how to sort a list of tuples with custom function

[Ho Yeung Lee]

https://gist.github.com/hoyeunglee/f371f66d55f90dda043f7e7fea38ffa2

I am near succeed in another way, please run above code

when so much black words, it will be very slow
so I only open notepad and maximum it without any content
then capture screen and save as roster.png

and run it, but I discover it can not circle all words with red rectangle
and only part of words


On Wednesday, August 2, 2017 at 3:06:40 PM UTC+8, Peter Otten wrote:
> Glenn Linderman wrote:
> 
> > On 8/1/2017 2:10 PM, Piet van Oostrum wrote:
> >> Ho Yeung Lee <jobmattcon at gmail.com> writes:
> >>
> >>> def isneighborlocation(lo1, lo2):
> >>>      if abs(lo1[0] - lo2[0]) < 7  and abs(lo1[1] - lo2[1]) < 7:
> >>>          return 1
> >>>      elif abs(lo1[0] - lo2[0]) == 1  and lo1[1] == lo2[1]:
> >>>          return 1
> >>>      elif abs(lo1[1] - lo2[1]) == 1  and lo1[0] == lo2[0]:
> >>>          return 1
> >>>      else:
> >>>          return 0
> >>>
> >>>
> >>> sorted(testing1, key=lambda x: (isneighborlocation.get(x[0]), x[1]))
> >>>
> >>> return something like
> >>> [(1,2),(3,3),(2,5)]
> 
> >> I think you are trying to sort a list of two-dimensional points into a
> >> one-dimensiqonal list in such a way thet points that are close together
> >> in the two-dimensional sense will also be close together in the
> >> one-dimensional list. But that is impossible.
> 
> > It's not impossible, it just requires an appropriate distance function
> > used in the sort.
> 
> That's a grossly misleading addition. 
> 
> Once you have an appropriate clustering algorithm
> 
> clusters = split_into_clusters(items) # needs access to all items
> 
> you can devise a key function
> 
> def get_cluster(item, clusters=split_into_clusters(items)):
>     return next(
>         index for index, cluster in enumerate(clusters) if item in cluster
>     )
> 
> such that
> 
> grouped_items = sorted(items, key=get_cluster)
> 
> but that's a roundabout way to write
> 
> grouped_items = sum(split_into_clusters(items), [])
> 
> In other words: sorting is useless, what you really need is a suitable 
> approach to split the data into groups. 
> 
> One well-known algorithm is k-means clustering:
> 
> https://docs.scipy.org/doc/scipy/reference/generated/scipy.cluster.vq.kmeans.html
> 
> Here is an example with pictures:
> 
> https://dzone.com/articles/k-means-clustering-scipy