Xtykutl

I have a tuple in Python like this -

a = ((-,-,x), (-,-,x), (-,-,y), (-,-,z), (-,-,z), (-,-,z))

Now, I want to group the tuples with the same third element. I have to convert this tuple,a into

b = (((-,-,x), (-,-,x)), ((-,-,y)), ((-,-,z), (-,-,z), (-,-,z)))

How do I write Python code for this? To convert a into b? Since tuples are immutable I'm not able to write the code successfully.

This can be done with itertools.groupby.

>>> from operator import itemgetter

>>> from itertools import groupby

>>> 

>>> a = (('-','-','x'), ('-','-','x'), ('-','-','y'), ('-','-','z'), ('-','-','z'), ('-','-','z'))

>>> tuple(tuple(grp) for _, grp in groupby(a, key=itemgetter(-1)))

>>> 

((('-', '-', 'x'), ('-', '-', 'x')),

 (('-', '-', 'y'),),

 (('-', '-', 'z'), ('-', '-', 'z'), ('-', '-', 'z')))

It's assumed that groups of equal last elements cannot be interrupted (like in your sample data). If they can, you need to sort a with key=itemgetter(-1) first.

...can you explain "interrupted"?

Sure. What I mean is that groupby lumps together groups of consecutive elements where key returns the same value. So if we had something like

b = (('-','-','z'), ('-', '-', 'x'), ('-','-','z'))

then the groupby from above would create three groups, not two.

>>> b = (('-','-','z'), ('-', '-', 'x'), ('-','-','z'))

>>> tuple(tuple(grp) for _, grp in groupby(b, key=itemgetter(-1)))

>>> ((('-', '-', 'z'),), (('-', '-', 'x'),), (('-', '-', 'z'),))

If you don't want that to happen, sort b first.

>>> last = itemgetter(-1)

>>> tuple(tuple(grp) for _, grp in groupby(sorted(b, key=last), key=last))

>>> ((('-', '-', 'x'),), (('-', '-', 'z'), ('-', '-', 'z')))

Use itertools.groupby:

from itertools import groupby



a = (('-','-','x'), ('-','-','x'), ('-','-','y'), ('-','-','z'), ('-','-','z'), ('-','-','z'))



print(tuple(tuple(x) for _, x in groupby(a)))

# ((('-', '-', 'x'), ('-', '-', 'x')), (('-', '-', 'y'),), (('-', '-', 'z'), ('-', '-', 'z'), ('-', '-', 'z')))

Given the elements are already ordered by the last element (well the tuples with x as last element, etc. are positioned in a sequence), we can use itertools.groupby:

from itertools import groupby

from operator import itemgetter



b = tuple(tuple(vs) for _, vs in groupby(a, itemgetter(-1)))