How to use a decimal range() step value?

Question

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.

You can 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, you can, with 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])

Python's range() can only do integers, not floating point. In your specific case, you can use a list comprehension instead:

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

(Replace the call to range with that expression.)

For the more general case, you may want to write a custom function or generator.

Building on 'xrange([start], stop[, step])', you can define a generator that accepts and produces any type you choose (stick to types supporting + and <):

>>> 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']
>>> 

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.

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).

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

from scipy import arange