Python函数定义，函数作为可选参数

Question

I would like to have a function as an optional argument of another function in python but it is not clear for me how I can do that.

For example I define the following function:

import os, time, datetime

def f(t=datetime.datetime.now()):
    return t.timetuple()

I have placed t=datetime.datetime.now() in order for the argument to be optional so to be able to call f() with no arguments.

Now whenever in time I execute f() I get the same datetime A (which is wrong according to what I expect), but whenever in time I execute f(datetime.datetime.now()) I get different datetimes (which is correct as expected).

For example

>>> f()
time.struct_time(tm_year=2015, tm_mon=6, tm_mday=20, tm_hour=15, tm_min=36, tm_sec=2, tm_wday=5, tm_yday=171, tm_isdst=-1)
>>> f()
time.struct_time(tm_year=2015, tm_mon=6, tm_mday=20, tm_hour=15, tm_min=36, tm_sec=2, tm_wday=5, tm_yday=171, tm_isdst=-1)
>>> f(datetime.datetime.now())
time.struct_time(tm_year=2015, tm_mon=6, tm_mday=20, tm_hour=15, tm_min=37, tm_sec=1, tm_wday=5, tm_yday=171, tm_isdst=-1)
>>> f()
time.struct_time(tm_year=2015, tm_mon=6, tm_mday=20, tm_hour=15, tm_min=36, tm_sec=2, tm_wday=5, tm_yday=171, tm_isdst=-1)

So why the fourth call returns me back to min 36 and sec 2 while the call was made before that? Why the first two calls give the same exact time even if I let plenty of time between them?

Answer 1

As mentioned by flask, the default value is evaluated when the function is parsed, so it will be set to one time.

The typical solution to this, is to not have the default a mutable value. You can do the followings:

def f(t=None):
    if not t:
        t = datetime.datetime.now()
    return t.timetuple()

BTW, for the readers' benefit, you should try to use meaningful method and variable names.

Answer 2

You can prevent the function from being evaluated by assigning the function when loading it into your second function.

import datetime

def f(t = datetime.datetime.now()):
    return t.timetuple()

def main(ff=f):
    print ff
    print ff()

>>> main()
<function f at 0x10a4e6938>
time.struct_time(tm_year=2015, tm_mon=6, tm_mday=20, tm_hour=15, tm_min=39, tm_sec=28, tm_wday=5, tm_yday=171, tm_isdst=-1)

edit: function is always evaluated in parameter. Solution: decouple it's assignment from the parameter

def f(t="now"):
    if t=="now":
        return datetime.datetime.now().timetuple()
    else:
        return t.timetuple()

def main(ff=f):
    print ff
    print ff()

import time
main()
time.sleep(3)
main()

Answer 3

Here, the optional parameter is the function for datetime now - no parentheses, as @jonsharpe was recommending. Calling f calls the default function, and calling it twice returns two different times:

>>> import datetime
>>> def f(t=datetime.datetime.now):
...     return t()
... 
>>> f()
datetime.datetime(2015, 6, 21, 0, 6, 10, 698000)

>>> f()
datetime.datetime(2015, 6, 21, 0, 6, 12, 269000)

Or you can pass in another function, here a test one, and override t . Calling f() calls the function passed in:

>>> def test():
...     return "hi"
... 
>>> f(t=test)
'hi'

Answer 4

定义函数时，默认参数值仅计算一次。