这一行代码不会使代码变得混乱。step参数的符号很重要。

def frange(start, stop, step):
    return [x*step+start for x in range(0,round(abs((stop-start)/step)+0.5001),
        int((stop-start)/step<0)*-2+1)]

Start和stop是包含的，而不是一个或另一个(通常不包括stop)，并且没有导入，并使用生成器

def rangef(start, stop, step, fround=5):
    """
    Yields sequence of numbers from start (inclusive) to stop (inclusive)
    by step (increment) with rounding set to n digits.

    :param start: start of sequence
    :param stop: end of sequence
    :param step: int or float increment (e.g. 1 or 0.001)
    :param fround: float rounding, n decimal places
    :return:
    """
    try:
        i = 0
        while stop >= start and step > 0:
            if i==0:
                yield start
            elif start >= stop:
                yield stop
            elif start < stop:
                if start == 0:
                    yield 0
                if start != 0:
                    yield start
            i += 1
            start += step
            start = round(start, fround)
        else:
            pass
    except TypeError as e:
        yield "type-error({})".format(e)
    else:
        pass


# passing
print(list(rangef(-100.0,10.0,1)))
print(list(rangef(-100,0,0.5)))
print(list(rangef(-1,1,0.2)))
print(list(rangef(-1,1,0.1)))
print(list(rangef(-1,1,0.05)))
print(list(rangef(-1,1,0.02)))
print(list(rangef(-1,1,0.01)))
print(list(rangef(-1,1,0.005)))
# failing: type-error:
print(list(rangef("1","10","1")))
print(list(rangef(1,10,"1")))

Python 3.6.2 (v3.6.2:5fd33b5, july 8 2017, 04:57:36) [MSC v.1900 64 . 位(AMD64)]

在'xrange([start]， stop[， step])'基础上，你可以定义一个生成器，它接受并生成你选择的任何类型(坚持支持+和<的类型):

>>> def drange(start, stop, step):
...     r = start
...     while r < stop:
...         yield r
...         r += step
...         
>>> i0=drange(0.0, 1.0, 0.1)
>>> ["%g" % x for x in i0]
['0', '0.1', '0.2', '0.3', '0.4', '0.5', '0.6', '0.7', '0.8', '0.9', '1']
>>> 

在步骤中添加可能出现错误符号的自动更正:

def frange(start,step,stop):
    step *= 2*((stop>start)^(step<0))-1
    return [start+i*step for i in range(int((stop-start)/step))]

为了精品的完整性，一个功能性的解决方案:

def frange(a,b,s):
  return [] if s > 0 and a > b or s < 0 and a < b or s==0 else [a]+frange(a+s,b,s)

增加循环的i的大小，然后在需要时减少它。

for i * 100 in range(0, 100, 10):
    print i / 100.0

编辑:老实说，我不记得为什么我认为这会在语法上工作

for i in range(0, 11, 1):
    print i / 10.0

这应该有期望的输出。