工业企业数据库处理——2.匹配样本_工业企业数据库匹配

样本匹配问题的由来：

对于一个多年企业数据来说，整理数据的第一步就是构建一个以企业ID和年份作为两维的面板数据。这通常并不是个问题，但对于中国工业企业数据库来说确是一个非常棘手的问题。因为在该数据库中，难以找到一个识别每一个样本企业的唯一特征来进行编码。通常的做法是，根据企业代码、企业名称、法人代表姓名、地址、邮编、电话、行业代码、主要产品名称、开业时间等基本信息来识别不同的样本点是否来自于同一家企业。但是因为这些基本信息在申报时并没有统一格式，在缺乏有效的智能模糊匹配手段的情况下，精确匹配的可操作性不强。这其中，企业代码和企业名称的准确性相对较高，可以作为我们进行匹配所依据的主要信息，比如Brandt 等(2012)就是先根据相同的企业代码识别同一家企业，然后再根据相同的企业名称进行识别，最后再参考其他信息。这种贯序识别法嘉定企业代码的准确性最高，企业名称准确性次之，也就是说共享同一企业代码的样本点必然被识别为同一企业，反过来，被识别为统一企业的样本点可能拥有不同的企业代码。在本数据库中，不但存在统一企业更改企业代码的情况（例如发生改制或者重组之后），更重要的是，还存在不同的企业共享企业代码的情况（可能是统计错误）。企业名称这一变量也存在类似问题。很多企业在改制、重组或者扩张时更改了企业名称。例如，中国有不少企业先是叫“XX厂”，然后改名叫“XX有限责任公司”，接着又叫“XX股份有限公司”。有时企业名称中的地理位置也略有差异，例如从“XX市机电厂”变成“XX机电厂”。如果按照企业名称进行精确匹配会错误地识别出“过多”的企业。

Stata程序

第一部分，从预处理的数据中保留下基本的用于match的变量。包括：法人代码、企业名称、法人、地址、省份等等信息。分别保存为m1998.dta~m2007.dta。

clear all
set more off

global PATH "/Volumes/TOSHIBA EXT/Projects/NBS/China Industry Business Performance Data/Match Over Years"
cd "$PATH"

******************************************************************************
* Part 1 
* Befor run this do-file, orignial_1998.dta ~ original_2007 must be already 
* generated. Whic means that do-files 1998.do ~ 2007.do have already been 
* runned.
*
* Generate a id variable (id_in_source) for further combining data set
* after match over years.
* 
* Only keep match variables and id_in_source for the next steps
*****************************************************************************

forvalues i = 1998/2007{
	disp "File `i'"
	use `"../original_`i'.dta"',clear
*	gen id_in_source = _n
	if `i'==2003{
		gen town = address
	}
	gen cic = cic_adj
	replace cic = real(industry_code) if cic == .
	/*
	if year <2003{
		gen cic = cic_adj
	}
	else{
		gen cic = cic03
	}
	*/

	if year<2004{
		gen revenue = sales_revenue
	}
	else{
		gen revenue = operating_revenue
	}
	gen profit = total_profit
	if year ==1999 | year == 2002{
		gen employment = staff
	}
	keep id_in_source firm_id firm_name legal_person town province ///
	     telephone zip product1 founding_year cic region_code revenue ///
		 employment profit
	destring founding_year revenue employment profit,replace force
	tostring cic,replace format(%04.0f)
	rename firm_id id
	rename firm_name name
	rename founding_year bdat
	rename region_code dq
	rename product1 product1_
	rename telephone phone
	foreach var of varlist *{
		rename `var' `var'`i'
	}
	compress
	saveold m`i'.dta,replace
}

第二部分，跨期匹配样本。

步骤1. 匹配连续两年的数据（1998-1999；1999-2000；...）

1.把id中的小写字母全部转换为大写。

forval i =1998/2007{
	use m`i'.dta,clear
	replace id`i' = strupper(id`i')
	compress
	saveold m`i'.10.dta,replace
}
forval i =1998/2007{
	use m`i'.10.dta,clear
	des,short
}

2.使用循环，匹配连续两年之间的数据。loca i 代表当年年份，local j=i+1 代表未来一年。

  Step 10.  根据法人代码 firm_id 匹配。

	*deal with duplicates of IDs (There are a few firms that have same IDs)
	disp "Step 10 "
	use m`i'.10.dta,clear
	bysort id`i': keep if _N>1
	compress
	saveold duplicates_ID`i'.dta,replace
	
	use m`i'.10.dta,clear
	bysort id`i': drop if _N>1
	rename id`i' id
	sort id
	keep *`i' id
	compress
	saveold match`i'.1.dta,replace
	
	use m`j'.10.dta,clear
	bysort id`j': keep if _N>1
	compress
	saveold duplicates_ID`j'.dta,replace
	
	use m`j'.10.dta,clear
	bysort id`j': drop if _N>1
	rename id`j' id
	keep *`j' id
	sort id
	compress
	saveold match`j'.1.dta,replace
	
	use match`i'.1.dta,clear
	merge 1:1 id using match`j'.1.dta
	keep if _m==3
	gen id`i' = id
	rename id id`j'
	drop _merge
	gen match_method_`i'_`j'="ID"
	gen match_status_`i'_`j'="3"
	compress
	saveold matched_by_ID`i'_`j'.dta,replace
	

  匹配后生成三类文件：

     *另外 step10之后可以注意到没有匹配成功的记录并不包括在上述三类文件中，在后续合并步骤中这些记录还会被重新匹配。

Step20 利用 firmname 继续匹配Step10中未匹配的记录。

	**step20: match by firm names**
	
	*match those unmatched firms in previous step by firm names*
	disp "Step 20 "	
	use match`i'.1.dta,clear
	merge 1:1 id using match`j'.1.dta
	keep if _m==1
	rename id id`i'
	append using duplicates_ID`i'.dta
	bysort name`i': keep if _N>1
	keep *`i'
	compress
	saveold duplicates_name`i'.dta,replace
	
	use match`i'.1.dta,clear
	merge 1:1 id using match`j'.1.dta
	keep if _m==1
	rename id id`i'
	append using duplicates_ID`i'.dta
	bysort name`i': drop if _N>1
	rename name`i' name
	sort name
	keep *`i' name
	compress
	saveold unmatched_by_ID`i'.dta,replace
	
	use match`i'.1.dta,clear
	merge 1:1 id using match`j'.1.dta
	keep if _m==2
	rename id id`j'
	append using duplicates_ID`j'.dta
	bysort name`j': keep if _N>1
	keep *`j'
	compress
	saveold duplicates_name`j'.dta,replace	

	use match`i'.1.dta,clear
	merge 1:1 id using match`j'.1.dta
	keep if _m==2
	rename id id`j'
	append using duplicates_ID`j'.dta
	bysort name`j': drop if _N>1
	rename name`j' name
	sort name
	keep *`j' name
	compress
	saveold unmatched_by_ID`j'.dta,replace
	
	use unmatched_by_ID`i'.dta,clear
	merge 1:1 name using unmatched_by_ID`j'.dta
	keep if _m==3
	gen name`i' = name
	rename name name`j'
	drop _m
	gen match_method_`i'_`j'="firm name"
	gen match_status_`i'_`j'="3"
	compress
	saveold matched_by_name`i'_`j'.dta,replace

匹配之后生成三类文件

	disp "Step 30 "
	use unmatched_by_ID`i'.dta,clear
	merge 1:1 name using unmatched_by_ID`j'.dta
	keep if _m == 1
	rename name name`i'
	append using duplicates_name`i'.dta
	replace legal_person`i' = "." if legal_person`i' == ""
	gen code1 = legal_person`i' + substr(dq`i',1,4)
	bysort code1: keep if _N>1
	keep *`i' 
	compress
	saveold duplicates_code1_`i'.dta,replace
	
	use unmatched_by_ID`i'.dta,clear
	merge 1:1 name using unmatched_by_ID`j'.dta
	keep if _m == 1
	rename name name`i'
	append using duplicates_name`i'.dta
	replace legal_person`i' = "." if legal_person`i' == ""
	gen code1 = legal_person`i' + substr(dq`i',1,4)
	bysort code1: drop if _N>1
	sort code1
	keep code1 *`i'
	compress
	saveold unmatched_by_ID_and_name`i'.dta,replace
	
	use unmatched_by_ID`i'.dta,clear
	merge 1:1 name using unmatched_by_ID`j'.dta
	keep if _m == 2
	rename name name`j'
	append using duplicates_name`j'.dta
*	replace legal_person`j' = "." if legal_person`j' == ""
	gen code1 = legal_person`j' + substr(dq`j',1,4)
	bysort code1: keep if _N>1
	keep *`j' 
	compress
	saveold duplicates_code1_`j'.dta,replace	
	
	use unmatched_by_ID`i'.dta,clear
	merge 1:1 name using unmatched_by_ID`j'.dta
	keep if _m == 2
	rename name name`j'
	append using duplicates_name`j'.dta
*	replace legal_person`j' = "." if legal_person`j' == ""
	gen code1 = legal_person`j' + substr(dq`j',1,4)
	bysort code1: drop if _N>1
	sort code1
	keep code1 *`j'
	compress
	saveold unmatched_by_ID_and_name`j'.dta,replace
	
	use unmatched_by_ID_and_name`i'.dta,clear
	disp _N
	merge 1:1 code1 using unmatched_by_ID_and_name`j'.dta
	keep if _m==3
	drop _m code1
	gen match_method_`i'_`j' = "legal_person"
	gen match_status_`i'_`j' = "3"
	compress
	saveold matched_by_legalperson`i'_`j'.dta,replace
	

此后，还继续使用了phone number 、foudingyear等信息，经过 Step40 Step50 Step60进一步匹配了信息。得到了如下几个文件：

将这六个文件合并，就得到了所有连续两年匹配后的样本

	disp "Step 60 "	
	use matched_by_ID`i'_`j'.dta,clear
	append using matched_by_name`i'_`j'.dta
	append using matched_by_legalperson`i'_`j'.dta
	append using matched_by_phone`i'_`j'.dta
	append using matched_by_code3_`i'_`j'.dta
	append using unmatched_by_ID_and_name_and_legalperson_and_phone_and_code2`i'.dta
	append using unmatched_by_ID_and_name_and_legalperson_and_phone_and_code2`j'.dta	
	compress
	saveold m`i'-m`j'.dta,replace	

匹配后的效果以m1998-m1999.dta为例：

这里只用开头的文件作为检查是不合适的，应该同时使用m2006-m2007.dta作为检查。

对比 Brandet(2012)中每年记录数目的表格：

可以验证合并过程没有问题，根据图1，1998年-1999年匹配的有140653条，未匹配成功1998年的共有24465条，相加刚好是165118条；其他年份的数据经检验后也是如此。至此，连续两年的跨期合并执行结束。

Step 70. 生成连续三年的平衡面板，并生成用于后续继续匹配的未匹配数据。年份用loca i， j， k 表示先后的连续三年。

	**Step 70: Create a three-year balanced sample
	disp "Step 70 "	
	use m`i'-m`j'.dta,clear
	keep if match_status_`i'_`j' == "1"
	keep *`i'
	compress
	saveold unmatched`i'.10.dta,replace

	use m`i'-m`j'.dta,clear
	drop if match_status_`i'_`j' == "1"
	gen code = id`j'+string(revenue`j')+string(employment`j')+string(profit`j')+province`j'
	sort code
	compress
	saveold m`i'-m`j'.10.dta,replace
	
	use m`j'-m`k'.dta,clear
	keep if match_status_`j'_`k' == "2"
	keep *`k'
	compress
	saveold unmatched`k'.10.dta,replace

	use m`j'-m`k'.dta,clear
	drop if match_status_`j'_`k' == "2"
	gen code = id`j'+string(revenue`j')+string(employment`j')+string(profit`j')+province`j'
	sort code
	compress
	saveold m`j'-m`k'.10.dta,replace

	use m`i'-m`j'.10.dta,clear
	merge 1:1 code using m`j'-m`k'.10.dta
	drop _m code
	keep if match_status_`i'_`j'=="3" & match_status_`j'_`k'=="3"
	gen match_status_`i'_`k'="3"
	gen match_method_`i'_`k'="`j'"
	compress
	saveold balanced.m`i'-m`j'-m`k'.dta,replace

执行之后生成了如下几类文件

unmatched`i'.10.dta - 来自于 m`i'-m`j'.dta 中，未成功匹配且仅在year i 中的记录。

m`i'-m`j'.10.dta - 来自于 m`i'-m`j'.dta中，包括成功匹配的和仅在year j中的记录。

unmatched`k'.10.dta - 来自于m`j'-m`k'.dta中，未成功匹配且仅在year k中的记录

m`j'-m`k'.10.dta - 来自于m`j'-m`k'.dta中，包括成功匹配和仅在year j中的记录

balanced.m`i'-m`j'-m`k'.dta - m`i'-m`j'.10.dta和m`j'-m`k'.10.dta merge之后，两边都match的记录，即连续三年均出现的记录。

Step 80. 生成未成功匹配的文件，用于后续的匹配。

Step 70 生成了连续三年的平衡面板数据，但还有部分记录未能match起来，这一部分就是提取出这部分数据以便用于进一步的匹配。

	**Step 80: Creat files for unmatched `i' firms and `k' firms**

	disp "Step 80"
	use m`i'-m`j'.10.dta,clear
	merge 1:1 code using m`j'-m`k'.10.dta
	drop _m code
	drop if match_status_`i'_`j'=="3" & match_status_`j'_`k'=="3"
	drop if id`i'==""
	gen code = id`i'+string(revenue`i')+string(employment`i')+string(profit`i')+province`i'
	sort code
	compress
	saveold unmatched`i'.15.dta,replace
	
	use unmatched`i'.15.dta,clear
	keep *`i' 
	append using unmatched`i'.10.dta
	compress
	saveold unmatched`i'.20.dta,replace
	
	
	use m`i'-m`j'.10.dta,clear
	merge 1:1 code using m`j'-m`k'.10.dta
	drop _m code
	drop if match_status_`i'_`j'=="3" & match_status_`j'_`k'=="3"
	drop if id`k'== ""
	gen code = id`k'+string(revenue`k')+string(employment`k')+string(profit`k')+province`k'
	sort code
	compress
	saveold unmatched`k'.15.dta,replace
	
	use unmatched`k'.15.dta,clear
	keep *`k' 
	append using unmatched`k'.10.dta
	compress
	saveold unmatched`k'.20.dta,replace
	
	
	use m`i'-m`j'.10.dta,clear
	merge 1:1 code using m`j'-m`k'.10.dta
	drop _m code
	drop if match_status_`i'_`j'=="3" & match_status_`j'_`k'=="3"
	gen code = id`j'+string(revenue`j')+string(employment`j')+string(profit`j')+province`j'
	sort code
	compress
	saveold unmatched`j'.15.dta,replace

生成的各个文件的含义

unmatched`i'.15.dta - m`i'-m`j'.10.dta和m`j'-m`k'.10.dta merge之后不在平衡面板中且仅仅来自于year i的记录。

unmatched`i'.20.dta - unmatched`i'.15.dta 与 unmatched`i'.10.dta append起来的记录，其中 10.dta 是由Step70生成，表示 m`i'-m`j'.dta 中仅仅来自于year i 的记录。二者合并之后得到的就是目前为止所有来自于i的未能匹配的记录。

unmatched`k'.15.dta - 同unmatched`i'.15.dta

unmatched`k'.20.dta - 同unmatched`i'.20.dta

unmatched`j'.15.dta -  m`i'-m`j'.10.dta和m`j'-m`k'.10.dta merge之后 不在平衡面板中所有记录（与 以i， k为下表的同类文件相比，由于这次merge肯定是完全merge，因此所有都是“无法按照j进行match的记录”，但显然这样存在一定的重叠。后续在合并的时候会对这里的重叠部分进行处理）

Step90 对Step80中得到的未匹配数据（即不在三年平衡面板中的记录：ummatched`i'.20.dta,unmatched`k'.20.dta）进行再一次匹配，运用 firm_id 和firm_name 进行匹配。

匹配思路与构建两年平衡面板时对应的处理思路是完全一样的：先确定有否duplicates的现象，区分后分别merge

	**Step 90: Match `i' firms and `k' firms by firm ID and name**
	
	
	*ID*
	disp "Step 90"
	use unmatched`i'.20.dta,clear
	bysort id`i': keep if _N>1
	compress
	saveold duplicates_ID`i'.dta,replace
	
	use unmatched`i'.20.dta,clear
	bysort id`i': drop if _N>1
	rename id`i' id
	keep *`i' id
	sort id
	compress
	saveold match`i'.1.dta,replace
	
	use unmatched`k'.20.dta,clear
	bysort id`k': keep if _N>1
	compress
	saveold duplicates_ID`k'.dta,replace
		
	use unmatched`k'.20.dta,clear
	bysort id`k': drop if _N>1
	rename id`k' id
	keep *`k' id
	sort id
	compress
	saveold match`k'.1.dta,replace
	
	use match`i'.1.dta,clear
	merge 1:1 id using match`k'.1.dta
	keep if _m==3
	gen id`i'=id
	rename id id`k'
	drop _m
	gen match_method_`i'_`k'="`j'"
	gen match_status_`i'_`k'="3"
	compress
	saveold matched_by_ID`i'_`k'.dta,replace
	
	
	
	*name*
	
	use match`i'.1.dta, clear
	merge 1:1 id using match`k'.1.dta
	keep if _merge==1
	rename id id`i'
	append using duplicates_ID`i'.dta
	bysort name`i': keep if _N>1
	keep *`i'
	compress
	saveold duplicates_name`i'.dta, replace
	
	use match`i'.1.dta, clear
	merge 1:1 id using match`k'.1.dta
	keep if _merge==1
	rename id id`i'
	append using duplicates_ID`i'.dta
	bysort name`i': drop if _N>1
	rename name`i' name
	sort name
	keep name *`i'
	compress
	saveold unmatched_by_ID`i'.dta, replace

	use match`i'.1.dta, clear
	merge 1:1 id using match`k'.1.dta
	keep if _merge==2
	rename id id`k'
	append using duplicates_ID`k'.dta
	bysort name`k': keep if _N>1
	keep *`k'
	compress
	saveold duplicates_name`k'.dta, replace

	use match`i'.1.dta, clear
	merge 1:1 id using match`k'.1.dta
	keep if _merge==2
	rename id id`k'
	append using duplicates_ID`k'.dta
	bysort name`k': drop if _N>1
	rename name`k' name
	sort name
	keep name *`k'
	compress
	saveold unmatched_by_ID`k'.dta, replace
	
	use unmatched_by_ID`i'.dta, clear
	merge 1:1 name using unmatched_by_ID`k'.dta
	keep if _merge==3
	gen name`i'=name
	rename name name`k'
	drop _merge
	gen match_method_`i'_`k'="firm name"
	gen match_status_`i'_`k'="3"
	compress
	saveold matched_by_name`i'_`k'.dta, replace
	

	use unmatched_by_ID`i'.dta, clear
	merge 1:1 name using unmatched_by_ID`k'.dta
	keep if _merge==1
	rename name name`i'
	keep *`i'
	append using duplicates_name`i'.dta
	gen match_method_`i'_`k'=""
	gen match_status_`i'_`k'="1"
	compress
	saveold unmatched_by_ID_and_name_`i'.dta, replace	
	
	
	use unmatched_by_ID`i'.dta, clear
	merge 1:1 name using unmatched_by_ID`k'.dta
	keep if _merge==2
	rename name name`k'
	keep *`k'
	append using duplicates_name`k'.dta
	gen match_method_`i'_`k'=""
	gen match_status_`i'_`k'="2"
	compress
	saveold unmatched_by_ID_and_name_`k'.dta, replace

执行之后得到了如下几个文件：

matched_by_ID`i'_`k'.dta, clear
matched_by_name`i'_`k'.dta
unmatched_by_ID_and_name_`i'.dta
unmatched_by_ID_and_name_`k'.dta

文件含义从名字上就已经很清楚。

Step 100 合并所有文件，处理冲突问题，生成连续三年的非平衡面板

	**step 100: merge the files**
	disp "Step 100"
	use matched_by_ID`i'_`k'.dta, clear
	append using matched_by_name`i'_`k'.dta
	append using unmatched_by_ID_and_name_`i'.dta
	append using unmatched_by_ID_and_name_`k'.dta
	compress
	saveold m`i'-m`k'.dta, replace
	
	use m`i'-m`k'.dta, clear
	gen code = id`i'+string(revenue`i')+string(employment`i')+string(profit`i')+province`i'
	sort code
	*drop if code == "..."
	merge code using unmatched`i'.15.dta
	drop code _merge
	sort id`i'
	compress
	compress
	saveold m`i'-m`k'.05.dta, replace
		
	*deal with disagreement (_merge==5 if "update" is used)*

	use m`i'-m`k'.05.dta, clear
	gen code = id`k'+string(revenue`k')+string(employment`k')+string(profit`k')+province`k'
	sort code
	
	merge code using unmatched`k'.15.dta, update
	keep if _merge==5
	drop *`k'
	drop code _merge 
	sort id`i'
	compress
	compress
	saveold m`i'-m`k'.disagree.dta, replace


	use m`i'-m`k'.05.dta, clear
	merge id`i' using m`i'-m`k'.disagree.dta
	drop if _merge==3
	drop _merge
	append using m`i'-m`k'.disagree.dta
	
	gen code = id`k'+string(revenue`k')+string(employment`k')+string(profit`k')+province`k'
	sort code
	merge code using unmatched`k'.15.dta, update
	drop code _merge 
	gen code = id`j'+string(revenue`j')+string(employment`j')+string(profit`j')+province`j'
	sort code
	merge code using unmatched`j'.15.dta, update
	drop code _merge
	compress
	saveold m`i'-m`k'.dta.10.dta, replace
	

	use m`i'-m`k'.dta.10.dta, clear
	append using balanced.m`i'-m`j'-m`k'.dta
	drop match_status_`i'_`j'
	drop match_status_`j'_`k'
	drop match_status_`i'_`k'
	drop match_method_`i'_`j'
	drop match_method_`j'_`k'
	drop match_method_`i'_`k'
	gen match_status_`i'_`j'_`k'="`i'-`j'-`k'" if id`i'!=""&id`j'!=""&id`k'!=""
	replace match_status_`i'_`j'_`k'="`i'-`j' only" if id`i'!=""&id`j'!=""&id`k'==""
	replace match_status_`i'_`j'_`k'="`j'-`k' only" if id`i'==""&id`j'!=""&id`k'!=""
	replace match_status_`i'_`j'_`k'="`i'-`k' only" if id`i'!=""&id`j'==""&id`k'!=""
	replace match_status_`i'_`j'_`k'="`i' no match" if id`i'!=""&id`j'==""&id`k'==""
	replace match_status_`i'_`j'_`k'="`j' no match" if id`i'==""&id`j'!=""&id`k'==""
	replace match_status_`i'_`j'_`k'="`k' no match" if id`i'==""&id`j'==""&id`k'!=""
	compress
	saveold unbalanced.`i'-`j'-`k'.dta, replace

执行完了之后 以unbalanced.1998-1999-2000.dta,检查匹配状况：

可以验证合并过程没有问题

*最初只用了unbalanced.1998-1999-2000.dta作为检查对象，没有用unbalanced.2005-2006-2007.dta 同时做检查，导致有一个小错误没有发现，所有检查应注意至少检测所有的情况。

第三部分，生成10年的非平衡面板。

首先，将1998-1999-2000的非平衡面板保存为test1.dta

use unbalanced.1998-1999-2000.dta, clear
tab match_status_1998_1999_2000
gen code=id2000+string(revenue2000)+string(employment2000)+string(profit2000)
sort code
save test1.dta, replace

Step 110. 将2011年的数据合并进来

**step 110: add 2001 from 1999-2000-2001**
 
use unbalanced.1999-2000-2001.dta, clear
tab match_status_1999_2000_2001
keep if match_status_1999_2000_2001=="1999-2000-2001"|match_status_1999_2000_2001=="2000-2001 only"
gen code=id2000+string(revenue2000)+string(employment2000)+string(profit2000)
sort code
save test2.dta, replace
 
use test1.dta, clear
merge code using test2.dta
tab _merge
drop _merge code
gen code=id1999+string(revenue1999)+string(employment1999)+string(profit1999)
sort code
save test3.dta, replace
 
use unbalanced.1999-2000-2001.dta, clear
tab match_status_1999_2000_2001
keep if match_status_1999_2000_2001=="1999-2001 only"
gen code=id1999+string(revenue1999)+string(employment1999)+string(profit1999)
sort code
save test4.dta, replace
 
use test3.dta, clear
merge code using test4.dta, update
tab _merge
drop code _merge
save test5.dta, replace
 
use test3.dta, clear
merge code using test4.dta, update replace
keep if _merge==5
keep id2001	bdat2001 cic2001 dq2001 e_HMT2001 e_collective2001 e_foreign2001 e_individual2001 e_legal_person2001 e_state2001 employment2001 export2001	fa_net2001 fa_original2001 a_dep2001 c_dep2001 input2001	name2001 new_product2001 output2001 profit2001 revenue2001 type2001 va2001 wage2001 legal_person2001 phone2001 product1_2001 street2001 town2001 village2001 zip2001       
save test6.dta, replace
 
use unbalanced.1999-2000-2001.dta, clear
keep if match_status_1999_2000_2001=="2001 no match"
display _N
save test7.dta, replace
 
use test5.dta, clear
append using test6.dta
dis _N
append using test7.dta
dis _N
gen code=id2001+string(revenue2001)+string(employment2001)+string(profit2001)
sort code
save test1.dta, replace

思路如下：

从 1999-2000-2001中挑选出 以2000年为合并基准的记录，保存为test2.dta

将test1.dta和test2.dta 依据 code2000 merge起来，存为test3

从 1999-2000-2001中挑选出 以1999年为合并基准的记录，保存为test4.dta

将test3.dta和test4.dta依据code1999 merge起来，存为test5.

将test3.dta和test4.dta依据code1999 merge起来,取值存在冲突的记录，存为test6. test3.dta和test4.dta中公共的部分是var2000-var2002。如果merge的时候这部分公共变量有不同，说明不是同一条，应作为两条数据处理。

将2001年未能匹配的记录存为test7.dta

append test5.dta, test6.dta, test7.dta。

随后，通过Step120将2002将2002合并进来，。。。，直到顺次将2007也合并进来，就得到了这样一个非平衡面板。

最后，通过下面的代码，得到面板和原始数据的对应关系

use "unbalanced.1998--2007.dta",clear
keep id_in_source*
gen id_in_panel=_n
reshape long id_in_source, i(id_in_panel) j(year)
drop if id_in_source == .
sort id_in_panel year
saveold "PanelID_1998-2007.dta",replace

转换后检查每一年的样本数量：

*由于数据库来源的问题，我使用的数据与Brandt（2012）的数据在某些年份上存在一定的误差。（主要是2004年，所用的数据来自于全国经济普查，这同时也导致了很多关键指标的缺失等问题）

至此，匹配程序处理完毕，完整版程序见另一篇博文： 工业企业数据库处理代码完整版本——2.匹配样本 。