list subsetting

Jervis Whitley jervisau at gmail.com
Wed Jan 21 23:20:23 CET 2009


On Thu, Jan 22, 2009 at 9:09 AM, Jeff McNeil <jeff at jmcneil.net> wrote:

> On Jan 21, 4:53 pm, culpritNr1 <ig2ar-s... at yahoo.co.uk> wrote:
> > Hello All,
> >
> > Say I have a list like this:
> >
> > a = [0 , 1, 3.14, 20, 8, 8, 3.14]
> >
> > Is there a simple python way to count the number of 3.14's in the list in
> > one statement?
> >
> > In R I do like this
> >
> > a = c(0 , 1, 3.14, 20, 8, 8, 3.14)
> >
> > length( a[ a[]==3.14 ] )
> >
> > How do I do that in standard python?
> >
> > (Note that this is just an example, I do not mean to use == in floating
> > point operations.)
> >
> > Thank you
> >
> > culpritNr1
> >
> > --
> > View this message in context:
> http://www.nabble.com/list-subsetting-tp21593123p21593123.html
> > Sent from the Python - python-list mailing list archive at Nabble.com.
>
> Just the number of occurrences? Count method?
>
> Python 2.6 (r26:66714, Oct 29 2008, 08:30:04)
> [GCC 4.1.2 20070925 (Red Hat 4.1.2-33)] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
> >>> [1,2,3,3.14,3.14,5,66].count(3.14)
> 2
> >>>
>
> Jeff
> --
> http://mail.python.org/mailman/listinfo/python-list
>
a = [3.14, 4, 3.15, 3.14 + 1E-12]
len([None for elem in a if abs(elem - 3.14) < 1E-9])
Just replace 1E-9 with the level of accuracy that you would like.

This creates a list of Nones, one None for each 3.14 in the original list,
then
counts the length of the list.
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