On Mon, 18 Aug 2008 18:04:53 +0000, Dan Lenski wrote: > On Mon, 18 Aug 2008 10:52:45 -0700, Alejandro wrote: > >> Hi: >> >> I need to find the multiples of a decimal number in a floating point >> list. For instance, if a have the list [0,0.01,0.02,...1], I want the >> multiples of 0.2: [0, 0.2,0.4,0.6,0.8,1]. >> >> With integers this problem is easy, just test for (i%n == 0), where i >> is the number I am testing, and n is the multiple. Given the finite >> resolution of floating point numbers, this is more complicated for >> float. >> >> I came with this solution: >> >> from numpy import arange >> from math import modf, fabs >> >> float_range = arange(0, 1, 0.01) >> for f in float_range: >> m = modf(f / 0.2)[0] >> if m<1e-13 or fabs(1-m)<1e-13: >> print f >> # Do something else >> >> This code works, however, I found it a little ugly. Is there a better >> way to do the same? >> >> Alejandro. > > Hi Alejandro, you can do the same thing more efficiently (both in terms > of lines of code and execution speed) by doing the whole array at once: > > from numpy import arange, absolute > from math import modf, fabs > > float_range = arange(0, 1, 0.01) > multiples = absolute(float_range % 0.2)<1e-13 # now multiples is a > boolean array > print float_range[multiples] > > HTH, > Dan
Oh, I forgot that you have to check the other case of it being slightly less than a multiple too! from numpy import arange, absolute from math import modf, fabs float_range = arange(0, 1, 0.01) mod = absolute(float_range % 0.2) multiples = (mod < 1e-13) + (mod > 0.2-1e-13) print float_range[multiples] That should do the trick! Dan -- http://mail.python.org/mailman/listinfo/python-list
