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?
In python, to sort list of tuples by the first element in descending order, we have to use the sort() method with the parameter ” (reverse=True) “ which will sort the elements in descending order.
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]])]
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)]
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