python: group elements of a tuple having the same first element

Question

i have a tuple like this

[
(379146591, \'it\', 55, 1, 1, \'NON ENTRARE\', \'NonEntrate\', 55, 1), 
(4746004, \'it\', 28, 2, 2, \'NON ENTRARE\', \'NonEntrate\         


        

Answer 1

It's pretty simple with defaultdict; You initialize the default value to be a list and then append the item to the value of the same key:

lst = [
    (379146591, 'it', 55, 1, 1, 'NON ENTRARE', 'NonEntrate', 55, 1), 
    (4746004, 'it', 28, 2, 2, 'NON ENTRARE', 'NonEntrate', 26, 2), 
    (4746004, 'it', 28, 2, 2, 'TheBestTroll Group', 'TheBestTrollGroup', 2, 3)
]

from collections import defaultdict    ​
d = defaultdict(list)

for k, *v in lst:
    d[k].append(v)

list(d.items())
#[(4746004,
#  [('it', 28, 2, 2, 'NON ENTRARE', 'NonEntrate', 26, 2),
#   ('it', 28, 2, 2, 'TheBestTroll Group', 'TheBestTrollGroup', 2, 3)]),
# (379146591, [('it', 55, 1, 1, 'NON ENTRARE', 'NonEntrate', 55, 1)])]

---

If order is important, use an OrderedDict which can remember the insertion orders:

from collections import OrderedDict
d = OrderedDict()
​
for k, *v in lst:
    d.setdefault(k, []).append(v)

list(d.items())
#[(379146591, [['it', 55, 1, 1, 'NON ENTRARE', 'NonEntrate', 55, 1]]),
# (4746004,
#  [['it', 28, 2, 2, 'NON ENTRARE', 'NonEntrate', 26, 2],
#   ['it', 28, 2, 2, 'TheBestTroll Group', 'TheBestTrollGroup', 2, 3]])]