获取由连字符分隔的值

Question

我在dataset中有一组值，这些值都是由2000-3222和10/1-10组成的。

我想将它们分开，以便它列出2000、2001等以及10/1、10/2等，它们都在各自的行中。

在Stata或R中有什么命令可以这样做吗？

编辑：

示例数据：

input int SRNo str200 SchemeName str30 CTSNo1 str4 CTSNo2
69 "Khimji Nagar SRA Co-op.Housing Society Ltd." "467" ""
70 "Jai Bhavani CHS Ltd. (Proposed)" "7 (Pt.)" ""
71 "Shivshakti SRA CHS Ltd." "364 ‘A’" ""
72 "Shree Ram CHS Ltd. (Prop.)" "96 (Pt.) -99(Pt.)" ""
end

Answer 1

根据您的示例数据(我在其中添加了一些更多的观察以使事情更具说明性)，您需要如下内容：

clear

input int SRNo str200 SchemeName str30 CTSNo1 str4 CTSNo2
69 "Khimji Nagar SRA Co-op.Housing Society Ltd." "467" ""
70 "Jai Bhavani CHS Ltd. (Proposed)" "7 (Pt.)" ""
71 "Bhavani Housing" "12(Pt.)-21(Pt.)" ""
72 "Shivshakti SRA CHS Ltd." "364 ‘A’" ""
73 "Shree Ram CHS Ltd. (Prop.)" "96 (Pt.)- 99(Pt.)" ""
74 "Ram CHS Ltd. (Prop.)" "107 (Pt.)- 114 (Pt.)" ""
end

generate tag = 0
replace tag = 1 if strmatch(CTSNo1, "*-*")

keep if tag == 1
generate part1 = regexs(0) if regexm(CTSNo1, "([0-9]+)")
generate part2 = substr(regexs(0), 2, .) if regexm(CTSNo1, "-.*([0-9])")

local obs = _N

forvalues i = 1 / `obs' {
       local xpa = abs(real(part1[`i']) - real(part2[`i'])) + 1
       expand `xpa' if _n == `i'        
}

bysort SRNo (CTSNo1): egen interim = seq()
bysort SRNo (CTSNo1): generate NCTSNo1 = real(part1) + interim - 1

drop tag part1 part2 interim
order SRNo SchemeName CTSNo1 NCTSNo1 CTSNo2

上面的代码片段产生了所需的结果：

list

     +-----------------------------------------------------------------------------+
     | SRNo                   SchemeName                 CTSNo1   NCTSNo1   CTSNo2 |
     |-----------------------------------------------------------------------------|
  1. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        12          |
  2. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        13          |
  3. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        14          |
  4. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        15          |
  5. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        16          |
     |-----------------------------------------------------------------------------|
  6. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        17          |
  7. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        18          |
  8. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        19          |
  9. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        20          |
 10. |   71              Bhavani Housing        12(Pt.)-21(Pt.)        21          |
     |-----------------------------------------------------------------------------|
 11. |   73   Shree Ram CHS Ltd. (Prop.)      96 (Pt.)- 99(Pt.)        96          |
 12. |   73   Shree Ram CHS Ltd. (Prop.)      96 (Pt.)- 99(Pt.)        97          |
 13. |   73   Shree Ram CHS Ltd. (Prop.)      96 (Pt.)- 99(Pt.)        98          |
 14. |   73   Shree Ram CHS Ltd. (Prop.)      96 (Pt.)- 99(Pt.)        99          |
 15. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       107          |
     |-----------------------------------------------------------------------------|
 16. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       108          |
 17. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       109          |
 18. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       110          |
 19. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       111          |
 20. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       112          |
     |-----------------------------------------------------------------------------|
 21. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       113          |
 22. |   74         Ram CHS Ltd. (Prop.)   107 (Pt.)- 114 (Pt.)       114          |
     +-----------------------------------------------------------------------------+

编辑：

我上面解决方案中的forvalues循环是不必要的。另一种避免重复观察的方法是：

bysort SRNo (CTSNo1): generate xpa = abs(real(part1) - real(part2)) + 1 
expand xpa

Answer 2

假设所有值都类似于您的示例，并且变量的类型为string：

. clear

. set obs 1
number of observations (_N) was 0, now 1

. 
. generate string1 = "2000-3222"

. generate new_string1 = substr(string1, 1, 4)

. 
. generate string2 = "10/1-10"

. generate new_string2 = substr(string2, 1, 4)

. 
. list

     +-------------------------------------------+
     |   string1   new_st~1   string2   new_st~2 |
     |-------------------------------------------|
  1. | 2000-3222       2000   10/1-10       10/1 |
     +-------------------------------------------+

如果您只需要原始变量的某一部分，则此解决方案非常有用。

编辑：

使用@Nick的优秀建议：

clear
set obs 1

generate string1 = "2000-3222"
generate string2 = "10/1-10"

split string1, parse("-") generate(split_string1)
split string2, parse("/") generate(split_string2)

list

     +-----------------------------------------------------------------+
     |   string1   string2   split~11   split~12   split~21   split~22 |
     |-----------------------------------------------------------------|
  1. | 2000-3222   10/1-10       2000       3222         10       1-10 |
     +-----------------------------------------------------------------+

如您所见，这个解决方案将为string1提供两个变量，为string2提供另外两个变量，每个变量包含原始变量的两个(单独)部分。