Range with step of type float [duplicate]_问答_开发者

This question already has answers here: How do I use a decimal step value for range()? (34 answers) Closed 6 years ago.

The documentation basically says that range must behave exactly as this implementation (for positive step):

def range(start, stop, step):
  x = start
  while True:
    if x >= stop: return
    yield x
    x += step

It also says that its arguments must be integers. Why is that? Isn't that definition also perfectly valid if step is a float?

In my case, I am esp. needing a range function which accepts a float type as its step argument. Is there any in Python or do I need to implement my own?

More specific: How would I translate this C code directly to Python in a nice way (i.e. not just doing开发者_JS百科 it via a while-loop manually):

for(float x = 0; x < 10; x += 0.5f) { /* ... */ }

You could use numpy.arange.

EDIT: The docs prefer numpy.linspace. _{Thanks @Droogans for noticing =)}

One explanation might be floating point rounding issues. For example, if you could call

range(0, 0.4, 0.1)

you might expect an output of

[0, 0.1, 0.2, 0.3]

but you in fact get something like

[0, 0.1, 0.2000000001, 0.3000000001]

due to rounding issues. And since range is often used to generate indices of some sort, it's integers only.

Still, if you want a range generator for floats, you can just roll your own.

def xfrange(start, stop, step):
    i = 0
    while start + i * step < stop:
        yield start + i * step
        i += 1

In order to be able to use decimal numbers in a range expression a cool way for doing it is the following: [x * 0.1 for x in range(0, 10)]

The problem with floating point is that you may not get the same number of items as you expected, due to inaccuracy. This can be a real problem if you are playing with polynomials where the exact number of items is quite important.

What you really want is an arithmetic progression; the following code will work quite happily for int, float and complex ... and strings, and lists ...

def arithmetic_progression(start, step, length):
    for i in xrange(length):
        yield start + i * step

Note that this code stands a better chance of your last value being within a bull's roar of the expected value than any alternative which maintains a running total.

>>> 10000 * 0.0001, sum(0.0001 for i in xrange(10000))
(1.0, 0.9999999999999062)
>>> 10000 * (1/3.), sum(1/3. for i in xrange(10000))
(3333.333333333333, 3333.3333333337314)

Correction: here's a competetive running-total gadget:

def kahan_range(start, stop, step):
    assert step > 0.0
    total = start
    compo = 0.0
    while total < stop:
        yield total
        y = step - compo
        temp = total + y
        compo = (temp - total) - y
        total = temp

>>> list(kahan_range(0, 1, 0.0001))[-1]
0.9999
>>> list(kahan_range(0, 3333.3334, 1/3.))[-1]
3333.333333333333
>>>

When you add floating point numbers together, there's often a little bit of error. Would a range(0.0, 2.2, 1.1) return [0.0, 1.1] or [0.0, 1.1, 2.199999999]? There's no way to be certain without rigorous analysis.

The code you posted is an OK work-around if you really need this. Just be aware of the possible shortcomings.

Here is a special case that might be good enough:

 [ (1.0/divStep)*x for x in range(start*divStep, stop*divStep)]

In your case this would be:

#for(float x = 0; x < 10; x += 0.5f) { /* ... */ } ==>
start = 0
stop  = 10
divstep = 1/.5 = 2 #This needs to be int, thats why I said 'special case'

and so:

>>> [ .5*x for x in range(0*2, 10*2)]
[0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5]

This is what I would use:

numbers = [float(x)/10 for x in range(10)]

rather than:

numbers = [x*0.1 for x in range(10)]
that would return :
[0.0, 0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6000000000000001, 0.7000000000000001, 0.8, 0.9]

hope it helps.

Probably because you can't have part of an iterable. Also, floats are imprecise.