问题
I am learning Stata and want to know how to generate random integers (without replacement). If I had 10 total rows, I would want each row to have a unique integer from 1 to 10 assigned to it. In R, one could simply do:
sample(1:10, 10)
But it seems more difficult to do in Stata. From this Stata page, I saw:
generate ui = floor((b-a+1)*runiform() + a)
If I substitute a=1 and b=10, I get something close to what I want, but it samples with replacement.
After getting that part figured out, how would I handle the following wrinkle: my data come in pairs. For example, in the 10 observations, there are 5 groups of 2. Each group of 2 has a unique identifier. How would I arrange the groups (and not the observations) in random order? The data would look something like this:
obs group mem value
1 A x 9345
2 A y 129
3 B x 251
4 B y 373
5 C x 788
6 C y 631
7 D x 239
8 D y 481
9 E x 224
10 E y 585
obs is the observation number. group is the group the observation (row) belongs to. mem is the member identifier in the group. Each group has one x and one y in it.
回答1:
First question:
You could just shuffle observation identifiers.
set obs 10
gen y = _n
gen rnd = runiform()
sort rnd
Or in Mata
jumble(1::10)
Second question: Several ways. Here's one.
gen rnd = runiform()
bysort group (rnd): replace rnd = rnd[1]
sort rnd
General comment: For reproducibility, set the random number seed beforehand.
set seed 2803
or whatever.
来源：https://stackoverflow.com/questions/24147881/stata-how-to-generate-random-integers