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', 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?

Sites

I think this would be a good use for a dictionary. You can use the first element as the key and the value can be a list of tuples. – Bipinnate

@Bipinnate ok thank you i was wondering if there was a library having a function like this or i have to write my own by me – Sites

You shouldn't need a library for this. I can write a sample here with dictionaries. Check out the answer by @Psidom – Bipinnate

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]])]

Lowborn

You don't need OrderedDict starting with CPython 3.6 and all other Python implementations starting with Python 3.7 because regular dicts maintain insertion order – Deliverance

Answer 2

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)]

Deliverance

Answer 3

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)))]

Mattingly

You don't need OrderedDict starting with CPython 3.6 and all other Python implementations starting with Python 3.7 because regular dicts maintain insertion order – Deliverance

Answer 4

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))

Tanh