range() for floats


Is there a range() equivalent for floats in Python?

>>> range(0.5,5,1.5)
[0, 1, 2, 3, 4]
>>> range(0.5,5,0.5)

Traceback (most recent call last):
  File "<pyshell#10>", line 1, in <module>
ValueError: range() step argument must not be zero
Asked By: Jonathan Livni



I don’t know a built-in function, but writing one like [this](https://stackoverflow.com/a/477610/623735) shouldn’t be too complicated.

def frange(x, y, jump):
  while x < y:
    yield x
    x += jump

As the comments mention, this could produce unpredictable results like:

>>> list(frange(0, 100, 0.1))[-1]

To get the expected result, you can use one of the other answers in this question, or as @Tadhg mentioned, you can use decimal.Decimal as the jump argument. Make sure to initialize it with a string rather than a float.

>>> import decimal
>>> list(frange(0, 100, decimal.Decimal('0.1')))[-1]

Or even:

import decimal

def drange(x, y, jump):
  while x < y:
    yield float(x)
    x += decimal.Decimal(jump)

And then:

>>> list(drange(0, 100, '0.1'))[-1]

[editor’s not: if you only use positive jump and integer start and stop (x and y) , this works fine. For a more general solution see here.]

Answered By: kichik

You can either use:

[x / 10.0 for x in range(5, 50, 15)]

or use lambda / map:

map(lambda x: x/10.0, range(5, 50, 15))
Answered By: Xaerxess

Eagerly evaluated (2.x range):

[x * .5 for x in range(10)]

Lazily evaluated (2.x xrange, 3.x range):

itertools.imap(lambda x: x * .5, xrange(10)) # or range(10) as appropriate


itertools.islice(itertools.imap(lambda x: x * .5, itertools.count()), 10)
# without applying the `islice`, we get an infinite stream of half-integers.
Answered By: Karl Knechtel

I used to use numpy.arange but had some complications controlling the number of elements it returns, due to floating point errors. So now I use linspace, e.g.:

>>> import numpy
>>> numpy.linspace(0, 10, num=4)
array([  0.        ,   3.33333333,   6.66666667,  10.        ])
Answered By: wim

Pylab has frange (a wrapper, actually, for matplotlib.mlab.frange):

>>> import pylab as pl
>>> pl.frange(0.5,5,0.5)
array([ 0.5,  1. ,  1.5,  2. ,  2.5,  3. ,  3.5,  4. ,  4.5,  5. ])
Answered By: Pat

i wrote a function that returns a tuple of a range of double precision floating point numbers without any decimal places beyond the hundredths. it was simply a matter of parsing the range values like strings and splitting off the excess. I use it for displaying ranges to select from within a UI. I hope someone else finds it useful.

def drange(start,stop,step):
    double_value_range = []
    while start<stop:
        a = str(start)
        start = float(str(a))
        start = start+step
    double_value_range_tuple = tuple(double_value_range)
   #print double_value_range_tuple
    return double_value_range_tuple
Answered By: chris mcinnis

using itertools: lazily evaluated floating point range:

>>> from itertools import count, takewhile
>>> def frange(start, stop, step):
        return takewhile(lambda x: x< stop, count(start, step))

>>> list(frange(0.5, 5, 1.5))
# [0.5, 2.0, 3.5]
Answered By: Ayush

A solution without numpy etc dependencies was provided by kichik but due to the floating point arithmetics, it often behaves unexpectedly. As noted by me and blubberdiblub, additional elements easily sneak into the result. For example naive_frange(0.0, 1.0, 0.1) would yield 0.999... as its last value and thus yield 11 values in total.

A bit more robust version is provided here:

def frange(x, y, jump=1.0):
    '''Range for floats.'''
    i = 0.0
    x = float(x)  # Prevent yielding integers.
    x0 = x
    epsilon = jump / 2.0
    yield x  # yield always first value
    while x + epsilon < y:
        i += 1.0
        x = x0 + i * jump
        if x < y:
          yield x

Because the multiplication, the rounding errors do not accumulate. The use of epsilon takes care of possible rounding error of the multiplication, even though issues of course might rise in the very small and very large ends. Now, as expected:

> a = list(frange(0.0, 1.0, 0.1))
> a[-1]
> len(a)

And with somewhat larger numbers:

> b = list(frange(0.0, 1000000.0, 0.1))
> b[-1]
> len(b)

The code is also available as a GitHub Gist.

Answered By: Akseli Palén

There is no such built-in function, but you can use the following (Python 3 code) to do the job as safe as Python allows you to.

from fractions import Fraction

def frange(start, stop, jump, end=False, via_str=False):
    Equivalent of Python 3 range for decimal numbers.

    Notice that, because of arithmetic errors, it is safest to
    pass the arguments as strings, so they can be interpreted to exact fractions.

    >>> assert Fraction('1.1') - Fraction(11, 10) == 0.0
    >>> assert Fraction( 0.1 ) - Fraction(1, 10) == Fraction(1, 180143985094819840)

    Parameter `via_str` can be set to True to transform inputs in strings and then to fractions.
    When inputs are all non-periodic (in base 10), even if decimal, this method is safe as long
    as approximation happens beyond the decimal digits that Python uses for printing.

    For example, in the case of 0.1, this is the case:

    >>> assert str(0.1) == '0.1'
    >>> assert '%.50f' % 0.1 == '0.10000000000000000555111512312578270211815834045410'

    If you are not sure whether your decimal inputs all have this property, you are better off
    passing them as strings. String representations can be in integer, decimal, exponential or
    even fraction notation.

    >>> assert list(frange(1, 100.0, '0.1', end=True))[-1] == 100.0
    >>> assert list(frange(1.0, '100', '1/10', end=True))[-1] == 100.0
    >>> assert list(frange('1', '100.0', '.1', end=True))[-1] == 100.0
    >>> assert list(frange('1.0', 100, '1e-1', end=True))[-1] == 100.0
    >>> assert list(frange(1, 100.0, 0.1, end=True))[-1] != 100.0
    >>> assert list(frange(1, 100.0, 0.1, end=True, via_str=True))[-1] == 100.0

    if via_str:
        start = str(start)
        stop = str(stop)
        jump = str(jump)
    start = Fraction(start)
    stop = Fraction(stop)
    jump = Fraction(jump)
    while start < stop:
        yield float(start)
        start += jump
    if end and start == stop:

You can verify all of it by running a few assertions:

assert Fraction('1.1') - Fraction(11, 10) == 0.0
assert Fraction( 0.1 ) - Fraction(1, 10) == Fraction(1, 180143985094819840)

assert str(0.1) == '0.1'
assert '%.50f' % 0.1 == '0.10000000000000000555111512312578270211815834045410'

assert list(frange(1, 100.0, '0.1', end=True))[-1] == 100.0
assert list(frange(1.0, '100', '1/10', end=True))[-1] == 100.0
assert list(frange('1', '100.0', '.1', end=True))[-1] == 100.0
assert list(frange('1.0', 100, '1e-1', end=True))[-1] == 100.0
assert list(frange(1, 100.0, 0.1, end=True))[-1] != 100.0
assert list(frange(1, 100.0, 0.1, end=True, via_str=True))[-1] == 100.0

assert list(frange(2, 3, '1/6', end=True))[-1] == 3.0
assert list(frange(0, 100, '1/3', end=True))[-1] == 100.0

Code available on GitHub

Answered By: marcotama
def Range(*argSequence):
    if len(argSequence) == 3:
        imin = argSequence[0]; imax = argSequence[1]; di = argSequence[2]
        i = imin; iList = []
        while i <= imax:
            i += di
        return iList
    if len(argSequence) == 2:
        return Range(argSequence[0], argSequence[1], 1)
    if len(argSequence) == 1:
        return Range(1, argSequence[0], 1)

Please note the first letter of Range is capital. This naming method is not encouraged for functions in Python. You can change Range to something like drange or frange if you want. The “Range” function behaves just as you want it to. You can check it’s manual here [ http://reference.wolfram.com/language/ref/Range.html ].

Answered By: Hua Congyi

Is there a range() equivalent for floats in Python?
Use this:

def f_range(start, end, step, coef=0.01):
    a = range(int(start/coef), int(end/coef), int(step/coef))
    var = []
    for item in a:
    return var
Answered By: Grigor Kolev

I helped add the function numeric_range to the package more-itertools.

more_itertools.numeric_range(start, stop, step) acts like the built in function range but can handle floats, Decimal, and Fraction types.

>>> from more_itertools import numeric_range
>>> tuple(numeric_range(.1, 5, 1))
(0.1, 1.1, 2.1, 3.1, 4.1)
Answered By: William Rusnack

I think that there is a very simple answer that really emulates all the features of range but for both float and integer. In this solution, you just suppose that your approximation by default is 1e-7 (or the one you choose) and you can change it when you call the function.

def drange(start,stop=None,jump=1,approx=7): # Approx to 1e-7 by default
  This function is equivalent to range but for both float and integer
  if not stop: # If there is no y value: range(x)
      stop= start
      start= 0
  valor= round(start,approx)
  while valor < stop:
      if valor==int(valor):
          yield int(round(valor,approx))
          yield float(round(valor,approx))
      valor += jump
  for i in drange(12):
Answered By: Jose Alavedra

Why Is There No Floating Point Range Implementation In The Standard Library?

As made clear by all the posts here, there is no floating point version of range(). That said, the omission makes sense if we consider that the range() function is often used as an index (and of course, that means an accessor) generator. So, when we call range(0,40), we’re in effect saying we want 40 values starting at 0, up to 40, but non-inclusive of 40 itself.

When we consider that index generation is as much about the number of indices as it is their values, the use of a float implementation of range() in the standard library makes less sense. For example, if we called the function frange(0, 10, 0.25), we would expect both 0 and 10 to be included, but that would yield a generator with 41 values, not the 40 one might expect from 10/0.25.

Thus, depending on its use, an frange() function will always exhibit counter intuitive behavior; it either has too many values as perceived from the indexing perspective or is not inclusive of a number that reasonably should be returned from the mathematical perspective. In other words, it’s easy to see how such a function would appear to conflate two very different use cases – the naming implies the indexing use case; the behavior implies a mathematical one.

The Mathematical Use Case

With that said, as discussed in other posts, numpy.linspace() performs the generation from the mathematical perspective nicely:

numpy.linspace(0, 10, 41)
array([  0.  ,   0.25,   0.5 ,   0.75,   1.  ,   1.25,   1.5 ,   1.75,
         2.  ,   2.25,   2.5 ,   2.75,   3.  ,   3.25,   3.5 ,   3.75,
         4.  ,   4.25,   4.5 ,   4.75,   5.  ,   5.25,   5.5 ,   5.75,
         6.  ,   6.25,   6.5 ,   6.75,   7.  ,   7.25,   7.5 ,   7.75,
         8.  ,   8.25,   8.5 ,   8.75,   9.  ,   9.25,   9.5 ,   9.75,  10.

The Indexing Use Case

And for the indexing perspective, I’ve written a slightly different approach with some tricksy string magic that allows us to specify the number of decimal places.

# Float range function - string formatting method
def frange_S (start, stop, skip = 1.0, decimals = 2):
    for i in range(int(start / skip), int(stop / skip)):
        yield float(("%0." + str(decimals) + "f") % (i * skip))

Similarly, we can also use the built-in round function and specify the number of decimals:

# Float range function - rounding method
def frange_R (start, stop, skip = 1.0, decimals = 2):
    for i in range(int(start / skip), int(stop / skip)):
        yield round(i * skip, ndigits = decimals)

A Quick Comparison & Performance

Of course, given the above discussion, these functions have a fairly limited use case. Nonetheless, here’s a quick comparison:

def compare_methods (start, stop, skip):

    string_test  = frange_S(start, stop, skip)
    round_test   = frange_R(start, stop, skip)

    for s, r in zip(string_test, round_test):
        print(s, r)

compare_methods(-2, 10, 1/3)

The results are identical for each:

-2.0 -2.0
-1.67 -1.67
-1.33 -1.33
-1.0 -1.0
-0.67 -0.67
-0.33 -0.33
0.0 0.0
8.0 8.0
8.33 8.33
8.67 8.67
9.0 9.0
9.33 9.33
9.67 9.67

And some timings:

>>> import timeit

>>> setup = """
... def frange_s (start, stop, skip = 1.0, decimals = 2):
...     for i in range(int(start / skip), int(stop / skip)):
...         yield float(("%0." + str(decimals) + "f") % (i * skip))
... def frange_r (start, stop, skip = 1.0, decimals = 2):
...     for i in range(int(start / skip), int(stop / skip)):
...         yield round(i * skip, ndigits = decimals)
... start, stop, skip = -1, 8, 1/3
... """

>>> min(timeit.Timer('string_test = frange_s(start, stop, skip); [x for x in string_test]', setup=setup).repeat(30, 1000))

>>> min(timeit.Timer('round_test = frange_r(start, stop, skip); [x for x in round_test]', setup=setup).repeat(30, 1000))

Looks like the string formatting method wins by a hair on my system.

The Limitations

And finally, a demonstration of the point from the discussion above and one last limitation:

# "Missing" the last value (10.0)
for x in frange_R(0, 10, 0.25):


Further, when the skip parameter is not divisible by the stop value, there can be a yawning gap given the latter issue:

# Clearly we know that 10 - 9.43 is equal to 0.57
for x in frange_R(0, 10, 3/7):


There are ways to address this issue, but at the end of the day, the best approach would probably be to just use Numpy.

Answered By: Greenstick

A simpler library-less version

Aw, heck — I’ll toss in a simple library-less version. Feel free to improve on it[*]:

def frange(start=0, stop=1, jump=0.1):
    nsteps = int((stop-start)/jump)
    dy = stop-start
    # f(i) goes from start to stop as i goes from 0 to nsteps
    return [start + float(i)*dy/nsteps for i in range(nsteps)]

The core idea is that nsteps is the number of steps to get you from start to stop and range(nsteps) always emits integers so there’s no loss of accuracy. The final step is to map [0..nsteps] linearly onto [start..stop].


If, like alancalvitti you’d like the series to have exact rational representation, you can always use Fractions:

from fractions import Fraction

def rrange(start=0, stop=1, jump=0.1):
    nsteps = int((stop-start)/jump)
    return [Fraction(i, nsteps) for i in range(nsteps)]

[*] In particular, frange() returns a list, not a generator. But it sufficed for my needs.

Answered By: fearless_fool

There several answers here that don’t handle simple edge cases like negative step, wrong start, stop etc. Here’s the version that handles many of these cases correctly giving same behaviour as native range():

def frange(start, stop=None, step=1):
  if stop is None:
    start, stop = 0, start
  steps = int((stop-start)/step)
  for i in range(steps):
    yield start
    start += step  

Note that this would error out step=0 just like native range. One difference is that native range returns object that is indexable and reversible while above doesn’t.

You can play with this code and test cases here.

Answered By: Shital Shah

This can be done with numpy.arange(start, stop, stepsize)

import numpy as np

>> [0.5, 2.0, 3.5, 5.0]

# OBS you will sometimes see stuff like this happening, 
# so you need to decide whether that's not an issue for you, or how you are going to catch it.
>> [0.50000001, 2.0, 3.5, 5.0]

Note 1:
From the discussion in the comment section here, “never use numpy.arange() (the numpy documentation itself recommends against it). Use numpy.linspace as recommended by wim, or one of the other suggestions in this answer”

Note 2:
I have read the discussion in a few comments here, but after coming back to this question for the third time now, I feel this information should be placed in a more readable position.

Answered By: mrk


# Counting up
drange(0, 0.4, 0.1)
[0, 0.1, 0.2, 0.30000000000000004, 0.4]

# Counting down
drange(0, -0.4, -0.1)
[0, -0.1, -0.2, -0.30000000000000004, -0.4]

To round each step to N decimal places

drange(0, 0.4, 0.1, round_decimal_places=4)
[0, 0.1, 0.2, 0.3, 0.4]

drange(0, -0.4, -0.1, round_decimal_places=4)
[0, -0.1, -0.2, -0.3, -0.4]


def drange(start, end, increment, round_decimal_places=None):
    result = []
    if start < end:
        # Counting up, e.g. 0 to 0.4 in 0.1 increments.
        if increment < 0:
            raise Exception("Error: When counting up, increment must be positive.")
        while start <= end:
            start += increment
            if round_decimal_places is not None:
                start = round(start, round_decimal_places)
        # Counting down, e.g. 0 to -0.4 in -0.1 increments.
        if increment > 0:
            raise Exception("Error: When counting down, increment must be negative.")
        while start >= end:
            start += increment
            if round_decimal_places is not None:
                start = round(start, round_decimal_places)
    return result

Why choose this answer?

  • Many other answers will hang when asked to count down.
  • Many other answers will give incorrectly rounded results.
  • Other answers based on np.linspace are hit-and-miss, they may or may not work due to difficulty in choosing the correct number of divisions. np.linspace really struggles with decimal increments of 0.1, and the order of divisions in the formula to convert the increment into a number of splits can result in either correct or broken code.
  • Other answers based on np.arange are deprecated.

If in doubt, try the four tests cases above.

Answered By: Contango

There will be of course some rounding errors, so this is not perfect, but this is what I use generally for applications, which don’t require high precision. If you wanted to make this more accurate, you could add an extra argument to specify how to handle rounding errors. Perhaps passing a rounding function might make this extensible and allow the programmer to specify how to handle rounding errors.

arange = lambda start, stop, step: [i + step * i for i in range(int((stop - start) / step))]

If I write:

arange(0, 1, 0.1)

It will output:

[0.0, 0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6000000000000001, 0.7000000000000001, 0.8, 0.9]
Answered By: MikeyE

As kichik wrote, this shouldn’t be too complicated. However this code:

def frange(x, y, jump):
  while x < y:
    yield x
    x += jump

Is inappropriate because of the cumulative effect of errors when working with floats.
That is why you receive something like:

>>>list(frange(0, 100, 0.1))[-1]

While the expected behavior would be:

>>>list(frange(0, 100, 0.1))[-1]

Solution 1

The cumulative error can simply be reduced by using an index variable. Here’s the example:

from math import ceil

    def frange2(start, stop, step):
        n_items = int(ceil((stop - start) / step))
        return (start + i*step for i in range(n_items))

This example works as expected.

Solution 2

No nested functions. Only a while and a counter variable:

def frange3(start, stop, step):
    res, n = start, 1

    while res < stop:
        yield res
        res = start + n * step
        n += 1

This function will work well too, except for the cases when you want the reversed range. E.g:

>>>list(frange3(1, 0, -.1))

Solution 1 in this case will work as expected. To make this function work in such situations, you must apply a hack, similar to the following:

from operator import gt, lt

def frange3(start, stop, step):
    res, n = start, 0.
    predicate = lt if start < stop else gt
    while predicate(res, stop):
        yield res
        res = start + n * step
        n += 1

With this hack you can use these functions with negative steps:

>>>list(frange3(1, 0, -.1))
[1, 0.9, 0.8, 0.7, 0.6, 0.5, 0.3999999999999999, 0.29999999999999993, 0.19999999999999996, 0.09999999999999998]

Solution 3

You can go even further with plain standard library and compose a range function for the most of numeric types:

from itertools import count
from itertools import takewhile

def any_range(start, stop, step):
    start = type(start + step)(start)
    return takewhile(lambda n: n < stop, count(start, step))

This generator is adapted from the Fluent Python book (Chapter 14. Iterables, Iterators and generators). It will not work with decreasing ranges. You must apply a hack, like in the previous solution.

You can use this generator as follows, for example:

>>>list(any_range(Fraction(2, 1), Fraction(100, 1), Fraction(1, 3)))[-1]
>>>list(any_range(Decimal('2.'), Decimal('4.'), Decimal('.3')))
[Decimal('2'), Decimal('2.3'), Decimal('2.6'), Decimal('2.9'), Decimal('3.2'), Decimal('3.5'), Decimal('3.8')]

And of course you can use it with float and int as well.

Be careful

If you want to use these functions with negative steps, you should add a check for the step sign, e.g.:

no_proceed = (start < stop and step < 0) or (start > stop and step > 0)
if no_proceed: raise StopIteration

The best option here is to raise StopIteration, if you want to mimic the range function itself.

Mimic range

If you would like to mimic the range function interface, you can provide some argument checks:

def any_range2(*args):
    if len(args) == 1:
        start, stop, step = 0, args[0], 1.
    elif len(args) == 2:
        start, stop, step = args[0], args[1], 1.
    elif len(args) == 3:
        start, stop, step = args
        raise TypeError('any_range2() requires 1-3 numeric arguments')

    # here you can check for isinstance numbers.Real or use more specific ABC or whatever ...

    start = type(start + step)(start)
    return takewhile(lambda n: n < stop, count(start, step))

I think, you’ve got the point. You can go with any of these functions (except the very first one) and all you need for them is python standard library.

Answered By: stolpa4

Talk about making a mountain out of a mole hill.
If you relax the requirement to make a float analog of the range function, and just create a list of floats that is easy to use in a for loop, the coding is simple and robust.

def super_range(first_value, last_value, number_steps):
    if not isinstance(number_steps, int):
        raise TypeError("The value of 'number_steps' is not an integer.")
    if number_steps < 1:
        raise ValueError("Your 'number_steps' is less than 1.")

    step_size = (last_value-first_value)/(number_steps-1)

    output_list = []
    for i in range(number_steps):
        output_list.append(first_value + step_size*i)
    return output_list

first = 20.0
last = -50.0
steps = 5

print(super_range(first, last, steps))

The output will be

[20.0, 2.5, -15.0, -32.5, -50.0]

Note that the function super_range is not limited to floats. It can handle any data type for which the operators +, -, *, and / are defined, such as complex, Decimal, and numpy.array:

import cmath
first = complex(1,2)
last = complex(5,6)
steps = 5

print(super_range(first, last, steps))

from decimal import *
first = Decimal(20)
last = Decimal(-50)
steps = 5

print(super_range(first, last, steps))

import numpy as np
first = np.array([[1, 2],[3, 4]])
last = np.array([[5, 6],[7, 8]])
steps = 5

print(super_range(first, last, steps))

The output will be:

[(1+2j), (2+3j), (3+4j), (4+5j), (5+6j)]
[Decimal('20.0'), Decimal('2.5'), Decimal('-15.0'), Decimal('-32.5'), Decimal('-50.0')]
[array([[1., 2.],[3., 4.]]),
 array([[2., 3.],[4., 5.]]),
 array([[3., 4.],[5., 6.]]),
 array([[4., 5.],[6., 7.]]),
 array([[5., 6.],[7., 8.]])]
Answered By: David Lambert

Whereas integer-based ranges are well defined in that "what you see is what you get", there are things that are not readily seen in floats that cause troubles in getting what appears to be a well defined behavior in a desired range.

There are two approaches that one can take:

  1. split a given range into a certain number of segment: the linspace approach in which you accept the large number of decimal digits when you select a number of points that does not divide the span well (e.g. 0 to 1 in 7 steps will give a first step value of 0.14285714285714285)

  2. give the desired WYSIWIG step size that you already know should work and wish that it would work. Your hopes will often be dashed by getting values that miss the end point that you wanted to hit.

Multiples can be higher or lower than you expect:

>>> 3*.1 > .3  # 0.30000000000000004

>>> 3*.3 < 0.9  # 0.8999999999999999

You will try to avoid accumulating errors by adding multiples of your step and not incrementing, but the problem will always present itself and you just won’t get what you expect if you did it by hand on paper — with exact decimals. But you know it should be possible since Python shows you 0.1 instead of the underlying integer ratio having a close approximation to 0.1:

>>> (3*.1).as_integer_ratio()
(1351079888211149, 4503599627370496)

In the methods offered as answers, the use of Fraction here with the option to handle input as strings is best. I have a few suggestions to make it better:

  1. make it handle range-like defaults so you can start from 0 automatically
  2. make it handle decreasing ranges
  3. make the output look like you would expect if you were using exact arithmetic

I offer a routine that does these same sort of thing but which does not use the Fraction object. Instead, it uses round to create numbers having the same apparent digits as the numbers would have if you printed them with python, e.g. 1 decimal for something like 0.1 and 3 decimals for something like 0.004:

def frange(start, stop, step, n=None):
    """return a WYSIWYG series of float values that mimic range behavior
    by excluding the end point and not printing extraneous digits beyond
    the precision of the input numbers (controlled by n and automatically
    detected based on the string representation of the numbers passed).


    non-WYSIWYS simple list-comprehension

    >>> [.11 + i*.1 for i in range(3)]
    [0.11, 0.21000000000000002, 0.31]

    WYSIWYG result for increasing sequence

    >>> list(frange(0.11, .33, .1))
    [0.11, 0.21, 0.31]

    and decreasing sequences

    >>> list(frange(.345, .1, -.1))
    [0.345, 0.245, 0.145]

    To hit the end point for a sequence that is divisibe by
    the step size, make the end point a little bigger by
    adding half the step size:

    >>> dx = .2
    >>> list(frange(0, 1 + dx/2, dx))
    [0.0, 0.2, 0.4, 0.6, 0.8, 1.0]

    if step == 0:
        raise ValueError('step must not be 0')
    # how many decimal places are showing?
    if n is None:
        n = max([0 if '.' not in str(i) else len(str(i).split('.')[1])
                for i in (start, stop, step)])
    if step*(stop - start) > 0:  # a non-null incr/decr range
        if step < 0:
            for i in frange(-start, -stop, -step, n):
                yield -i
            steps = round((stop - start)/step)
            while round(step*steps + start, n) < stop:
                steps += 1
            for i in range(steps):
                yield round(start + i*step, n)
Answered By: smichr

I do not know if the question is old but there is a arange function in the NumPy library, it could work as a range.



array([0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
Answered By: Maciek Woźniak
Categories: questions Tags: , , ,
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