problem making equally spaced value array with linspace
Jean Dubois
jeandubois314 at gmail.com
Tue Dec 18 14:37:18 CET 2012
On 18 dec, 14:09, Peter Otten <__pete... at web.de> wrote:
> Jean Dubois wrote:
> > I have trouble with the code beneath to make an array with equally
> > spaced values
> > When I enter 100e-6 as start value, 700e-6 as end value and 100e-6 I
> > get the following result:
> > [ 0.0001 0.00022 0.00034 0.00046 0.00058 0.0007 ]
> > But I was hoping for:
> > [ 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007]
> > It works correctly for other values like 1,7,1 but not for 0.1,0.7,0.1
> > then again for 0.01,0.07,0.01
>
> > What I find strange is that for the 1st example "1+abs(float(endvalue)-
> > float(startvalue))/float(incr)" gives 7.0 but int() of this value
> > gives 6
> > can someone provide help with this issue?
> > thanks
> > jean
>
> > #!/usr/bin/python
> > import math
> > import numpy as np
> > print "Enter start value as a float (e.g. 0.001) or in scientific
> > notation (e.g. 1e-3): ",
> > startvalue = raw_input()
> > print "Enter end value: ",
> > endvalue = raw_input()
> > print "Enter step: ",
> > incr = raw_input()
> > #nom = number of measurements
> > nom=int(1+abs(float(endvalue)-float(startvalue))/float(incr))
> > array=np.linspace(float(startvalue), float(endvalue), float(nom))
> > print "Array with current values: ",array
>
> If you repeat the calculation of the number of intervals in the interpreter
> you get
>
> >>> 1 + abs(0.0007-0.0001)/0.0001
>
> 6.999999999999999
>
> Many numbers cannot be represented exactly as float (that's the price you
> have to pay for covering a wide range with just a few (8) bytes), and you
> have introduced such a small error. The subsequent int() call will round
> that float to the integer below it:
>
> >>> int(_)
>
> 6
>
> While applying round() would work here
>
> >>> int(round(1 + abs(0.0007-0.0001)/0.0001))
>
> 7
>
> there is no once-and-for-all solution to the underlying problem. E. g.
>
> >>> x = 2.**53
> >>> x == x + 1
>
> True
thanks
jean
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