2014-10-09 99 views
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我有一个包含三个独立变量的线性模型。我有下面的最后一个模型。我想使用increase_po和decrease_po变量重新估计回归模型中的y值。如何拟合SAS中使用不同参数的线性回归模型

Dependent Variable Estimate increase_po decrease_po 
Rate  Rate_lag1 0.54  0.60   0.49 
Rate  UN   0.07  0.08   0.06 
Rate  SQ   0.03  0.03   0.02 

我想要做的就是写一个DO循环生产的六个可能的组合:

comb1 comb2 comb3 comb4 comb5 comb6 
0.60 0.49 0.08 0.06 0.03 0.02 
0.07 0.07 0.54 0.54 0.54 0.54 
0.03 0.03 0.03 0.03 0.07 0.07 

我想用这些参数来重新拟合模型,并得到估计ÿ值。

fitted y1= b0 + 0.60*Rate_lag1 + 0.07*UN + 0.03*SQ (only Rate_lag1 change parameter) 

fitted y2= b0 + 0.49*Rate_lag1 + 0.07*UN + 0.03*SQ (only Rate_lag1 change parameter) 

fitted y3= b0 + 0.54*Rate_lag1 + 0.08*UN + 0.03*SQ (only UN change parameter) 

....................

因此,如果不使用宏循环是很难甚至是不可能的。

+0

你的问题是重复的,仍然不完全清楚。看看proc分数。 – Reeza 2014-10-09 04:18:58

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第一个数据集来自哪里?什么是增加 - 减少和减少?我不能说这是否简单地尝试用不同的x值或改装模型来估计y以获得不同的参数估计值。 – Reeza 2014-10-09 05:50:35

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我简化了这里的问题:从SAS中的不同变量生成不同的组合,我的意思是你可以在Stackoverflow中搜索。当问题解决后,我会关闭这个问题。你可以看看这个问题吗? – 2014-10-09 05:52:34

回答

1

我看到你的其他线程,但想在这里发布。

如果我理解正确: 您已经为您的数据拟合了3个模型,并且每个模型使用相同的三个独立变量/预测变量。 您已经显示了与3个模型的每个预测变量相对应的beta参数。 您想从3个原始模型开始创建6个新模型,并且一次只更改一个β参数。

下面是我认为会做你想做的一些SAS代码。

但是,您的3个原始模型中的测试版估计都非常相似!所以我不知道这个练习揭示了什么是“最佳”模型。

祝你好运!

data have; 
    infile cards; 
    input Dependent $ Variable $ Estimate increase_po decrease_po; 
    cards; 
Rate  Rate_lag1 0.54  0.60   0.49 
Rate  UN   0.07  0.08   0.06 
Rate  SQ   0.03  0.03   0.02 
; 
run; 

*** TRANSPOSE SO EACH PREDICTOR VARIABLE IS A COLUMN AND EACH MODEL IS A ROW ***; 
*** NOTE: RATE_LAG1 CHANGES TO RATE_LAG IN THE TRANSPOSE ***; 
proc transpose data=have out=have_transpose; 
    id variable; 
    var Estimate increase_po decrease_po; 
run; 

*** CREATE VARIABLE FOR MODEL NUMBERS 1-3 ***; 
data have_transpose; 
    set have_transpose; 
    modelnum=_N_; 
run;  

proc print data=have_transpose; 
run; 

*** PUT EACH COLUMN INTO A SEPARATE DATASET AND KEEP ORIGINAL MODEL NUMBER IN EACH DATASET ***; 
data col1(keep=rate_lag modelnum rename=(modelnum=rate_lag_num)) 
     col2(keep=un modelnum rename=(modelnum=un_num)) 
     col3(keep=sq modelnum rename=(modelnum=sq_num)) 
    ; 
    set have_transpose; 
run; 

*** USE SQL TO DO A MANY-TO-MANY MERGE FOR ALL THREE DATASETS ***; 
*** THE CREATED DATASET WILL CONTAIN ALL POSSIBLE COMBINATIONS OF PARAMETER ESTIMATES FROM ALL MODELS ***; 
*** IN THIS CASE THERE WILL BE 3x3x3 = 27 RECORDS ***; 
proc sql; 
    create table col123 as 
    select * 
    from col1, col2, col3 
; 
quit; 

data almost; 
    set col123; 
    *** FOR EACH POSSIBLE COMBINATION, COUNT HOW MANY PARAMETERS ARE UNCHANGED FROM MODEL 1 ***; 
    flag = (rate_lag_num=1) + (un_num=1) + (sq_num=1); 
run; 

proc print data=almost; 
run; 

*** WANT TO ONLY KEEP MODELS WHERE TWO PARAMETERS ARE UNCHANGED ESTIMATES (WHERE FLAG=2) ***; 
data want; 
    set almost; 
    if flag=2; 
    keep rate_lag un sq ; 
run; 

*** THIS DATASET CONTAINS 6 RECORDS ***; 
proc print data=want; 
run; 


*** FROM HERE YOU CAN USE SQL TO DO A MANY-TO-MANY MERGE WITH YOUR 6 NEW MODELS AND DATASET WITH THE PREDICTOR VARIABLES ***; 
*** AND THEN CREATE YOUR NEW ESTIMATES FOR EACH MODEL ***; 
*** SOMETHING LIKE THIS BELOW ***; 
/* 

*** RENAME VARIABLES AND CREATE NEW MODEL NUMBER ***; 
data betas; 
    set want; 
    new_model_num = _N_; 
    rename rate_lag=rate_lag_beta un=un_beta sq=sq_beta ; 
run; 

proc sql; 
    create table all as 
    select * 
    from betas, ORIGINAL_DATA; *** CHANGE "ORIGINAL_DATA" TO YOUR DATA SET NAME ***; 
run; 

data new_model; 
    set all; 
    fitted = b0 + (Rate_lag_beta * Rate_lag1) + (un_beta * un) + (sq_beta * sq); 
run; 
*/ 

*** NOTE: IN YOUR EXAMPLE, YOU ASSUME THE SAME B0 FOR ALL MODELS 
*** HOWEVER, YOUR THREE STARTER MODELS MAY HAVE DIFFERENT B0 ESTIMATES ***; 
*** SO YOU WILL HAVE TO THINK ABOUT THE BEST WAY TO HANDLE THAT ***; 
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