I'm creating a program which will eventually have like 500 calls for print function, and some others too. Each of these functions will take the exact same parameters every time, like this:
print(a, end='-', sep='.')
print(b, end='-', sep='.')
print(c, end='-', sep='.')
print(..., end='-', sep='.')
Is there a way to change the default values of print function's parameters? So that I wouldn't have to type end='-', sep='.' every time?
You can define a special version of print() using functools.partial() to give it default arguments:
from functools import partial
myprint = partial(print, end='-', sep='.')
and myprint() will then use those defaults throughout your code:
myprint(a)
myprint(b)
myprint(c)
__defaults__ (which apparently doesn't exist on builtin functions anyway ...)
def myprint(*args): print(*args, end='-', sep'.') or using lambda like Lee suggested? Just curious, is this better in some way?
functools.partial codifies defining your own wrapper function into one line. It's better in that it is shorter and clearer (in my view); it signals you are still just calling print() but with default arguments.You can also use a lambda function:
my_print = lambda x: print(x, end='-', sep='-')
my_print(a)
my_print(b)
my_print(c)
my_print can only accept one argument, which makes the function far less interesting. And of course, it defeats the purpose of the sep='-' .There is also a method that allows multiple parameters and works with lambdas:
my_print = lambda *args: print(*args, end="-", sep=".")
