快速代碼寫入循環

Question

我想創建一個快速代碼寫入的循環。 我希望x1- x30的每一個可能性都相等：x平方i，當i是x1的索引（即1）時。

例如，x7將是x7 = x ** 7;

我寫了一個代碼，但它不起作用。我不知道如何解決他。我會爲你的幫助人員感到高興。

DATA maarah (drop = i e); 
e = constant("e"); 
do i = -10 to 10 by 0.01; 
x=i; 
y=e**x; 
output; 
end; 
length x1-x30 $2001; 
do i =1 to 30 by 1; 
x i=x**i; 
output; 
end; 
run; 

Answer 1

你很近。你需要聲明一個數組。你不解釋前半部分是什麼（e ** i部分），所以目前還不清楚你想要的是什麼 - 你想要幾千行e的權力，然後一些行x1-x30？你爲什麼每次輸出第二個循環？要回答的核心問題，在這裏：

DATA maarah (drop = i e); 
e = constant("e"); 
do i = -10 to 10 by 0.01; 
x=i; 
y=e**x; 
output; 
end; 

*length x1-x30 $2001; *what is this? Why do you want it 2001 characters, instead of numeric?; 
array xs x1-x30; *you would need a $ after this if you truly wanted character; 
do i =1 to 30 by 1; 
xs[i]=x**i; 
*output; *You probably do not want this. Output is probably outside of the loop.; 
end; 
run; 

我猜你真正想要的是這樣的：

DATA maarah (drop = i e); 
e = constant("e"); 
do i = -10 to 10 by 0.01; 
x=i; 
y=e**x; 
*length x1-x30 $2001; *what is this? Why do you want it 2001 characters, instead of numeric?; 
array xs x1-x30; *you would need a $ after this if you truly wanted character; 
do j =1 to 30; 
    xs[j]=x**j; 
end; *the x1-x30 loop; 
output; 
end; *the outer loop; 
run;