Adjust date and time

Note

The Macro to Python converter is available from Metview version 5.22.0

Date/time literals and dates initialised from numbers in Macro are automatically converted to Python datetime objects:

Date/time literals converted to Python

Macro code

Generated Python code

 
d = 2019-03-21
t = 03:11:32
d1 = 2021-03-21 05:11:32
d2 = date(20220321)

d = datetime.date(2019, 3, 21)
t = datetime.time(3, 11, 32)
d1 = datetime.datetime(2021, 3, 21, 5, 11, 32)
d2 = mv.date(20220321)

However, a problem arises when it comes down to date arithmetics. In Macro we can add floating point numbers to a date (1=1 day, 0.5=12 hours etc). However, in Python only a datetime.timedelta object can be added to a date. The converter is not able to resolve this, therefore every occurrence has to be checked and manually adjusted in the generated Python code. This is true for cases when:

  • we directly add a number to a date

  • we add the result of the year(), month(), day(), hour(), minute() or second() Macro functions to a date (see function reference here)

  • we add a number stored in a variable or generated by an expression to a date

The following examples demonstrate how the adjustment can be done in various use cases:

Adding days to a date

Macro

Generated Python code

Adjusted Python code

 
d = 2019-03-21
# add 1 day
d1 = d + 1

d = datetime.date(2019, 3, 21)
# not supported in datetime
d1 = d + 1

d = datetime.date(2019, 3, 21)
# correct
d1 = d + datetime.timedelta(days=1)
Adding fractional days to a date

Macro

Generated Python code

Adjusted Python code

 
d = 2019-03-21
# add 1.5 day
d1 = d + 1.5

d = datetime.date(2019, 3, 21)
# error: not supported in datetime
d1 = d + 1.5

d = datetime.date(2019, 3, 21)
# correct
d1 = d + datetime.timedelta(days=1, hours=12)
Adding hours to a date

Macro

Generated Python code

Adjusted Python code

 
d = 2019-03-21
# add 1 hours
d1 = d + hour(1)

d = datetime.date(2019, 3, 21)
# error: not supported in datetime
d1 = d + mv.hour(1)

d = datetime.date(2019, 3, 21)
# correct
d1 = d + datetime.timedelta(hours=1)