，這是舊版SAS的用法 DATA AGE_; z=100;put z; array item(z) age1989 age1993 age1996 age1999 age2003 age2007 age2011 age2015(1989 1993 1996 1999 2003 2007 2011 2015); do over item; put item z ; item=item+1000; put item z ; end; RUN;
INVALID INDEX RANGE DATA AGE_; array item() age1989 age1993 age1996 age1999 age2003 age2007 age2011 age2015 (1989 1993 1996 1999 2003 2007 2011 2015); array X(8); ii=1; do until ( ii gt 8); / do until ( ii ge 8); 不行到8就停了 X8就做不出來*/
put "from 1st put " ii;
x(ii)=item(ii)+1000;
ii=ii+1;
put "from 2nd put " ii;
end; RUN;
FUNCTION NAME AS AN ARRAY NAME ARRAY REFERENCED IN MULTIPLE DATA STEPS, BUT DEFINED IN ONLY ONE
ref: Arrays Made Easy An Introduction to Arrays and Array Processing
/Chapter 15 Processing Variables with Arrays/
/*
ARRAY array-name
elements 也可用 NUMERIC CHARACTER ALL Temporary 。
資料檔中沒有 element 列出 的變項，會自動建立該變項
不寫element，系統會自行產生新變項。
*/
data test_array; Array L[4] spring summer fall winter; Array K[5] ; run; /Arrays of Character Variables 各變項預設為$8. / array Chars {5} $ 24; /Assigning Initial Values to Arrays/ data test_array; Array L[5] spring summer fall winter all (1 2 3 4 5); Array K[4] Temporary (11 12 13 14 ) ; Array Z [4]; Do i=1 to dim(z); Z[i]= 1000L[i+1] +100L[i] + K[i] ; End; run; /Multidimensional Arrays/ DATA test_array2 ; array new{3,4} x1-x12; /3 rows, 4 columns/ new(2,3)=0; run; /* x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 */
/Referencing Elements of a Two-Dimensional Array/ data finance.quarters(drop=i j); set finance.monthly; array m{4,3} month1-month12; array Qtr{4}; do i=1 to 4; /*當 i=1，M1j 就全進到 PDV了 */ qtr=0; do j=1 to 3; /*j=1，M11加總入qtr1 / qtr+m{i,j}; end; end; run; / month1 month2 month3 month4 month4 month4 month7 month7 month7 month10 month10 month10 / /資料轉秩/ DATA trans; input id spring summer fall winter ; datalines; 100 0 5 10 15 101 20 25 30 35 102 40 45 50 55 103 60 65 70 75 ; data trans2 ; set trans; array contrib{4} spring summer fall winter ; do i =1 to 4; Value=contrib[i]; output; end; run; proc print ; var id i value ;run; / id i Value 100 1 0 100 2 5 100 3 10 100 4 15 101 1 20 101 2 25 101 3 30 101 4 35 */ /quiz/ data coat; input category high1-high3 / low1-low3; array compare{2,3} high1-high3 low1-low3; do i=1 to 2; do j=1 to 3; compare{i,j}=round( compare{i,j} * 1.12 ); end; end; datalines; 5555 9 8 7 6 4 3 2 1 8888 21 12 34 64 13 14 15 16 ; run;
/* 篩選多個變項，是否有某範圍的值 */ data k ; input a1 a2 a3 ; cards; 10 20 99 1 2 20 99 99 99 5 99 99 ;
data k2 ; set k ; Array k (*) a1-a3; DO i = 1 to 3 ; if 0<k(i) <10 then x1=1; if 0<k(i) <10 then x2=sum(X2 , 1 ); if 0<k(i) <10 then x3+1; end; run;
問題： 由大到小 作 iteration
正確：要加 by 數值
do i3 =dim(AGREE) to 2 by -1; end;
一時忘了，用以下就不行，無法反向
do i3 =dim(AGREE) to 2 ; end;
