如下小程序应该能满足你的要求：

// 生成模拟数据
clear
set obs 400
gen id = 600000 + _n
gen industry = 1+int(20*uniform())
gen x1 = uniform()
gen x2 = uniform()
gen y = industry + industry*x1 + 2*industry*x2 + invnormal(uniform())
move y x1
gen con = .
gen b1 = .
gen b2 = .
gen N = .
gen r2adj = .

// 回归
quietly forval i = 1/400 {
 regress y x1 x2 if industry == industry[`i'] & _n != `i'
 matrix eb = e(b)
 replace con = eb[1,3] in `i' // 回归常数项
 replace b1 = eb[1,1] in `i' // 回归系数1
 replace b2 = eb[1,2] in `i' // 回归系数2
 replace N = `e(N)' in `i' // 样本数
 replace r2adj = `e(r2_a)' in `i' // 调整后r2
}

// 生成未预期值
gen resid = y - con - b1*x1 - b2*x2

// 生成行业系数平均值
sort industry
by industry: egen m_b1 = mean(b1)
by industry: egen m_b2 = mean(b2)

// 保存
save result, replace

也请论坛高手就该问题继续支招，让我也学习学习！

[此贴子已经被作者于2009-4-30 10:56:03编辑过]

 
g a=.
g b1=.
g b2=.
g yp=.
g e=.

forv i=1/400{
reg y x1 x2 if industry==industry[`i']&_n!=`i'
mat eb=e(b)
predict p in `i', xb
replace yp=p in `i'
replace e=y-yp in `i`
replace a=eb[1,3] in `i'
replace b1=eb[1,1] in `i'
replace b2=eb[1,2] in `i'
drop p
}

bys industry: egen m_b1=mean(b1)
bys industry: egen m_b2=mean(b2)

*完全按voodoo的思路。

发现voodoo程序中的
    matrix eb = e(b)
    replace con = eb[1,3] in `i' // 回归常数项
    replace b1 = eb[1,1] in `i' // 回归系数1
    replace b2 = eb[1,2] in `i' // 回归系数2

可以
    replace con = _b[_con] in `i' // 回归常数项
    replace b1 = _b[x1] in `i' // 回归系数1
    replace b2 = _b[x2] in `i' // 回归系数2

替换之，似乎简单一点点。

发现sungmoo程序中的
      predict p in `i', xb
      replace yp = p in `i'
      replace e = y - yp in `i`
      ....
      drop p

可以
      predict resid, residuals
      replace e = resid in `i`
      ...
      drop resid

替换之，似乎也简单一点。

[此贴子已经被作者于2009-5-1 18:23:09编辑过]

运行成功！！！太感谢了！解决了我好久几来的难题！太感谢了，好开心。谢谢各位