CODEWITHSUNDEEP
pythonlisttuples
91DarioDev
91DarioDev

Reputation: 1690

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?

Upvotes: 10

Views: 11613

Answers (4)

user3064538
user3064538

Reputation:

Use itertools.groupby (and operator.itemgetter to get the first item). The only thing is that your data needs to already be sorted so that the groups appear one after the other (if you've used the uniq and sort bash commands, same idea), you can use sorted() for this

import operator
from itertools import groupby

data = [
    (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),
]

data = sorted(data, key=operator.itemgetter(0))  # this might be unnecessary
for k, g in groupby(data, operator.itemgetter(0)):
    print(k, list(g))

Will output

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

In your case, you also need to remove the first element from your lists of values. Change the last two lines of the above to:

for k, g in groupby(data, operator.itemgetter(0)):
    print(k, [item[1:] for item in g])

Output:

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)]

Upvotes: 4

Ming.Xu
Ming.Xu

Reputation: 51

u can use collection.defaultdict:

data = [
    (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
a = defaultdict(list)
a = defaultdict(list)


from collections import defaultdict
a = defaultdict(list)

for d in data:
    a[d[0]].append(d[1:])

for k,v in a.items():
    a[k] = tuple(a[k])

print(dict(a))

Upvotes: 0

akuiper
akuiper

Reputation: 214927

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]])]

Upvotes: 12

Ajax1234
Ajax1234

Reputation: 71451

You can use Python3 variable unpacking and OrderedDict to retain order:

from collections import OrderedDict
d = OrderedDict()
l = [
  (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)
]

for a, *b in l:
  if a in d:
     d[a].append(b)
  else:
     d[a] = [b]

final_data = [(a, tuple(map(tuple, b))) for a, b in d.items()]

Output:

[(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)))]

Upvotes: 3

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