Second argument of three mandatory

问题

I have a function that mimics range(). I am stuck at one point. I need to be able to make the first (x) and third (step) arguments optional, but the middle argument (y) mandatory. In the code below, everything works except the two commented out lines.

If I am only passing in one argument, how do I construct the function to accept the single passed in argument as the mandatory (y) argument?

I cannot do this: def float_range(x=0, y, step=1.0):

Non-default parameter cannot follow a default parameter.

def float_range(x, y, step=1.0):
    if x < y:
        while x < y:
            yield x
            x += step
    else:
        while x > y:
            yield x
            x += step


for n in float_range(0.5, 2.5, 0.5):
    print(n)

print(list(float_range(3.5, 0, -1)))

for n in float_range(0.0, 3.0):
    print(n)

# for n in float_range(3.0):
#     print(n)

Output:

0.5 1.0 1.5 2.0 [3.5, 2.5, 1.5, 0.5] 0.0 1.0 2.0

回答1:

You have to use sentinel values:

def float_range(value, end=None, step=1.0):
    if end is None:
        start, end = 0.0, value
    else:
        start = value

    if start < end:
        while start < end:
            yield start
            start += step
    else:
        while start > end:
            yield start
            start += step

for n in float_range(0.5, 2.5, 0.5):
    print(n)
#  0.5
#  1.0
#  1.5
#  2.0

print(list(float_range(3.5, 0, -1)))
#  [3.5, 2.5, 1.5, 0.5]

for n in float_range(0.0, 3.0):
    print(n)
#  0.0
#  1.0
#  2.0

for n in float_range(3.0):
    print(n)
#  0.0
#  1.0
#  2.0

By the way, numpy implements arange which is essentially what you are trying to reinvent, but it isn't a generator (it returns a numpy array)

import numpy

print(numpy.arange(0, 3, 0.5))
# [0.  0.5 1.  1.5 2.  2.5]

来源：https://stackoverflow.com/questions/51250476/second-argument-of-three-mandatory