为范围之间的列生成值
Generating values for columns between a range
我有以下数据集
A B begin_yr end_yr
asset brown 2007 2010
asset blue 2008 2008
basics caramel 2015 2015
cows dork 2004 2006
我希望 A
和 B
具有代表每一年的行。
我每年扩展:
gen x = end_yr - begin_yr
expand x +1
这给了我以下信息:
A B begin_yr end_yr x
asset brown 2007 2010 3
asset brown 2007 2010 3
asset brown 2007 2010 3
asset brown 2007 2010 3
asset blue 2008 2008 0
basics caramel 2015 2015 0
cows dork 2004 2006 2
最终,我想要以下数据集:
A B begin_yr end_yr x year
asset brown 2007 2010 3 2007
asset brown 2007 2010 3 2008
asset brown 2007 2010 3 2009
asset brown 2007 2010 3 2010
asset blue 2008 2008 0 2008
basics caramel 2015 2015 0 2015
cows dork 2004 2006 2 2004
cows dork 2004 2006 2 2005
cows dork 2004 2006 2 2006
这是我目前拥有的:
gen year = begin_yr if begin_yr!=end_yr
如何填充变量的其余部分 year
?
以下对我有用:
clear
input strL A strL B begin_yr end_yr
asset brown 2007 2010
basics caramel 2015 2015
cows dork 2004 2006
end
generate id = _n
expand 2
clonevar year = begin_yr
bysort id: replace year = end_yr[2] if _n == _N
drop if _n == 3
tsset id year
tsfill
foreach var in A B begin_yr end_yr {
bysort id: replace `var' = `var'[1]
}
list
+--------------------------------------------------+
| A B begin_yr end_yr id year |
|--------------------------------------------------|
1. | asset brown 2007 2010 1 2007 |
2. | asset brown 2007 2010 1 2008 |
3. | asset brown 2007 2010 1 2009 |
4. | asset brown 2007 2010 1 2010 |
5. | basics caramel 2015 2015 2 2015 |
|--------------------------------------------------|
6. | cows dork 2004 2006 3 2004 |
7. | cows dork 2004 2006 3 2005 |
8. | cows dork 2004 2006 3 2006 |
+--------------------------------------------------+
这是在@Pearly Spencer 的代码基础上进行的转折:
clear
input strL A strL B begin_yr end_yr
asset brown 2007 2010
basics caramel 2015 2015
cows dork 2004 2006
end
gen toexpand = end - begin + 1
expand toexpand
bysort A : gen year = begin + _n - 1
list, sepby(A)
+--------------------------------------------------------+
| A B begin_yr end_yr toexpand year |
|--------------------------------------------------------|
1. | asset brown 2007 2010 4 2007 |
2. | asset brown 2007 2010 4 2008 |
3. | asset brown 2007 2010 4 2009 |
4. | asset brown 2007 2010 4 2010 |
|--------------------------------------------------------|
5. | basics caramel 2015 2015 1 2015 |
|--------------------------------------------------------|
6. | cows dork 2004 2006 3 2004 |
7. | cows dork 2004 2006 3 2005 |
8. | cows dork 2004 2006 3 2006 |
+--------------------------------------------------------+
没有反对 tsset
或 tsfill
但两者都不需要。
我有以下数据集
A B begin_yr end_yr
asset brown 2007 2010
asset blue 2008 2008
basics caramel 2015 2015
cows dork 2004 2006
我希望 A
和 B
具有代表每一年的行。
我每年扩展:
gen x = end_yr - begin_yr
expand x +1
这给了我以下信息:
A B begin_yr end_yr x
asset brown 2007 2010 3
asset brown 2007 2010 3
asset brown 2007 2010 3
asset brown 2007 2010 3
asset blue 2008 2008 0
basics caramel 2015 2015 0
cows dork 2004 2006 2
最终,我想要以下数据集:
A B begin_yr end_yr x year
asset brown 2007 2010 3 2007
asset brown 2007 2010 3 2008
asset brown 2007 2010 3 2009
asset brown 2007 2010 3 2010
asset blue 2008 2008 0 2008
basics caramel 2015 2015 0 2015
cows dork 2004 2006 2 2004
cows dork 2004 2006 2 2005
cows dork 2004 2006 2 2006
这是我目前拥有的:
gen year = begin_yr if begin_yr!=end_yr
如何填充变量的其余部分 year
?
以下对我有用:
clear
input strL A strL B begin_yr end_yr
asset brown 2007 2010
basics caramel 2015 2015
cows dork 2004 2006
end
generate id = _n
expand 2
clonevar year = begin_yr
bysort id: replace year = end_yr[2] if _n == _N
drop if _n == 3
tsset id year
tsfill
foreach var in A B begin_yr end_yr {
bysort id: replace `var' = `var'[1]
}
list
+--------------------------------------------------+
| A B begin_yr end_yr id year |
|--------------------------------------------------|
1. | asset brown 2007 2010 1 2007 |
2. | asset brown 2007 2010 1 2008 |
3. | asset brown 2007 2010 1 2009 |
4. | asset brown 2007 2010 1 2010 |
5. | basics caramel 2015 2015 2 2015 |
|--------------------------------------------------|
6. | cows dork 2004 2006 3 2004 |
7. | cows dork 2004 2006 3 2005 |
8. | cows dork 2004 2006 3 2006 |
+--------------------------------------------------+
这是在@Pearly Spencer 的代码基础上进行的转折:
clear
input strL A strL B begin_yr end_yr
asset brown 2007 2010
basics caramel 2015 2015
cows dork 2004 2006
end
gen toexpand = end - begin + 1
expand toexpand
bysort A : gen year = begin + _n - 1
list, sepby(A)
+--------------------------------------------------------+
| A B begin_yr end_yr toexpand year |
|--------------------------------------------------------|
1. | asset brown 2007 2010 4 2007 |
2. | asset brown 2007 2010 4 2008 |
3. | asset brown 2007 2010 4 2009 |
4. | asset brown 2007 2010 4 2010 |
|--------------------------------------------------------|
5. | basics caramel 2015 2015 1 2015 |
|--------------------------------------------------------|
6. | cows dork 2004 2006 3 2004 |
7. | cows dork 2004 2006 3 2005 |
8. | cows dork 2004 2006 3 2006 |
+--------------------------------------------------------+
没有反对 tsset
或 tsfill
但两者都不需要。