我正在学习Stata编程的书Stata programming Introduction, Second Edition。
在第4章中,generate
是一个变量,用于测试其他变量是否满足逻辑条件,代码如下:
foreach v of varlist child1-child12{
local n_school "`n_school' + inrange(`v', 1, 5)"
}
gen n_school = `n_school'
当我修改这段代码以适应我自己的数据时,
foreach v of varlist qp605_s_1-qp605_s_5 {
local n_med "`n_med' + inrange(`v', 1, 5)"
}
gen n_med = `n_med'
其中qp605_s_1
的取值范围是1到17,则Stata返回:
. foreach v of varlist qp605_s_1-qp605_s_5 {
2. local n_med "`n_med' + inrange(`v', 1, 5)"
3. }
. gen n_med = `n_med'
unknown function +inrange()
r(133);
你知道这段代码有什么问题吗?
我知道我错在哪里了
本地n_med
以+
开头,因此我将其更改为:
local n_med 0
foreach v of varlist qp605_s_1-qp605_s_5{
local n_med "`n_med' + inrange(`v', 1, 5)"
}
gen n_med = `n_med',after(qp605_s_5)
,它工作!
BTW,根据Stata Programming Introduction to Stata Programming,这种方法比先generate
一个全为零的变量,然后通过循环replace
它要快,因为replace
命令比generate
慢,所以最好避免replace
。
这是另一种方法。
* Example generated by -dataex-. For more info, type help dataex
clear
input float(var1 var2)
1 5
2 6
3 7
4 8
end
gen wanted = .
mata :
data = st_data(., "var*")
st_store(., "wanted", rowsum(data :>= 1 :& data :<= 5))
end
list
+----------------------+
| var1 var2 wanted |
|----------------------|
1. | 1 5 2 |
2. | 2 6 1 |
3. | 3 7 1 |
4. | 4 8 1 |
+----------------------+