Oh, and there's an FAQ that is still instructive:
FAQ . Handling date information when numbers are in a continuous
sequence
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.
Reichel
8/07 How do I convert date variables into Stata elapsed
dates when the numbers run together, like 4151999?
http://www.stata.com/support/faqs/data/dateseq.html
Nick Cox
Phil is right that -substr()- is the main tool for parsing (string)
variables positionally. -split- is only for situations in which you have
parsing characters.
I once tried to think up a command that that would be a wrapper for
repeated -substr()- and had just as much flexibility and simpler syntax,
but decided that the quest was futile.
But given integer variables say -monthday- and -year-, you have other
methods open to you.
I guess that dates like 15 October and 31 December are held as 1015 and
1231. (That being so, you do not want to parse after the first
character.)
1. Concatenate -monthday- and -year- (using -egen, concat()-, for
example) and apply -date()-.
Or
2. gen dailydate = mdy(floor(monthday/100), mod(monthday, 100), year)
For the last, consider 1231. The monthly part is 12, and the daily part
is 31. The monthly part is thus floor(monthday/100) or int(monthday/100)
and the daily part is the remainder mod(monthday, 100).
Philip Ryan
the substr() string function might be useful.
Rufus
>I'm trying to convert a variable into a date. Ordinarily, this would
>be simple using Stata's date() functions, but one of my variables
>contains both month and date (i.e. 403 is April 3rd), while another
>contains year. I'm trying to split the month+date variable into two
>variables by parsing after the first character. I tried converting
>the variable to a string and using the "split" command, but it does
>not allow me to parse positionally; it merely allows me to specify a
>parsing character. Is there a way to parse by position?
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