I'm trying to make a new variable which will be the age at a time point (clinic date and dob). I have used this syntax to get a numeric age: compute agechr=(aw2 - dob) / (365.25 * time.days(1)).

However, I need the format of the age in years, months and days between the two dates. So I made 3 new variables from each date variable made up of the day, month and year. I then minused these variables from each other to find the time between. I thought I'd cracked it until realising I had minus answers from dates that had months or days smaller for the clinic date e.g. clinic in janurary and dob in july (1-7=-6). 

The chronological age variable I need assumes that all months have 30 days in them. So I thought maybe I could add 30 days to the day variable and minus 1 for the month when the day is a minus answer, or, add 12 to the month variable and minus 1 for the year when the month is a minus answer. Does anyone know how I would write the syntax to get this to work?... or any other suggestions are welcome.

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Beth
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#SPSSStatistics

Hi, Bethan. Consider the DATEDIFF function. Here is an excerpt from the Command Syntax Reference manual:

Date differences

The DATEDIFF function calculates the difference between two date values and returns an integer (with any fraction component truncated) in the specified date/time units. The general form of the expression is DATEDIFF(datetime2, datetime1, "unit") where datetime2 and datetime1 are both date or time format variables (or numeric values that represent valid date/time values), and "unit" is one of the following string literal values, enclosed in quotes:

• Years
• Quarters
• Months
• Weeks
• Days
• Hours
• Minutes
• Seconds
  

Example 1

DATA LIST FREE /date1 date2 (2ADATE10).
BEGIN DATA
1/1/2004 2/1/2005 1/1/2004 2/15/2005 1/30/2004 1/29/2005
END DATA.
COMPUTE years=DATEDIFF(date2, date1, "years").


• The result will be the integer portion of the number of years between the two dates, with any fractional component truncated.
• One "year" is defined as the same month and day, one year before or after the second date argument.
• For the first two cases, the result is 1, since in both cases the number of years is greater than or equal to 1 and less than 2.
• For the third case, the result is 0, since the difference is one day short of a year based on the definition of year.

Example 2

DATA LIST FREE /date1 date2 (2ADATE10).
BEGIN DATA
1/1/2004 2/1/2004 1/1/2004 2/15/2004 1/30/2004 2/1/2004
END DATA.
COMPUTE months=DATEDIFF(date2, date1, "months").

• The result will be the integer portion of the number of months between the two dates, with any fractional component truncated.
• One "month" is defined as the same day of the month, one calendar month before or after the second date argument.

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Rick Marcantonio
Quality Assurance
IBM
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Hello

you can use DATEdiff function: 

compute agechr=datediff(aw2, dob, 'days'). 

agechr is the days between aw2 and dob.

best regards,

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xq
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DATEDIFF will give you the exact difference in the requested units with truncation.  So that matches the age calculation that applies if you are trying to buy a drink or get a driver's license.  However, that is not itself a date: it is a duration and, therefore, would not have a date format.

If you want a datelike form, you can compute the parts using the mod function, e.g.,
compute years = mod(delta, 365).
compute resid1 = delta - years * 365.
compute months = mod(resid1, 31).
etc.
where delta is the datediff value with "days" as the unit.
But you have to decide how long a month is.  DATEDIFF takes the calendar into account, but the subsequent calculations use a fixed number of days for years and months.

You might also be interested in the DATESUM function, which adds a fixed number of units - days, months, whatever, to a date.  It takes into account the actual unit lengths according to the calendar but still uses a constant for the number of units to add.

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Jon Peck
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Original Message:
Sent: Tue November 30, 2021 10:16 AM
From: xiaoqin Oezener-Wan
Subject: Chronological age

Hello

you can use DATEdiff function: 

compute agechr=datediff(aw2, dob, 'days'). 

agechr is the days between aw2 and dob.

best regards,

------------------------------
xq
------------------------------

Original Message:
Sent: 11/30/2021 12:11:00 PM
From: Jon Peck
Subject: RE: Chronological age

DATEDIFF will give you the exact difference in the requested units with truncation.  So that matches the age calculation that applies if you are trying to buy a drink or get a driver's license.  However, that is not itself a date: it is a duration and, therefore, would not have a date format.

If you want a datelike form, you can compute the parts using the mod function, e.g.,
compute years = mod(delta, 365).
compute resid1 = delta - years * 365.
compute months = mod(resid1, 31).
etc.
where delta is the datediff value with "days" as the unit.
But you have to decide how long a month is.  DATEDIFF takes the calendar into account, but the subsequent calculations use a fixed number of days for years and months.

You might also be interested in the DATESUM function, which adds a fixed number of units - days, months, whatever, to a date.  It takes into account the actual unit lengths according to the calendar but still uses a constant for the number of units to add.

------------------------------
Jon Peck
------------------------------

Even though this thread is a few years old, it still ranks highly for SPSS date-math queries. If anyone else lands here trying to get the exact "Years, Months, and Days" breakdown without their syntax breaking on leap years or unequal month lengths, you don't actually have to write complex MOD formulas.

You can cascade DATEDIFF and DATESUM together in SPSS to let the system dynamically handle the calendar math:

COMPUTE years = DATEDIFF(aw2, dob, "years").
COMPUTE temp_date = DATESUM(dob, years, "years").
COMPUTE months = DATEDIFF(aw2, temp_date, "months").
COMPUTE temp_date2 = DATESUM(temp_date, months, "months").
COMPUTE days = DATEDIFF(aw2, temp_date2, "days").
EXECUTE.

This respects the exact length of each calendar month. When writing scripts like this for clinical datasets, I usually test a few complex boundary dates (like leap days) against an external tool like calculate age to make sure the SPSS output matches real-world chronological dates perfectly.

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Mustafa Al-Harbi
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