DATEDIF Function
Calculates the difference between two valid date values for the specified units of measure.
Inputs must be column references.
The first value is used as the baseline to compare the date values.
Results are calculated to the integer value that is closest to and lower than the exact total; remaining decimal values are dropped.
Wrangle vs. SQL: This function is part of Wrangle, a proprietary data transformation language. Wrangle is not SQL. For more information, see Wrangle Language.
Basic Usage
datedif(StartDate, EndDate, month)
Output: Returns the number of full months that have elapsed between StartDate
and EndDate
.
Syntax and Arguments
datedif(date1,date2,date_units)
Argument | Required? | Data Type | Description |
---|---|---|---|
date1 | Y | datetime | Starting date to compare |
date2 | Y | datetime | Ending date to compare |
date_units | Y | string | String literal representing the date units to use in the comparison |
For more information on syntax standards, see Language Documentation Syntax Notes.
date1, date2
Date values to compare using the date_units
units. If date2
> date1
, then results are positive.
Date values must be column references.
If
date1
anddate2
have a specified time zone offset, the function calculates the difference including the timezone offsets.If
date1
does not have a specified time zone butdate2
does, the function uses the local time in the same time zone asdate2
to calculate the difference. The functions returns the difference without the time zone offset.
Usage Notes:
Required? | Data Type | Example Value |
---|---|---|
Yes | String (Date column reference) | LastContactDate |
date_units
Unit of date measurement to calculate between the two valid dates.
Usage Notes:
Required? | Data Type | Example Value |
---|---|---|
Yes | String | year |
Accepted Value for date units:
year
quarter
month
dayofyear
week
day
hour
minute
second
millisecond
Examples
Dica
For additional examples, see Common Tasks.
Example - aged orders
This example illustrates how to calculate the number of days that have elapsed between the order date and today.
Function:
Item | Description |
---|---|
DATEDIF Function | Calculates the difference between two valid date values for the specified units of measure. |
TODAY Function | Derives the value for the current date in UTC time zone. You can specify a different time zone by optional parameter. |
IF Function | The |
Source:
For the orders in the following set, you want to charge interest for those ones that are older than 90 days.
OrderId | OrderDate | Amount |
---|---|---|
1001 | 1/31/16 | 1000 |
1002 | 11/15/15 | 1000 |
1003 | 12/18/15 | 1000 |
1004 | 1/15/16 | 1000 |
Transformation:
The first step is to create a column containing today's (03/03/16) date value:
Transformation Name | |
---|---|
Parameter: Formula type | Single row formula |
Parameter: Formula | TODAY() |
Parameter: New column name | 'Today' |
You can now use this value as the basis for computing the number of elapsed days for each invoice:
Transformation Name | |
---|---|
Parameter: Formula type | Single row formula |
Parameter: Formula | DATEDIF(OrderDate, Today, day) |
The age of each invoice in days is displayed in the new column. Now, you want to add a little bit of information to this comparison. Instead of just calculating the number of days, you could write out the action to undertake. Replace the above with the following:
Transformation Name | |
---|---|
Parameter: Formula type | Single row formula |
Parameter: Formula | IF((DATEDIF(OrderDate, Today, day) > 90),'Charge interest','no action') |
Parameter: New column name | 'TakeAction' |
To be fair to your customers, you might want to issue a notice at 45 days that the invoice is outstanding. You can replace the above with the following:
Transformation Name | |
---|---|
Parameter: Formula type | Single row formula |
Parameter: Formula | IF(DATEDIF(OrderDate, Today, day) > 90,'Charge interest',IF(DATEDIF(OrderDate, Today, day) > 45),'Send letter','no action')) |
Parameter: New column name | 'TakeAction' |
By using nested instances of the IF
function, you can generate multiple results in the TakeAction
column.
For the items that are over 90 days old, you want to charge 5% interest. You can do the following:
Transformation Name | |
---|---|
Parameter: Columns | Amount |
Parameter: Formula | IF(TakeAction == 'Charge interest',Amount * 1.05,Amount) |
The above sets the value in the Amount
column based on the conditional of whether the TakeAction
column value is Charge interest
. If so, apply 5% interest to the value in the Amount
column.
Results:
OrderId | OrderDate | Amount | Today | TakeAction |
---|---|---|---|---|
1001 | 1/31/16 | 1000 | 03/03/16 | no action |
1002 | 11/15/15 | 1050 | 03/03/16 | Charge interest |
1003 | 12/18/15 | 1000 | 03/03/16 | Send letter |
1004 | 1/15/16 | 1000 | 03/03/16 | Send letter |
Example - dayofyear Calculations
This example demonstrates how dayofyear
is calculated using the DATEDIF
function, specifically how leap years and leap days are handled. Below, you can see some example dates. The year 2012 was a leap year.
Source:
dateId | d1 | d2 | Notes |
---|---|---|---|
1 | 1/1/10 | 10/10/10 | Same year; no leap year |
2 | 1/1/10 | 10/10/11 | Different years; no leap year |
3 | 10/10/11 | 1/1/10 | Reverse dates of previous row |
4 | 2/28/11 | 4/1/11 | Same year; no leap year; |
5 | 2/28/12 | 4/1/12 | Same year; leap year; spans leap day |
6 | 2/29/12 | 4/1/12 | Same year; leap year; d1 = leap day |
7 | 2/28/11 | 2/29/12 | Diff years; d2 = leap day; converted to March 1 in d1 year |
Transformation:
In this case, the transform is simple:
Transformation Name | |
---|---|
Parameter: Formula type | Single row formula |
Parameter: Formula | datedif(d1,d2,dayofyear) |
Parameter: New column name | 'datedifs' |
Results:
dateId | d1 | d2 | datedifs | Notes |
---|---|---|---|---|
1 | 1/1/10 | 10/10/10 | 282 | Same year; no leap year |
2 | 1/1/10 | 10/10/11 | 282 | Different years; no leap year |
3 | 10/10/11 | 1/1/10 | -282 | Reverse dates of previous row |
4 | 2/28/11 | 4/1/11 | 32 | Same year; no leap year; |
5 | 2/28/12 | 4/1/12 | 33 | Same year; leap year; spans leap day |
6 | 2/29/12 | 4/1/12 | 32 | Same year; leap year; d1 = leap day |
7 | 2/28/11 | 2/29/12 | 1 | Diff years; d2 = leap day; converted to March 1 in d1 year |
Rows 1 - 3:
Row 1 provides the baseline calc.
In Row 2, the same days of the year are used, but the year is different by a count of 1. However, since we are computing
dayofyear
the result is the same as for Row 1.Nota
When computing
dayofyear
, the year value ford2
is converted to the year ofd1
. The difference is then computed.Row 3 represents the reversal of dates in Row 2.
Nota
Negative values for a
dayofyear
calculation indicate thatd2
occurs earlier in the calendar thand1
, ignoring year.
Rows 4 - 7: Leap years
Row 4 provides a baseline calculation for a non-leap year.
Row 5 uses the same days of year as Row 4, but the year (2012) is a leap year. Dates span a leap date (February 29). Note that the
DATEDIF
result is 1 more than the value in the previous row.Nota
When the two dates span a leap date and the year for
d1
is a leap year, then February 29 is included as part of the calculated result.Row 6 moves date 1 forward by one day, so it is now on a leap day date. Result is one less than the previous row, which also spanned leap date.
Row 7 switches the leap date to
d2
. In this case,d2
is converted to the year ofd1
. However, since it was a leap day originally, in the year ofd1
, this value is March 1. Thus, the difference between the two dates is1
.Nota
If
d2
is a leap date and the year ford1
is not a leap year, the date used in ford2
is March 1 in the year ofd1
.