还有,如果将日期型数据转化为数值型或者字符型该怎么能保证显示跟之前的日期型数据是一致的。我用你提到的类似的方法将月份型数据2012022转化为字符型的，g begin=string(year(yeartime)*10^2+month(yeartime,"%6.0f"),结果出来的是196109，不知道怎么回事啊！ http://cache.soso.com/img/img/e109.gif 你第一个问题，把日数据变成月度数据：
还是假设时间变量名为v1
gen ym=mofd(v1)
format ym %tm
如果变成年度数据；
gen Year=year(DateAnnounced)
如果变成季度数据：
yq=qofd(DateAnnounced)
From Nick Cox <<A href="mailto:njcoxstata@gmail.com">njcoxstata@gmail.com>
To statalist@hsphsun2.harvard.edu
Subject Re: st: RE: dates
Date Sun, 9 Oct 2011 14:37:00 +0100
As you say, and as I implied, "month" here is in the sense of Stata
monthly date, as returmed by -mofd()-, not month of year alone. On Sun, Oct 9, 2011 at 1:48 PM, Steven Samuels <<A href="mailto:sjsamuels@gmail.com">sjsamuels@gmail.com> wrote: > The one-step solution is very neat! I wasn't even aware of of -dofm- . But substituting the month for -mofd- won't do because information on year is missing In the original one-step formula, substitute for "visdate" any date in the visit month, e.g. the first:
> day(dofm(1 + mofd(mdy(vmonth,1,vyear)))-1)
> Steve
> On Oct 7, 2011, at 8:01 AM, Nick Cox wrote:
> In this particular problem you don't have a daily date, but just use the month instead of the result of -mofd()-.
> n.j.cox@durham.ac.uk
> Nick Cox
> Suppose -visdate- as here is a daily date variable.
> Then the length of the current month is given by the last day of the
> current month, which is given by the first day of the next month less
> day(dofm(1 + mofd(visdate)) - 1)
> In steps:
> 1. current month is mofd(visdate)
> 2. next month is 1 + mofd(visdate)
> 3. first day of next month is dofm(1 + mofd(visdate))
> 4. last day of this month is dofm(1 + mofd(visdate)) - 1
> 5. day of last day ... you got it long since.
> But I never remember most of the function names and always have to
> look them up.
> It's key _never_ to type in rules about 28/31 or leap years, because
> Stata already knows.
> On Thu, Oct 6, 2011 at 11:55 PM, Steven Samuels <<A href="mailto:sjsamuels@gmail.com">sjsamuels@gmail.com> wrote:
>> Oops! The original algorithm assigned days only from 1 to 15. The correction is below. A better version would assign days according to whether the month has 28, 29, 30, or 31 days, but I'll leave that to others.
>> Steve
>> With enough missing dates it might be better to randomly assign a day of the month, or you risk distorting the distribution of inter-visit intervals.
>> gen visdate = date(date, "YMD")
>> tempvar day
>> gen str2 `day' = string(ceil(30*runiform())) if length(date)==6
>> replace `day' = "0"+`day' if real(`day')<10
>> gen fakeday = (length(date)==6)
>> replace visdate = date(date + `day', "YMD") if length(date)==6
>> format visdate %td
>> list date visdate fakeday
>> *****************************
>> On Oct 6, 2011, at 5:46 PM, Michael Eisenberg wrote:
>> Thanks so much.
>> On Thu, Oct 6, 2011 at 8:23 AM, Nick Cox <<A href="mailto:n.j.cox@durham.ac.uk">n.j.cox@durham.ac.uk> wrote:
>>> You don't say what "without success" means precisely.
>>> "200801" does not match either date pattern. If there is no information on day of month, Stata can only return missing for a daily date.
>>> -date("200801" + "15", "YMD")- seems to be the most common fudge. I would always tag such guessed dates with an indicator variable.
>>> n.j.cox@durham.ac.uk
>>> Michael Eisenberg
>>> I have a list of visit dates for patients. Unfortunately, the format
>>> is not constant.
>>> Most are listed with the year, month, day such as 20080105 for Jan 5,
>>> 2008 but some are listed only with the year and month 200801 for Jan
>>> 2008.
>>> I attempted to convert them into stata dates with the commands below
>>> without success.
>>> gen ndate = date(dx_date, "YMD")
>>> gen ndate = date(dx_date, "CCYYNNDD")
>>> Can stata handle such inconsistent data? gen visdate = date(date, "YMD")
tempvar day
gen str2 `day' = string(ceil(30*runiform())) if length(date)==6
replace `day' = "0"+`day' if real(`day')<10
gen fakeday = (length(date)==6)
replace visdate = date(date + `day', "YMD") if length(date)==6
format visdate %td
list date visdate fakeday gen ym=mofd(visdate)
format ym %tm
Stata has many tools for working with dates. This article will introduce you to some of the most useful and easy to use features.
A Stata date is simply a number, but with the %td format applied Stata will interpret that number as "number of days since January 1, 1960." You can then use that number in a variety of ways. Stata has similar tools that measure time in terms of milliseconds, months, quarters, years and more. This article will focus on days, but if you know how to work with days you can quickly learn the others.
Often the first task is to convert the data you've been given into official Stata dates. Converting Strings to Dates If you've been given a date in string form, such as "November 3, 2010", "11/3/2010" or "2010-11-03 08:35:12" it can be converted using the date function. The date function takes two arguments, the string to be converted, and a series of letters called a "mask" that tells Stata how the string is structured. In a date mask, Y means year, M means month, D means day and # means an element should be skipped.
Thus the mask MDY means "month, day, year" and can be used to convert both "November 3, 2010" and "11/3/2010". A date like "2010-11-03 08:35:12" requires the mask YMD### so that the last three numbers are skipped. If you are interested in tracking the time of day you need to switch to the clock function and the %tc format so time is measured in milliseconds rather than days, but they are very similar.
To see this in action, type (or copy and paste) the following into Stata:
use http://www.ssc.wisc.edu/sscc/pubs/files/dates.dta
This is an example data set containing the above dates as dateString1 , dateString2 and dateString3 . To convert them to Stata dates do the following:
gen date1=date(dateString1,"MDY")
gen date2=date(dateString2,"MDY")
gen date3=date(dateString3,"YMD###")
Note that the mask goes in quotes. Converting Numbers to Dates Another common scenario gives you dates as three separate numeric variables, one for the year, one for the month and one for the day. The year , month and day variables in the example data set contain the same date as the others but in this format. To convert such dates to Stata dates, use the mdy function. It takes three numeric arguments: the month, day and year to be converted.
gen date4=mdy(month,day,year) Formatting Date Variables While the four date variables you've created are perfectly functional dates as far as Stata is concerned, they're difficult for humans to interpret. However, the %td format tells Stata to print them out as human readable dates:
format date1 %td
format date2 %td
format date3 %td
format date4 %td
This turns the 18569 now stored in all four variables into 03nov2010 (18,569 days since January 1, 1960) in all output. Try a list to see the result. If you remember your varlist syntax , you can do them all at once with:
format date? %td
You can have Stata output dates in different formats as well. For instructions type help dates and then click on the link Formatting date and time values . Using Dates Often your goal in creating a Stata date will be to create a time variable that can be included in a statistical command. If so, you can probably use it with no further modification. However, there are some common data preparation tasks involving dates. Date Constants If you need to refer to a particular date in your code, then in principle you could refer to it by number. However, it's usually more convenient to use the same functions used to import date variables. For example, the following are all equivalent ways of referring to November 3, 2010:
18569
date("November 3, 2010","MDY")
mdy(11,3,2010)
The td pseudofunction was designed for tasks like this and is somewhat more convenient to use. It takes a single argument (which cannot be a variable name) and converts it to a date on the assumption that the argument is a string containing a date in the format day, month, year. This matches the output of the %td format, e.g. 3nov2010 . Thus the following is also equivalent:
td(3nov2010)
However, the following is not:
td(11/3/2010)
This will be interpreted as March 11, 2010, not November 3, 2010. Extracting Date Components Sometimes you need to pull out the components of a date. You can do so with the year , month and day functions:
gen year1=year(date1)
gen month1=month(date1)
gen day1=day(date1) Since dates are just numbers, before and after are equivalent to less than and greater than. Thus:
gen before2010=(date1 Durations in days can be found using simple subtraction. The example data set contains the dates beginning and ending , and you can find out the duration of the interval between them with:

gen duration=ending-beginning

Durations in months are more difficult because months vary in length. One common approach is to ignore days entirely and calculate the duration solely from the year and month components of the dates involved:

durationInMonths=(year(ending)-year(beginning))*12+month(ending)-month(beginning)

Just keep in mind that this approach says January 31 and February 1 are one month apart, while January 1 and January 31 are zero months apart.

Date Arithmetic If you need to add (or subtract) a period measured in days to a date, it is straightforward to do so. Just remember to format all new date variables as dates with %td :

gen tenDaysLater=date1+10
gen yesterday=date1-1
format %td tenDaysLater yesterday

If the period is measured in weeks, just multiply by 7. Months are again problematic since different months have different lengths. Years have the same problem if you need to be precise enough to care about leap years.

You can avoid this by building a new date based on the components of the old one, modified as required. The only trick is that you must handle year changes properly. For example, the following works properly:

gen oneMonthLater=mdy(month(date1)+1,day(date1),year(date1))
format %td oneMonthLater

oneMonthLater is now December 3, 2010. But the following does not:

twoMonthsLaterBad=mdy(month(date1)+2,day(date1),year(date1))
format %td twoMonthsLaterBad

This tries to set the month component of the new date to 13, which is invalid. It needs to be January of the next year instead. The following code will do allow you to add or subtract any number of months (just change the final number in the first line and the name of the new variable):

gen newMonth=month(date1)+2
gen newYear=year(date1)+floor((newMonth-1)/12)
replace newMonth=mod((newMonth-1),12)+1
gen twoMonthsLater=mdy(newMonth,day(date1),newYear)
format %td twoMonthsLater
drop newMonth newYear

If you need to do such things frequently you might want to turn this bit of code into a program, or even an ado file.

Learning More To read the full documentation on Stata dates, type help dates and then click on the dates and times link at the top (the PDF documentation is much easier to read in this case). There you'll learn to:

Work with times Use intervals other than days, such as months, quarters or years Create your own date format for output (e.g. November 3rd, 2010 rather than 3nov2010 ) Track leap seconds, in case you need to be extremely precise--you'll also find an explanation of why such things exist Last Revised: 11/9/2010

http://dss.princeton.edu/online_help/stats_packages/stata/time_series_data.htm

Time Series Data in Stata series data and tsset formats in Stata series variable lists Time series (guide) Time series data and tsset To use Stata's time-series functions and analyses, you must first make sure that your data are, indeed, time-series. First, you must have a date variable that is in Stata date format. Secondly, you must make sure that your data are sorted by this date variable. If you have panel data, then your data must be sorted by the date variable within the variable that identifies the panel. Finally, you must use the tsset command to tell Stata that your data are time-series:

sort datevar tsset datevar sort panelvar datevar tsset panelvar datevar The first example tells Stata that you have simple time-series data, and the second tells Stata that you have panel data.

Stata Date Format functions for single string variables functions for partial date variables Converting a date variable stored as a single number series date formats For additional tips click here Stata stores dates as the number of elapsed days since January 1, 1960. There are different ways to create elapsed Stata dates that depend on how dates are represented in your data. If your original dataset already contains a single date variable, then use the date() function or one of the other string-date commands. If you have separate variables storing different parts of the date (month, day and year; year and quarter, etc.) then you will need to use the partial date variable functions.

Date functions for a single string date variable Sometimes, your data will have the dates in string format. (A string variable is simply a variable containing anything other than just numbers.) Stata provides a way to convert these to time-series dates. The first thing you need to know is that the string must be easily separated into its components. In other words, strings like "01feb1990" "February 1, 1990" "02/01/90" are acceptable, but "020190" is not.

For example, let's say that you have a string variable "sdate" with values like "01feb1990" and you need to convert it to a daily time-series date:

gen daily=date(sdate,"DMY") Note that in this function, as with the other functions to convert strings to time-series dates, the "DMY" portion indicates the order of the day, month and year in the variable. Had the values been coded as "February 1, 1990" we would have used "MDY" instead. What if the original date only has two digits for the year? Then we would use:

gen daily=date(sdate,"DM19Y") Whenever you have two digit years, simply place the century before the "Y." If you have the last two digit years mixed, such as 1/2/98 and 1/2/00, use:

gen daily=date(sdate,"DMY",2020) where 2020 is the largest year you have in your data set. Here are the other functions:

weekly(stringvar,"wy")
monthly(stringvar,"my")
quarterly(stringvar,"qy")
halfyearly(stringvar,"hy")
yearly(stringvar,"y")

Note: Stata 10 uses upper case letters as DMY whereas earlier version of Stata uses lower case, dmy.

Date functions for partial date variables Often you will have separate variables for the various components of the date; you need to put them together before you can designate them as proper time-series dates. Stata provides an easy way to do this with numeric variables. If you have separate variables for month, day and year then use the mdy() function to create an elapsed date variable. Once you have created an elapsed date variable, you will probably want to format it, as described below.

Use the mdy() function to create an elapsed Stata date variable when your original data contains separate variables for month, day and year. The month, day and year variables must be numeric. For example, suppose you are working with these data:

where mydate is an elapsed date varible, mdy() is the Stata function, and month, day, and year are the names of the variables that contain data for month, day and year, respectively.

If you have two variables, "year" and "quarter" use the "yq()" function:

gen qtr=yq(year,quarter) gen qtr=yq(1990,3) The other functions are:

Converting a date variable stored as a single number If you have a date variable where the date is stored as a single number of the form yyyymmdd (for example, 20041231 for December 31, 2004) the following set of functions will convert it into a Stata elapsed date.

gen year = int(date/10000)
gen month = int((date-year*10000)/100)
gen day = int((date-year*10000-month*100))
gen mydate = mdy(month,day,year)
format mydate %d

Time series date formats Use the format command to display elapsed Stata dates as calendar dates. In the example given above, the elapsed date variable, mydate, has the following values, which represent the number of days before or after January 1, 1960.

You can use the format command to display elapsed dates in a more customary way. For example:

format mydate %d where mydate is an elapsed date variable and %d is the format which will be used to display values for that variable.

Time-series dates in Stata have their own formats similar to regular date formats. The main difference is that for a regular date format a "unit" or single "time period" is one day. For time series formats, a unit or single time period can be a day, week, month, quarter, half-year or year. There is a format for each of these time periods:

You should note that in the weekly format, the year is divided into 52 weeks. The first week is defined as the first seven days, regardless of what day of the week it may be. Also, the last week, week 52, may have 8 or 9 days. For the quarterly format, the first quarter is January through March. For the half-yearly format, the first half of the year is January through June.

It's even more important to note that you cannot jump from one format to another by simply re-issuing the format command because the units are different in each format. Here are the corresponding results for January 1, 1999, which is an elapsed date of 14245:

These dates are so different because the elapsed date is actually the number of weeks, quarters, etc., from the first week, quarter, etc of 1960. The value for %ty is missing because it would be equal to the year 14,245 which is beyond what Stata can accept.

Any of these time units can be translated to any of the others. Stata provides functions to translate any time unit to and from %td daily units, so all that is needed is to combine these functions.

These functions translate to %td dates:

Often we need to consuct a particular analysis only on observations that fall on a certain date. To do this, we have to use something called a date literal. A date literal is simply a way of entering a date in words and have Stata automatically convert it to an elapsed date. As with the d() literal to specify a regular date, there are the w(), m(), q(), h(), and y() literals for entering weekly, monthly, quarterly, half-yearly, and yearly dates, respectively. Here are some examples:

reg x y if w(1995w9) sum income if q(1988-3) tab gender if y(1999) If you want to specify a range of dates, you can use the tin() and twithin() functions:

reg y x if tin(01feb1990,01jun1990) sum income if twithin(1988-3,1998-3) The difference between tin() and twithin() is that tin() includes the beginning and end dates, whereas twithin() excludes them. Always enter the beginning date first, and write them out as you would for any of the d(), w(), etc. functions.

Time Series Variable Lists Often in time-series analyses we need to "lag" or "lead" the values of a variable from one observation to the next. If we have many variables, this can be cumbersome, especially if we need to lag a variable more than once. In Stata, we can specify which variables are to be lagged and how many times without having to create new variables, thus saving alot of disk space and memory. You should note that the tsset command must have been issued before any of the "tricks" in this section will work. Also, if you have defined your data as panel data, Stata will automatically re-start the calculations as it comes to the beginning of a panel so you need not worry about values from one panel being carried over to the next.

L.varname and F.varname If you need to lag or lead a variable for an analysis, you can do so by using the L.varname (to lag) and F.varname (to lead). Both work the same way, so we'll just show some examples with L.varname. Let's say you want to regress this year's income on last year's income:

reg income L.income would accomplish this. The "L." tells Stata to lag income by one time period. If you wanted to lag income by more than one time period, you would simply change the L. to something like "L2." or "L3." to lag it by 2 and 3 time periods, respectively. The following two commands will produce the same results:

reg income L.income L2.income L3.income
reg income L(1/3).income D.varname Another useful shortcut is D.varname, which takes the difference of income in time 1 and income in time 2. For example, let's say a person earned $20 yesterday and $30 today.

So, you can see that D.=(income-income _t-1 ) and D2=(income-income _t-1 )-(income _t-1 -income _t-2 )

S.varname S.varname refers to seasonal differences and works like D.varname, except that the difference is always taken from the current observation to the n ^th observation:

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