how to sort a list of tuples with custom function

[Ho Yeung Lee]

I remove red line 
and capture another version

https://gist.github.com/hoyeunglee/99bbe7999bc489a79ffdf0277e80ecb6

it can capture words in windows, but since window words some are black
and some gray, some are not exactly black, 
so I only choose notepad , since it is using black words

but some words are splitted, I have already sorted by x[0] and x[1]

can it improver to a consecutively a few words

"檔案" <- File is succeed
but "另存新檔"  failed since words are splitted

On Thursday, August 3, 2017 at 3:54:13 PM UTC+8, Ho Yeung Lee wrote:
> https://gist.github.com/hoyeunglee/3d340ab4e9a3e2b7ad7307322055b550
> 
> I updated again
> 
> how to do better because some words are stored in different files
> 
> On Thursday, August 3, 2017 at 10:02:01 AM UTC+8, Ho Yeung Lee wrote:
> > 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