Stata中R样本函数的等价物是什么
我有一个关于 Stata 的基本问题.我有 R 编程经验,但我开始了一份新工作,其中 Stata 是主要语言.我目前正在自己深入研究 Stata,有时很难理解如何做简单的事情.
I have a basic question about Stata. I have programming experience in R but I've started a new job where Stata is the main language. I'm currently diving into Stata on my own and sometimes it's hard to understand how to do simple things.
我试图获得 3 到 50 之间的 5 个随机数,但没有成功.
I've trying to get 5 random numbers between 3 and 50 but without success.
在 R 中,任何这些都可以工作:
In R, any of these would work:
floor(runif(5, min=3, max=50))
16 39 11 11 5 # output
sample(3:50, 5, replace=TRUE)
28 13 5 36 19 # output
但我不确定如何在 Stata 中执行此操作,特别是如何返回所需范围(3:50)内的随机数.任何指针将不胜感激.我找到了 runiform()
函数,但我认为我无法获得相同的输出.
But I'm not sure how to do this in Stata, specifically how to return random numbers within the desired range (3:50). Any pointers would be appreciated. I found the runiform()
function but I don't think I can get the same output.
这是您想要的吗?
set obs 5
generate rnum = runiform(3, 50)
您基本上是先创建数据集,然后生成具有所需属性的变量.
You are basically creating a dataset first and then generating a variable with the desired properties.