Reputation: 10203
I have data on two groups, A and B, that looks like this:
group key val
A 0.00 1.23
A 0.25 2.31
A 0.50 3.10
A 0.75 4.21
A 1.00 2.51
A 0.00 1.43
B 0.25 1.31
B 0.50 5.10
B 0.75 2.21
B 1.00 8.51
I would like a fourth column, call it Bval
that looks as follows:
val
for the observation in group B that shares the same key.I'm sure this can't be too hard, but I'm a total Stata newbie, have been Googling for an hour now and can't seem to figure it out.
Upvotes: 1
Views: 196
Reputation: 363
How about something like this?
clear
input str1 group key val
A 0.00 1.23
A 0.25 2.31
A 0.50 3.10
A 0.75 4.21
A 1.00 2.51
A 0.00 1.43
B 0.25 1.31
B 0.50 5.10
B 0.75 2.21
B 1.00 8.51
end
preserve
keep if group == "B"
rename val Bval
tempfile b
save `b'
restore
merge m:1 key using `b', nogenerate keep(1 3)
replace Bval = . if group == "B"
Upvotes: 2
Reputation: 37208
This works too for your example.
. clear
. input str1 group key val
group key val
1. A 0.00 1.23
2. A 0.25 2.31
3. A 0.50 3.10
4. A 0.75 4.21
5. A 1.00 2.51
6. A 0.00 1.43
7. B 0.25 1.31
8. B 0.50 5.10
9. B 0.75 2.21
10. B 1.00 8.51
11. end
. sort key group
. gen Bval = val[_n+1] if group == "A" & group[_n+1] == "B" & key == key[_n+1]
(6 missing values generated)
. l
+---------------------------+
| group key val Bval |
|---------------------------|
1. | A 0 1.23 . |
2. | A 0 1.43 . |
3. | A .25 2.31 1.31 |
4. | B .25 1.31 . |
5. | A .5 3.1 5.1 |
|---------------------------|
6. | B .5 5.1 . |
7. | A .75 4.21 2.21 |
8. | B .75 2.21 . |
9. | A 1 2.51 8.51 |
10. | B 1 8.51 . |
+---------------------------+
Upvotes: 2