Question

我有一个如下所示的SAS数据集。我想使用第一列值作为minuend，即第2列 - 第1列，第3列 - 第1列，第4列 - 第1列......哪种在SAS中进行这种差异的最有效方法是什么？使用数组还是执行循环？

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Answer 1

使用arrays，如下所示

data _NULL_;
set test;
array all{*} expense2-expense5;
array diff{*} diff_exp2-diff_exp5;
do i=1 to dim(all);
diff[i]= expense1 - all{i};
end;
put _ALL_;
run;

在SAS中执行列之间差异的最有效方法

1 个答案: