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Re: st: Create Timeline based on Dates
From
Nick Cox <[email protected]>
To
[email protected]
Subject
Re: st: Create Timeline based on Dates
Date
Fri, 26 Oct 2012 10:18:41 +0100
On weekends:
It seems that you have Mondays to Fridays only in your data, and you
want time to run in sequence so that Monday follows Friday.
The -dow()- function returns 1 to 5 for Mondays to Fridays. A bit of
messing around yields a mapping to sequential "dates" that omit
Saturdays and Sundays. (If you want another origin, you can just shift
it.)
Again, I need a sandpit.
clear
set obs 14
gen date = mdy(10, 20, 2012) + _n
format date %td
gen dow = dow(date)
. l
+-----------------+
| date dow |
|-----------------|
1. | 21oct2012 0 |
2. | 22oct2012 1 |
3. | 23oct2012 2 |
4. | 24oct2012 3 |
5. | 25oct2012 4 |
|-----------------|
6. | 26oct2012 5 |
7. | 27oct2012 6 |
8. | 28oct2012 0 |
9. | 29oct2012 1 |
10. | 30oct2012 2 |
|-----------------|
11. | 31oct2012 3 |
12. | 01nov2012 4 |
13. | 02nov2012 5 |
14. | 03nov2012 6 |
+-----------------+
gen seqdate = (5 * (date - dow(date) - 2) / 7) + dow(date)
replace seqdate = . if inlist(dow(date), 0, 6)
. l
+---------------------------+
| date dow seqdate |
|---------------------------|
1. | 21oct2012 0 . |
2. | 22oct2012 1 13776 |
3. | 23oct2012 2 13777 |
4. | 24oct2012 3 13778 |
5. | 25oct2012 4 13779 |
|---------------------------|
6. | 26oct2012 5 13780 |
7. | 27oct2012 6 . |
8. | 28oct2012 0 . |
9. | 29oct2012 1 13781 |
10. | 30oct2012 2 13782 |
|---------------------------|
11. | 31oct2012 3 13783 |
12. | 01nov2012 4 13784 |
13. | 02nov2012 5 13785 |
14. | 03nov2012 6 . |
If you are now going to tell me that the schools have
{holidays|vacations} too, then you really need a business calendar.
On Fri, Oct 26, 2012 at 9:36 AM, Nick Cox <[email protected]> wrote:
> For problems like this I need a sandpit to play in. Here is one I made:
>
> list, sepby(id)
>
> +-------------------+
> | id date event |
> |-------------------|
> 1. | 1 13 0 |
> 2. | 1 13 0 |
> 3. | 1 14 1 |
> 4. | 1 15 0 |
> 5. | 1 17 0 |
> 6. | 1 18 1 |
> 7. | 1 19 0 |
> |-------------------|
> 8. | 2 14 0 |
> 9. | 2 15 0 |
> 10. | 2 15 1 |
> 11. | 2 17 0 |
> 12. | 2 18 0 |
> 13. | 2 19 1 |
> 14. | 2 20 0 |
> +-------------------+
>
> I see this as follows.
>
> 1. There is a date looking forward, which is (present date - previous
> event date), and is thus zero or positive
>
> 1'. There is a twist on 1: There can be multiple observations with the
> same date.
>
> 2. There is a date looking backward which is (present date - next
> event date), and is thus zero or negative
>
> 2'. As 1'.
>
> 3. The wanted date is the smaller in absolute value. If there is a tie
> in absolute value, I choose the positive value.
>
> 4. For dates before the first event, no previous date can be
> identified. But this is not a problem, as the backward date will be
> the solution for these dates.
>
> 4. For dates after the last event, no next date can be identified. But
> this is not a problem, as the forward date will be the solution for
> these dates.
>
> To get "forward dates", we just copy previous values as needed, after
> spreading each event to all dates that are the same:
>
> gen prev = date if event == 1
> bysort id date (prev) : replace prev = prev[1] if prev[1] == 1
> bysort id (date) : replace prev = prev[_n-1] if mi(prev)
> gen forward = date - prev
>
> To get "backward dates", we can use the trick of reversing time.
>
> gen negdate = -date
> gen next = date if event == 1
> bysort id negdate (next) : replace next = next[1] if next[1] == 1
> bysort id (negdate) : replace next = next[_n-1] if mi(next)
> gen backward = date - next
>
> Now can we do the comparison:
> .
> gen timeline = cond(abs(forward) <= abs(backward), forward, backward)
> sort id date
>
> list id date forw backw timeline, sepby(id)
>
> +-------------------------------------------+
> | id date forward backward timeline |
> |-------------------------------------------|
> 1. | 1 13 . -1 -1 |
> 2. | 1 13 . -1 -1 |
> 3. | 1 14 0 0 0 |
> 4. | 1 15 1 -3 1 |
> 5. | 1 17 3 -1 -1 |
> 6. | 1 18 0 0 0 |
> 7. | 1 19 1 . 1 |
> |-------------------------------------------|
> 8. | 2 14 . -1 -1 |
> 9. | 2 15 0 0 0 |
> 10. | 2 15 0 0 0 |
> 11. | 2 17 2 -2 2 |
> 12. | 2 18 3 -1 -1 |
> 13. | 2 19 0 0 0 |
> 14. | 2 20 1 . 1 |
> +-------------------------------------------+
>
> For the complication with weekends, Stata offers business calendars as
> a complete solution. I have never used them.
>
> On Fri, Oct 26, 2012 at 1:02 AM, Lisa Wang <[email protected]> wrote:
>
>> I would like to create a timeline based on some event date (ie. ...-5,
>> -4, -3, -2, -1, 0, +1, +2, +3...etc). I have different students names
>> in a variable named "as" (column 1) and also a set of dates (column
>> 2) as well as another variable 'edate' (column 3) which has the event
>> dates and . everywhere else if it didn't match with column 2. What I
>> would like to know is how to create the timeline with the event date
>> being 0 for each student.
>>
>> This is the code I have run so far:
>>
>> - bysort as: generate rank =_n
>>
>> . bysort as: generate erank = rank if date==edate
>>
>> . bysort as: egen erank_pop = min(erank)
>>
>> . bysort as: generate t = rank -erank_pop -
>>
>> There are three problems which have me now stuck.
>>
>> 1. I might have multiple observations for a particular student on the
>> same date as well. Therefore, when I run the first line of code, it's
>> already erroneous as Stata will treat it as being different dates. I
>> tried also -bysort as(date): generate rank =_n - instead but it
>> returns an error: "factor variables and time-series operators not
>> allowed".
>>
>> 2. Sometimes I have multiple event dates for a particular student - I
>> would like Stata to guess which event date the date is closer to and
>> then do the time differences from that.
>>
>> 3. The dates in column 2 have all weekdays but no weekends (as the
>> students don't need to go to school on those days), so if I do a
>> timeline then it will skip some dates (eg. -5 then to -2,-1 etc. as a
>> result of the weekend). How would I overcome this, so that it actually
>> is -3,-2,-1 etc?
>>
>>
>> Thank you so much. My email is a bit long but just wanted everyone to
>> understand what I wanted to achieve.
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