how to sort a list of tuples with custom function
Peter Otten
__peter__ at web.de
Wed Aug 2 03:05:56 EDT 2017
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
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