CODEWITHSUNDEEP
pythonlistsplit
Link_tester
Link_tester

Reputation: 1081

How to split a list into sublists based on unique values of one column?

I have a list and want to split it based on unique values of its last column in python. This is my list:

pnt=[[1.,2.,4.,'AA'], [0.,0.,0.,'AA'], [2.,1.,0.,'AA'],\
     [0.,-3.,1.,'BB'], [2.,5.,8.,'BB'], [.1,3.,0.,'CC']]

I want to get it as:

pnt=[[[1.,2.,4.,'AA'], [0.,0.,0.,'AA'], [2.,1.,0.,'AA']],\
     [[0.,-3.,1.,'BB'], [2.,5.,8.,'BB']], [[.1,3.,0.,'CC']]]

I read this solution but still cannot solve my issue.

Upvotes: 1

Views: 1290

Answers (2)

Jean-François Fabre
Jean-François Fabre

Reputation: 140307

itertools.groupby is the function you need. It needs to be customized to check key from the fourth element of each list.

Then elements and the return value itself needs to be forced to list to see the actual result (else it's optimized as generator functions). We discard the key (with _) because we're only interested in the elements, not the value that was used to group them.

import itertools

pnt=[[1.,2.,4.,'AA'], [0.,0.,0.,'AA'], [2.,1.,0.,'AA'],\
     [0.,-3.,1.,'BB'], [2.,5.,8.,'BB'], [.1,3.,0.,'CC']]

print([list(x) for _,x in itertools.groupby(pnt,lambda x:x[3])])

result:

[[[1.0, 2.0, 4.0, 'AA'], [0.0, 0.0, 0.0, 'AA'], [2.0, 1.0, 0.0, 'AA']],
 [[0.0, -3.0, 1.0, 'BB'], [2.0, 5.0, 8.0, 'BB']],
 [[0.1, 3.0, 0.0, 'CC']]]

note that this method will group the consecutive groups only.

Upvotes: 2

Rakesh
Rakesh

Reputation: 82795

You can use a dict to get the desired output.

Ex:

pnt=[[1.,2.,4.,'AA'], [0.,0.,0.,'AA'], [2.,1.,0.,'AA'],\
     [0.,-3.,1.,'BB'], [2.,5.,8.,'BB'], [.1,3.,0.,'CC']]

result = {}
for i in pnt:
    result.setdefault(i[-1], []).append(i)  # Form Dict with key as last value from list and value as the list

print(list(result.values()))    # Get Values of Dict

Output:

[[[1.0, 2.0, 4.0, 'AA'], [0.0, 0.0, 0.0, 'AA'], [2.0, 1.0, 0.0, 'AA']],
 [[0.0, -3.0, 1.0, 'BB'], [2.0, 5.0, 8.0, 'BB']],
 [[0.1, 3.0, 0.0, 'CC']]]

Upvotes: 2

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