Can You help me to convert Python list:
[(1, 'a'), (2, 'b'), (2, 'c'), (3, 'd'), (3, 'e')]
so that:
(1, 'a') is index 0
(2, 'b'), (2, 'c') are both index 1
(3, 'd'), (3, 'e') are both index 2
Simply, all tuples which element[0] is equal, have same index.
Thank You,
itertools.groupby to the rescue!:
lst = [(1, 'a'), (2, 'b'), (2, 'c'), (3, 'd'), (3, 'e')]
lst.sort(key=lambda x: x[0]) #Only necessary if your list isn't sorted already.
new_lst = [list(v) for k,v in itertools.groupby(lst,key=lambda x:x[0])]
You could use operator.itemgetter(0) instead of the lambda if you wanted...
demo:
>>> import itertools
>>> lst = [(1, 'a'), (2, 'b'), (2, 'c'), (3, 'd'), (3, 'e')]
>>> lst.sort(key=lambda x: x[0])
>>> new_lst = [list(v) for k,v in itertools.groupby(lst,key=lambda x:x[0])]
>>> new_lst
[[(1, 'a')], [(2, 'b'), (2, 'c')], [(3, 'd'), (3, 'e')]]
It's not clear what you want, but this will group the items into lists according ot their first element.
groups = collections.defaultdict(list)
for x,y in items:
groups[x].append(y)
