# 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> range(0.5,5,0.5) ValueError: range() step argument must not be zero

## Answers

I don't know a built-in function, but writing one like this 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] 99.9999999999986

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] Decimal('99.9')

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] 99.9

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))

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. ])

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. ])

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

Alternately:

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

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]

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)

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 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 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] 0.9 > len(a) 10

And with somewhat larger numbers:

> b = list(frange(0.0, 1000000.0, 0.1)) > b[-1] 999999.9 > len(b) 10000000

The code is also available as a GitHub Gist.

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: yield(float(start))

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

**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 vector with 41 values.

Thus, an frange() function depending on its use 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.

**The Mathematical Use Case**

With that said, as discussed, numpy.linspace() performs the generation with 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)) 0.024284090992296115 >>> min(timeit.Timer('round_test = frange_r(start, stop, skip); [x for x in round_test]', setup=setup).repeat(30, 1000)) 0.025324633985292166

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): print(x) 0.25 0.5 0.75 1.0 ... 9.0 9.25 9.5 9.75

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): print(x) 0.0 0.43 0.86 1.29 ... 8.14 8.57 9.0 9.43

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

##### 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].

##### edit

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.

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) a.split('.')[1].split('0')[0] start = float(str(a)) double_value_range.append(start) start = start+step double_value_range_tuple = tuple(double_value_range) #print double_value_range_tuple return double_value_range_tuple

def Range(*argSequence): if len(argSequence) == 3: imin = argSequence[0]; imax = argSequence[1]; di = argSequence[2] i = imin; iList = [] while i <= imax: iList.append(i) 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 ].

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)) else: yield float(round(valor,approx)) valor += jump for i in drange(12): print(i)

Is there a range() equivalent for floats in Python? NO Use this:

def f_range(start, end, step): a = range(int(start/0.01), int(end/0.01), int(step/0.01)) var = [] for item in a: var.append(item*0.01) return var

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.