How do I use a decimal step value for range()?

Question

How do I iterate between 0 and 1 by a step of 0.1?

This says that the step argument cannot be zero:

for i in range(0, 1, 0.1):
    print(i)

Answer 1

This will behave like range but for floating step.

sign = lambda x: (1, -1)[x < 0]
def frange(start, stop, step):
    i = 0
    r=len(str(step).split('.')[-1])
    args=(start,stop,step)
    if not step :return []
    if all(int(i)==float(i) for i in args):
        start,stop,step=map(int,args)
    if sign(step)==1:
        while start + i * step < stop:
            yield round(start + i * step,r)
            i += 1
    else:
        while start + i * step > stop:
            yield round(start + i * step,r)
            i += 1

Answer 2

Pure python generator version of numpy's linspace, specifying the number of points instead of step size:

def linspace(start, end, n):
    step = (end - start) / (n-1)
    return (start + i*step for i in range(n))

Doesn't suffer from rounding error from repeated summation.

Answer 3

np.linspace is a popular choice, however, its numerical stability is less than ideal, resulting in values being too far away from the correct decimal value, which often causes unexpected behaviours.

Here are a couple of alternative implementations of linspace function, coded with a goal to minimize numerical instabilities:

import numpy as np
import pandas as pd


def linspace1(first, last, n_points):
    """
    Interpolate n_points points from first to last inclusively.
    """
    points = np.arange(n_points)
    return (first * points[::-1] + last * points) / (n_points - 1)


def linspace1_gen(first, last, n_points):
    """
    Interpolate n_points points from first to last inclusively.
    """
    last_point = n_points - 1
    for point in range(n_points):
        yield (first * (last_point - point) + last * point) / last_point


def test_compare_linspace():
    beg, end, steps = -1, 1, 21
    s = pd.DataFrame({
        "basic": np.arange(beg * 10, end * 10 + 1) / 10,
        "np.linspace": np.linspace(beg, end, steps),
        "linspace1": linspace1(beg, end, steps),
        "linspace1_gen": np.fromiter(linspace1_gen(beg, end, steps), dtype=float),
    })
    with pd.option_context("display.float_format", "{:.18f}".format):
        print("values:", s, "", "sum error:", s.sum(), sep='\n')

Outputs:

values:
                   basic           np.linspace             linspace1         linspace1_gen
0  -1.000000000000000000 -1.000000000000000000 -1.000000000000000000 -1.000000000000000000
1  -0.900000000000000022 -0.900000000000000022 -0.900000000000000022 -0.900000000000000022
2  -0.800000000000000044 -0.800000000000000044 -0.800000000000000044 -0.800000000000000044
3  -0.699999999999999956 -0.699999999999999956 -0.699999999999999956 -0.699999999999999956
4  -0.599999999999999978 -0.599999999999999978 -0.599999999999999978 -0.599999999999999978
5  -0.500000000000000000 -0.500000000000000000 -0.500000000000000000 -0.500000000000000000
6  -0.400000000000000022 -0.399999999999999911 -0.400000000000000022 -0.400000000000000022
7  -0.299999999999999989 -0.299999999999999933 -0.299999999999999989 -0.299999999999999989
8  -0.200000000000000011 -0.199999999999999956 -0.200000000000000011 -0.200000000000000011
9  -0.100000000000000006 -0.099999999999999978 -0.100000000000000006 -0.100000000000000006
10  0.000000000000000000  0.000000000000000000  0.000000000000000000  0.000000000000000000
11  0.100000000000000006  0.100000000000000089  0.100000000000000006  0.100000000000000006
12  0.200000000000000011  0.200000000000000178  0.200000000000000011  0.200000000000000011
13  0.299999999999999989  0.300000000000000044  0.299999999999999989  0.299999999999999989
14  0.400000000000000022  0.400000000000000133  0.400000000000000022  0.400000000000000022
15  0.500000000000000000  0.500000000000000000  0.500000000000000000  0.500000000000000000
16  0.599999999999999978  0.600000000000000089  0.599999999999999978  0.599999999999999978
17  0.699999999999999956  0.700000000000000178  0.699999999999999956  0.699999999999999956
18  0.800000000000000044  0.800000000000000044  0.800000000000000044  0.800000000000000044
19  0.900000000000000022  0.900000000000000133  0.900000000000000022  0.900000000000000022
20  1.000000000000000000  1.000000000000000000  1.000000000000000000  1.000000000000000000

sum error:
basic           0.000000000000000000
np.linspace     0.000000000000000888
linspace1       0.000000000000000000
linspace1_gen   0.000000000000000000

In the above output, values generated by np.linspace are further away from the correct decimal value than one unit of least precision, and these errors accumulate, rather than cancel out.

---

A couple of neat side-effects of implementing linspace1 and linspace1_gen as linear combinations of first and last:

1. No order requirement for first <= last.
2. first and last can be vectors, coordinates of points in N-dimensional space. Scalar inputs are coordinates of points in 1D-space, from this perspective.

Answer 4

range() can only do integers, not floating point.

Use a list comprehension instead to obtain a list of steps:

[x * 0.1 for x in range(0, 10)]

More generally, a generator comprehension minimizes memory allocations:

xs = (x * 0.1 for x in range(0, 10))
for x in xs:
    print(x)

Answer 5

Rather than using a decimal step directly, it's much safer to express this in terms of how many points you want. Otherwise, floating-point rounding error is likely to give you a wrong result.

Use the linspace function from the NumPy library (which isn't part of the standard library but is relatively easy to obtain). linspace takes a number of points to return, and also lets you specify whether or not to include the right endpoint:

>>> np.linspace(0,1,11)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9,  1. ])
>>> np.linspace(0,1,10,endpoint=False)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

If you really want to use a floating-point step value, use numpy.arange:

>>> import numpy as np
>>> np.arange(0.0, 1.0, 0.1)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

Floating-point rounding error will cause problems, though. Here's a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers:

>>> numpy.arange(1, 1.3, 0.1)
array([1. , 1.1, 1.2, 1.3])

Answer 6

f = lambda x,y,z: (x+i*z for i in range(int((y-x)/z)))

Above can do fractional step without resorting to any library.

Answer 7

To counter the float precision issues, you could use the Decimal module.

This demands an extra effort of converting to Decimal from int or float while writing the code, but you can instead pass str and modify the function if that sort of convenience is indeed necessary.

from decimal import Decimal


def decimal_range(*args):

    zero, one = Decimal('0'), Decimal('1')

    if len(args) == 1:
        start, stop, step = zero, args[0], one
    elif len(args) == 2:
        start, stop, step = args + (one,)
    elif len(args) == 3:
        start, stop, step = args
    else:
        raise ValueError('Expected 1 or 2 arguments, got %s' % len(args))

    if not all([type(arg) == Decimal for arg in (start, stop, step)]):
        raise ValueError('Arguments must be passed as <type: Decimal>')

    # neglect bad cases
    if (start == stop) or (start > stop and step >= zero) or \
                          (start < stop and step <= zero):
        return []

    current = start
    while abs(current) < abs(stop):
        yield current
        current += step

Sample outputs -

from decimal import Decimal as D

list(decimal_range(D('2')))
# [Decimal('0'), Decimal('1')]
list(decimal_range(D('2'), D('4.5')))
# [Decimal('2'), Decimal('3'), Decimal('4')]
list(decimal_range(D('2'), D('4.5'), D('0.5')))
# [Decimal('2'), Decimal('2.5'), Decimal('3.0'), Decimal('3.5'), Decimal('4.0')]
list(decimal_range(D('2'), D('4.5'), D('-0.5')))
# []
list(decimal_range(D('2'), D('-4.5'), D('-0.5')))
# [Decimal('2'),
#  Decimal('1.5'),
#  Decimal('1.0'),
#  Decimal('0.5'),
#  Decimal('0.0'),
#  Decimal('-0.5'),
#  Decimal('-1.0'),
#  Decimal('-1.5'),
#  Decimal('-2.0'),
#  Decimal('-2.5'),
#  Decimal('-3.0'),
#  Decimal('-3.5'),
#  Decimal('-4.0')]

Answer 8

Best Solution: no rounding error

>>> step = .1
>>> N = 10     # number of data points
>>> [ x / pow(step, -1) for x in range(0, N + 1) ]

[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

Or, for a set range instead of set data points (e.g. continuous function), use:

>>> step = .1
>>> rnge = 1     # NOTE range = 1, i.e. span of data points
>>> N = int(rnge / step
>>> [ x / pow(step,-1) for x in range(0, N + 1) ]

[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

To implement a function: replace x / pow(step, -1) with f( x / pow(step, -1) ), and define f.
For example:

>>> import math
>>> def f(x):
        return math.sin(x)

>>> step = .1
>>> rnge = 1     # NOTE range = 1, i.e. span of data points
>>> N = int(rnge / step)
>>> [ f( x / pow(step,-1) ) for x in range(0, N + 1) ]

[0.0, 0.09983341664682815, 0.19866933079506122, 0.29552020666133955, 0.3894183423086505, 
 0.479425538604203, 0.5646424733950354, 0.644217687237691, 0.7173560908995228,
 0.7833269096274834, 0.8414709848078965]

Answer 9

Lots of the solutions here still had floating point errors in Python 3.6 and didnt do exactly what I personally needed.

Function below takes integers or floats, doesnt require imports and doesnt return floating point errors.

def frange(x, y, step):
    if int(x + y + step) == (x + y + step):
        r = list(range(int(x), int(y), int(step)))
    else:
        f = 10 ** (len(str(step)) - str(step).find('.') - 1)
        rf = list(range(int(x * f), int(y * f), int(step * f)))
        r = [i / f for i in rf]

    return r

Answer 10

I know I'm late to the party here, but here's a trivial generator solution that's working in 3.6:

def floatRange(*args):
    start, step = 0, 1
    if len(args) == 1:
        stop = args[0]
    elif len(args) == 2:
        start, stop = args[0], args[1]
    elif len(args) == 3:
        start, stop, step = args[0], args[1], args[2]
    else:
        raise TypeError("floatRange accepts 1, 2, or 3 arguments. ({0} given)".format(len(args)))
    for num in start, step, stop:
        if not isinstance(num, (int, float)):
            raise TypeError("floatRange only accepts float and integer arguments. ({0} : {1} given)".format(type(num), str(num)))
    for x in range(int((stop-start)/step)):
        yield start + (x * step)
    return

then you can call it just like the original range()... there's no error handling, but let me know if there is an error that can be reasonably caught, and I'll update. or you can update it. this is StackOverflow.

Answer 11

Here's a solution using itertools:

import itertools

def seq(start, end, step):
    if step == 0:
        raise ValueError("step must not be 0")
    sample_count = int(abs(end - start) / step)
    return itertools.islice(itertools.count(start, step), sample_count)

Usage Example:

for i in seq(0, 1, 0.1):
    print(i)

Answer 12

Suprised no-one has yet mentioned the recommended solution in the Python 3 docs:

See also:

• The linspace recipe shows how to implement a lazy version of range that suitable for floating point applications.

Once defined, the recipe is easy to use and does not require numpy or any other external libraries, but functions like numpy.linspace(). Note that rather than a step argument, the third num argument specifies the number of desired values, for example:

print(linspace(0, 10, 5))
# linspace(0, 10, 5)
print(list(linspace(0, 10, 5)))
# [0.0, 2.5, 5.0, 7.5, 10]

I quote a modified version of the full Python 3 recipe from Andrew Barnert below:

import collections.abc
import numbers

class linspace(collections.abc.Sequence):
    """linspace(start, stop, num) -> linspace object

    Return a virtual sequence of num numbers from start to stop (inclusive).

    If you need a half-open range, use linspace(start, stop, num+1)[:-1].
    """
    def __init__(self, start, stop, num):
        if not isinstance(num, numbers.Integral) or num <= 1:
            raise ValueError('num must be an integer > 1')
        self.start, self.stop, self.num = start, stop, num
        self.step = (stop-start)/(num-1)
    def __len__(self):
        return self.num
    def __getitem__(self, i):
        if isinstance(i, slice):
            return [self[x] for x in range(*i.indices(len(self)))]
        if i < 0:
            i = self.num + i
        if i >= self.num:
            raise IndexError('linspace object index out of range')
        if i == self.num-1:
            return self.stop
        return self.start + i*self.step
    def __repr__(self):
        return '{}({}, {}, {})'.format(type(self).__name__,
                                       self.start, self.stop, self.num)
    def __eq__(self, other):
        if not isinstance(other, linspace):
            return False
        return ((self.start, self.stop, self.num) ==
                (other.start, other.stop, other.num))
    def __ne__(self, other):
        return not self==other
    def __hash__(self):
        return hash((type(self), self.start, self.stop, self.num))

Answer 13

start and stop are inclusive rather than one or the other (usually stop is excluded) and without imports, and using generators

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, Jul 8 2017, 04:57:36) [MSC v.1900 64 bit (AMD64)]

Answer 14

more_itertools is a third-party library that implements a numeric_range tool:

import more_itertools as mit


for x in mit.numeric_range(0, 1, 0.1):
    print("{:.1f}".format(x))

Output

0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9

This tool also works for Decimal and Fraction.

Answer 15

Similar to R's seq function, this one returns a sequence in any order given the correct step value. The last value is equal to the stop value.

def seq(start, stop, step=1):
    n = int(round((stop - start)/float(step)))
    if n > 1:
        return([start + step*i for i in range(n+1)])
    elif n == 1:
        return([start])
    else:
        return([])

Results

seq(1, 5, 0.5)

[1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0]

seq(10, 0, -1)

[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

seq(10, 0, -2)

[10, 8, 6, 4, 2, 0]

seq(1, 1)

[ 1 ]

Answer 16

NumPy is a bit overkill, I think.

[p/10 for p in range(0, 10)]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]

Generally speaking, to do a step-by-1/x up to y you would do

x=100
y=2
[p/x for p in range(0, int(x*y))]
[0.0, 0.01, 0.02, 0.03, ..., 1.97, 1.98, 1.99]

(1/x produced less rounding noise when I tested).

Answer 17

My answer is similar to others using map(), without need of NumPy, and without using lambda (though you could). To get a list of float values from 0.0 to t_max in steps of dt:

def xdt(n):
    return dt*float(n)
tlist  = map(xdt, range(int(t_max/dt)+1))

Answer 18

It can be done using Numpy library. arange() function allows steps in float. But, it returns a numpy array which can be converted to list using tolist() for our convenience.

for i in np.arange(0, 1, 0.1).tolist():
   print i

Answer 19

For completeness of boutique, a functional solution:

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)

Answer 20

The trick to avoid round-off problem is to use a separate number to move through the range, that starts and half the step ahead of start.

# floating point range
def frange(a, b, stp=1.0):
  i = a+stp/2.0
  while i<b:
    yield a
    a += stp
    i += stp

Alternatively, numpy.arange can be used.

Answer 21

Increase the magnitude of i for the loop and then reduce it when you need it.

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

EDIT: I honestly cannot remember why I thought that would work syntactically

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

That should have the desired output.

Answer 22

frange(start, stop, precision)

def frange(a,b,i):
    p = 10**i
    sr = a*p
    er = (b*p) + 1
    p = float(p)
    return map(lambda x: x/p, xrange(sr,er))

In >frange(-1,1,1)

Out>[-1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

Answer 23

scipy has a built in function arange which generalizes Python's range() constructor to satisfy your requirement of float handling.

from scipy import arange

Answer 24

This is my solution to get ranges with float steps.
Using this function it's not necessary to import numpy, nor install it.
I'm pretty sure that it could be improved and optimized. Feel free to do it and post it here.

from __future__ import division
from math import log

def xfrange(start, stop, step):

    old_start = start #backup this value

    digits = int(round(log(10000, 10)))+1 #get number of digits
    magnitude = 10**digits
    stop = int(magnitude * stop) #convert from 
    step = int(magnitude * step) #0.1 to 10 (e.g.)

    if start == 0:
        start = 10**(digits-1)
    else:
        start = 10**(digits)*start

    data = []   #create array

    #calc number of iterations
    end_loop = int((stop-start)//step)
    if old_start == 0:
        end_loop += 1

    acc = start

    for i in xrange(0, end_loop):
        data.append(acc/magnitude)
        acc += step

    return data

print xfrange(1, 2.1, 0.1)
print xfrange(0, 1.1, 0.1)
print xfrange(-1, 0.1, 0.1)

The output is:

[1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]
[0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1]
[-1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0.0]

Answer 25

This one liner will not clutter your code. The sign of the step parameter is important.

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)]

Answer 26

I am only a beginner, but I had the same problem, when simulating some calculations. Here is how I attempted to work this out, which seems to be working with decimal steps.

I am also quite lazy and so I found it hard to write my own range function.

Basically what I did is changed my xrange(0.0, 1.0, 0.01) to xrange(0, 100, 1) and used the division by 100.0 inside the loop. I was also concerned, if there will be rounding mistakes. So I decided to test, whether there are any. Now I heard, that if for example 0.01 from a calculation isn't exactly the float 0.01 comparing them should return False (if I am wrong, please let me know).

So I decided to test if my solution will work for my range by running a short test:

for d100 in xrange(0, 100, 1):
    d = d100 / 100.0
    fl = float("0.00"[:4 - len(str(d100))] + str(d100))
    print d, "=", fl , d == fl

And it printed True for each.

Now, if I'm getting it totally wrong, please let me know.

Answer 27

Here is my solution which works fine with float_range(-1, 0, 0.01) and works without floating point representation errors. It is not very fast, but works fine:

from decimal import Decimal

def get_multiplier(_from, _to, step):
    digits = []
    for number in [_from, _to, step]:
        pre = Decimal(str(number)) % 1
        digit = len(str(pre)) - 2
        digits.append(digit)
    max_digits = max(digits)
    return float(10 ** (max_digits))


def float_range(_from, _to, step, include=False):
    """Generates a range list of floating point values over the Range [start, stop]
       with step size step
       include=True - allows to include right value to if possible
       !! Works fine with floating point representation !!
    """
    mult = get_multiplier(_from, _to, step)
    # print mult
    int_from = int(round(_from * mult))
    int_to = int(round(_to * mult))
    int_step = int(round(step * mult))
    # print int_from,int_to,int_step
    if include:
        result = range(int_from, int_to + int_step, int_step)
        result = [r for r in result if r <= int_to]
    else:
        result = range(int_from, int_to, int_step)
    # print result
    float_result = [r / mult for r in result]
    return float_result


print float_range(-1, 0, 0.01,include=False)

assert float_range(1.01, 2.06, 5.05 % 1, True) ==\
[1.01, 1.06, 1.11, 1.16, 1.21, 1.26, 1.31, 1.36, 1.41, 1.46, 1.51, 1.56, 1.61, 1.66, 1.71, 1.76, 1.81, 1.86, 1.91, 1.96, 2.01, 2.06]

assert float_range(1.01, 2.06, 5.05 % 1, False)==\
[1.01, 1.06, 1.11, 1.16, 1.21, 1.26, 1.31, 1.36, 1.41, 1.46, 1.51, 1.56, 1.61, 1.66, 1.71, 1.76, 1.81, 1.86, 1.91, 1.96, 2.01]

Answer 28

My versions use the original range function to create multiplicative indices for the shift. This allows same syntax to the original range function. I have made two versions, one using float, and one using Decimal, because I found that in some cases I wanted to avoid the roundoff drift introduced by the floating point arithmetic.

It is consistent with empty set results as in range/xrange.

Passing only a single numeric value to either function will return the standard range output to the integer ceiling value of the input parameter (so if you gave it 5.5, it would return range(6).)

Edit: the code below is now available as package on pypi: Franges

## frange.py
from math import ceil
# find best range function available to version (2.7.x / 3.x.x)
try:
    _xrange = xrange
except NameError:
    _xrange = range

def frange(start, stop = None, step = 1):
    """frange generates a set of floating point values over the 
    range [start, stop) with step size step

    frange([start,] stop [, step ])"""

    if stop is None:
        for x in _xrange(int(ceil(start))):
            yield x
    else:
        # create a generator expression for the index values
        indices = (i for i in _xrange(0, int((stop-start)/step)))  
        # yield results
        for i in indices:
            yield start + step*i

## drange.py
import decimal
from math import ceil
# find best range function available to version (2.7.x / 3.x.x)
try:
    _xrange = xrange
except NameError:
    _xrange = range

def drange(start, stop = None, step = 1, precision = None):
    """drange generates a set of Decimal values over the
    range [start, stop) with step size step

    drange([start,] stop, [step [,precision]])"""

    if stop is None:
        for x in _xrange(int(ceil(start))):
            yield x
    else:
        # find precision
        if precision is not None:
            decimal.getcontext().prec = precision
        # convert values to decimals
        start = decimal.Decimal(start)
        stop = decimal.Decimal(stop)
        step = decimal.Decimal(step)
        # create a generator expression for the index values
        indices = (
            i for i in _xrange(
                0, 
                ((stop-start)/step).to_integral_value()
            )
        )  
        # yield results
        for i in indices:
            yield float(start + step*i)

## testranges.py
import frange
import drange
list(frange.frange(0, 2, 0.5)) # [0.0, 0.5, 1.0, 1.5]
list(drange.drange(0, 2, 0.5, precision = 6)) # [0.0, 0.5, 1.0, 1.5]
list(frange.frange(3)) # [0, 1, 2]
list(frange.frange(3.5)) # [0, 1, 2, 3]
list(frange.frange(0,10, -1)) # []

Answer 29

My solution:

def seq(start, stop, step=1, digit=0):
    x = float(start)
    v = []
    while x <= stop:
        v.append(round(x,digit))
        x += step
    return v

Answer 30

You can use this function:

def frange(start,end,step):
    return map(lambda x: x*step, range(int(start*1./step),int(end*1./step)))