clusters of numbers

[Vincent Davis]

Why not start with a histogram.

Vincent

On Sat, Dec 15, 2018 at 6:46 PM Marc Lucke <marc at marcsnet.com> wrote:

> hey guys,
>
> I have a hobby project that sorts my email automatically for me & I want
> to improve it.  There's data science and statistical info that I'm
> missing, & I always enjoy reading about the pythonic way to do things too.
>
> I have a list of percentage scores:
>
>
> (1,11,1,7,5,7,2,2,2,10,10,1,2,2,1,7,2,1,7,5,3,8,2,6,3,2,7,2,12,3,1,2,19,3,5,1,1,7,8,8,1,5,6,7,3,14,6,1,6,7,6,15,6,3,7,2,6,23,2,7,1,21,21,8,8,3,2,20,1,3,12,3,1,2,10,16,16,15,6,5,3,2,2,11,1,14,6,3,7,1,5,3,3,14,3,7,3,5,8,3,6,17,1,1,7,3,1,2,6,1,7,7,12,6,6,2,1,6,3,6,2,1,5,1,8,10,2,6,1,7,3,5,7,7,5,7,2,5,1,19,19,1,12,5,10,2,19,1,3,19,6,1,5,11,2,1,2,5,2,5,8,2,2,2,5,3,1,21,2,3,7,10,1,8,1,3,17,17,1,5,3,10,14,1,2,14,14,1,15,6,3,2,17,17,1,1,1,2,2,3,3,2,2,7,7,2,1,2,8,2,20,3,2,3,12,7,6,5,12,2,3,11,3,1,1,8,16,10,1,6,6,6,11,1,6,5,2,5,11,1,2,10,6,14,6,3,3,5,2,6,17,15,1,2,2,17,5,3,3,5,8,1,6,3,14,3,2,1,7,2,8,11,5,14,3,19,1,3,7,3,3,8,8,6,1,3,1,14,14,10,3,2,1,12,2,3,1,2,2,6,6,7,10,10,12,24,1,21,21,5,11,12,12,2,1,19,8,6,2,1,1,19,10,6,2,15,15,7,10,14,12,14,5,11,7,12,2,1,14,10,7,10,3,17,25,10,5,5,3,12,5,2,14,5,8,1,11,5,29,2,7,20,12,14,1,10,6,17,16,6,7,11,12,3,1,23,11,10,11,5,10,6,2,17,15,20,5,10,1,17,3,7,15,5,11,6,19,14,15,7,1,2,17,8,15,10,26,6,1,2,10,6,14,12,6,1,16,6,12,10,10,14,1,6,1,6,6,12,6,6,1,2,5,10,8,10,1,6,8,17,11,6,3,6,5,1,2,1,2,6,6,12,14,7,1,7,1,8,2,3,14,11,6,3,11,3,1,6,17,12,8,2,10,3,12,12,2,7,5,5,17,2,5,10,12,21,15,6,10,10,7,15,11,2,7,10,3,1,2,7,10,15,1,1,6,5,5,3,17,19,7,1,15,2,8,7,1,6,2,1,15,19,7,15,1,8,3,3,20,8,1,11,7,8,7,1,12,11,1,10,17,2,23,3,7,20,20,3,11,5,1,1,8,1,6,2,11,1,5,1,10,7,20,17,8,1,2,10,6,2,1,23,11,11,7,2,21,5,5,8,1,1,10,12,15,2,1,10,5,2,2,5,1,2,11,10,1,8,10,12,2,12,2,8,6,19,15,8,2,16,7,5,14,2,1,3,3,10,16,20,5,8,14,8,3,14,2,1,5,16,16,2,10,8,17,17,10,10,11,3,5,1,17,17,3,17,5,6,7,7,12,19,15,20,11,10,2,6,6,5,5,1,16,16,8,7,2,1,3,5,20,20,6,7,5,23,14,3,10,2,2,7,10,10,3,5,5,8,14,11,14,14,11,19,5,5,2,12,25,5,2,11,8,10,5,11,10,12,10,2,15,15,15,5,10,1,12,14,8,5,6,2,26,15,21,15,12,2,8,11,5,5,16,5,2,17,3,2,2,3,15,3,8,10,7,10,3,1,14,14,8,8,8,19,10,12,3,8,2,20,16,10,6,15,6,1,12,12,15,15,8,11,17,7,7,7,3,10,1,5,19,11,7,12,8,12,7,5,10,1,11,1,6,21,1,1,10,3,8,5,6,5,20,25,17,5,2,16,14,11,1,17,10,14,5,16,5,2,7,3,8,17,7,19,12,6,5,1,3,12,43,11,8,11,5,19,10,5,11,7,20,6,12,35,5,3,17,10,2,12,6,5,21,24,15,5,10,3,15,1,12,6,3,17,3,2,3,5,5,14,11,8,1,8,10,5,25,8,7,2,6,3,11,1,11,7,3,10,7,12,10,8,6,1,1,17,3,1,1,2,19,6,10,2,2,7,5,16,3,2,11,10,7,10,21,3,5,2,21,3,14,6,7,2,24,3,17,3,21,8,5,11,17,5,6,10,5,20,1,12,2,3,20,6,11,12,14,6,6,1,14,15,12,15,6,20,7,7,19,3,7,5,16,12,6,7,2,10,3,2,11,8,6,6,5,1,11,1,15,21,14,6,3,2,2,5,6,1,3,5,3,6,20,1,15,12,2,3,3,7,1,16,5,24,10,7,1,12,16,8,26,16,15,10,19,11,6,6,5,6,5)
>
>   & I'd like to know know whether, & how the numbers are clustered.  In
> an extreme & illustrative example, 1..10 would have zero clusters;
> 1,1,1,2,2,2,7,7,7 would have 3 clusters (around 1,2 & 7);
> 17,22,20,45,47,51,82,84,83  would have 3 clusters. (around 20, 47 &
> 83).  In my set, when I scan it, I intuitively figure there's lots of
> numbers close to 0 & a lot close to 20 (or there abouts).
>
> I saw info about k-clusters but I'm not sure if I'm going down the right
> path.  I'm interested in k-clusters & will teach myself, but my priority
> is working out this problem.
>
> Do you know the name of the algorithm I'm trying to use?  If so, are
> there python libraries like numpy that I can leverage?  I imagine that I
> could iterate from 0 to 100% using that as an artificial mean, discard
> values that are over a standard deviation away, and count the number of
> scores for that mean; then at the end of that I could set a threshold
> for which the artificial mean would be kept something like (no attempt
> at correct syntax:
>
> means={}
> deviation=5
> threshold=int(0.25*len(list))
> for i in range 100:
>    count=0
>    for j in list:
>      if abs(j-i) > deviation:
>        count+=1
>    if count > threshold:
>      means[i]=count
>
> That algorithm is entirely untested & I think it could work, it's just I
> don't want to reinvent the wheel.  Any ideas kindly appreciated.
>
>
> --
> https://mail.python.org/mailman/listinfo/python-list
>