91DarioDev 91DarioDev - 3 years ago 182
Python Question

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

i have a tuple like this

[
(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)
]


i would like to get instead this:

[
(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)))
]


so the for any element, anything that is not the first element is inside a sub-tuple of it, and if the following element has the same element as first element, it will be set as another sub-tuple of the previous one.

so i can do:

for i in data:
# getting the first element of the list
for sub_i in i[1]:
# i access all the tuples inside


are there some functions to do this?

Answer Source

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]])]
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