[email protected] started a thread with
>
> I am working with panel data and I found the program
> xtpattern extremely useful.
> Basically, it creates a string variable for each
> observation with the patterns
> you would obtain by "xtdes" (describing) the data, sp say:
> 1.1..1
> indicates that a unit of observation (say, a household) is
> present on time
> periods 1,3 and 6.
> What I need now is to know how many contiguous observations
> there are, so for
> instance:
> 1.1..1 would be zero
> 111111 would be six
> 111.11 would be (perhaps) three (the max).
>
> The obvious thing would be to parse through the string
> generated by xtpattern...
> but I couldn't find an elegant way around this problem. Any
> ideas on this?
>
I had forgotten about -xtpattern-, but it's not obvious to me
that it's a good starting point. I tried using -spell-, but
couldn't get that to work... Incidentally, -spell- was
written by Richard Goldstein as well as myself.
So here's another approach.
As I understand it, Guillermo wants the length of the longest
run of consecutive observations.
Here's a made-up panel data set:
. l
panel time
1. 1 1
2. 1 2
3. 1 3
4. 1 5
5. 1 7
6. 2 1
7. 2 3
8. 2 4
9. 2 7
10. 2 8
11. 2 10
. tsset panel time
panel variable: panel, 1 to 2
time variable: time, 1 to 10, but with gaps
A spell of consecutive observations ends when F.time is missing:
. bysort panel (time) : gen gapfollows = F.time
(7 missing values generated)
. l
panel time gapfoll~s
1. 1 1 2
2. 1 2 3
3. 1 3 .
4. 1 5 .
5. 1 7 .
6. 2 1 .
7. 2 3 4
8. 2 4 .
9. 2 7 8
10. 2 8 .
11. 2 10 .
Let's chop into spells using those markers. The
white magic here is that gap[0] is also missing.
. bysort panel : gen spell = sum(gap[_n-1] == .)
. l
panel time gapfoll~s spell
1. 1 1 2 1
2. 1 2 3 1
3. 1 3 . 1
4. 1 5 . 2
5. 1 7 . 3
6. 2 1 . 1
7. 2 3 4 2
8. 2 4 . 2
9. 2 7 8 3
10. 2 8 . 3
11. 2 10 . 4
For each spell, we want the number of observations:
. bysort panel spell : gen freq = _N
. l
panel time gapfoll~s spell freq
1. 1 1 2 1 3
2. 1 2 3 1 3
3. 1 3 . 1 3
4. 1 5 . 2 1
5. 1 7 . 3 1
6. 2 1 . 1 1
7. 2 3 4 2 2
8. 2 4 . 2 2
9. 2 7 8 3 2
10. 2 8 . 3 2
11. 2 10 . 4 1
For each panel, we want the longest run:
. bysort panel (freq) : gen longestrun = freq[_N]
. l
panel time gapfoll~s spell freq longest~n
1. 1 7 . 3 1 3
2. 1 5 . 2 1 3
3. 1 1 2 1 3 3
4. 1 3 . 1 3 3
5. 1 2 3 1 3 3
6. 2 10 . 4 1 2
7. 2 1 . 1 1 2
8. 2 8 . 3 2 2
9. 2 3 4 2 2 2
10. 2 4 . 2 2 2
11. 2 7 8 3 2 2
Nick
[email protected]
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