The Strange Arguments Of Range
Solution 1:
range()
takes 1 positional argument and two optional arguments, and interprets these arguments differently depending on how many arguments you passed in.
If only one argument was passed in, it is assumed to be the stop
argument, otherwise that first argument is interpreted as the start instead.
In reality, range()
, coded in C, takes a variable number of arguments. You could emulate that like this:
def foo(*params):
if 3 < len(params) < 1:
raise ValueError('foo takes 1 - 3 arguments')
elif len(params) == 1
b = params[0]
elif:
a, b = params[:2]
c = params[2] iflen(params) > 2 else 1
but you could also just swap arguments:
def range(start, stop=None, step=1):
if stop isNone:
start, stop =0, start
Solution 2:
range
does not take keyword arguments:
range(start=0,stop=10)
TypeError: range() takes no keyword arguments
it takes 1, 2 or 3 positional arguments, they are evaluated according to their number:
range(stop) # 1 argumentrange(start, stop) # 2 argumentsrange(start, stop, step) # 3 arguments
i.e. it is not possible to create a range with defined stop
and step
and default start
.
Solution 3:
def foo(first, second=None, third=1):
if secondisNone:
start, stop, step =0, first, 1else:
start, stop, step =first, second, third
Solution 4:
As you know by now the real answer is that range
is a C function which for some reason does not have the same rules of python (would be nice to know why).
People might hate me for suggesting this but I've being doing this for range since I have a terrible memory of what the order of things are. Imo this shouldn't be a problem so I'm fixing it:
range(*{'start':0,'stop':10,'step':2}.values())
the reason I made it a one liner is because I don't want to have to define a range function that needs to be defined everywhere or imported everywhere. This is pure python.
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