在网上找了半天也么找到如何从回归中提取P值

之后自己动手了

上code：

reg NII BANKHI NIEAA
 return list
 local i "BANKHI"
 di _b[`i']
 di _se[`i']
 di _b[`i']/_se[`i']
 di e(df_r)
 scalar t=_b[`i']/_se[`i']
scalar pvalue=2*ttail(e(df_r),abs(t))
di pvalue
上结果：

. reg NII BANKHI NIEAA

      Source |       SS       df       MS              Number of obs =     728
-------------+------------------------------           F(  2,   725) =   10.49
       Model |  .396349751     2  .198174876           Prob > F      =  0.0000
    Residual |  13.6996586   725  .018896081           R-squared     =  0.0281
-------------+------------------------------           Adj R-squared =  0.0254
       Total |  14.0960083   727  .019389282           Root MSE      =  .13746

------------------------------------------------------------------------------
         NII |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
      BANKHI |    .152789   .1184969     1.29   0.198    -.0798491    .3854271
       NIEAA |  -.0432607    .010102    -4.28   0.000    -.0630933   -.0234281
       _cons |   .4703797   .0144333    32.59   0.000     .4420437    .4987157
------------------------------------------------------------------------------

.  return list

scalars:
              r(level) =  95

matrices:
              r(table) :  9 x 3

.  local i "BANKHI"

.  di _b[`i']
.15278902

.  di _se[`i']
.11849693

.  di _b[`i']/_se[`i']
1.2893922

.  di e(df_r)
725

.  scalar t=_b[`i']/_se[`i']

. scalar pvalue=2*ttail(e(df_r),abs(t))

. di pvalue
.19767298

参考链接：

1.http://www.stata.com/statalist/archive/2005-02/msg00481.html

2.http://repec.org/bocode/e/estout/esttab.html（很牛逼的链接，有时间好好消化一下）

3.http://repec.org/bocode/e/estout/advanced.html

4.http://www.ats.ucla.edu/stat/stata/faq/returned_results.htm(UCLA的返回链接，有实际列子可以参考)