Question

我有这样的元组

[
(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)
]

我想改为：

[
(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)))
]

所以对于任何元素，任何不是第一个元素的东西都在它的子元组内，如果下面的元素与第一个元素具有相同的元素，它将被设置为前一个元素的另一个子元组之一。

所以我能做到：

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

是否有一些功能可以做到这一点？

Answer 1

defaultdict非常简单;您将默认值初始化为列表，然后将该项追加到相同键的值：

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)])]

如果订单很重要，请使用可以记住广告订单的OrderedDict：

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]])]

Answer 2

您可以使用Python3变量解压缩和OrderedDict来保留顺序：

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()]

输出：

[(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)))]

Answer 3

使用itertools.groupby（和operator.itemgetter获取第一项）。唯一的事情是您的数据需要已经排序，以便这些组一个接一个地显示（如果您使用过uniq和sort bash命令，则是相同的想法），则可以使用{ {3}}为此

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))

将输出

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)]

对于您而言，还需要从值列表中删除第一个元素。将上面的最后两行更改为：

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

输出：

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)]

Answer 4

你可以使用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))

python：具有相同第一个元素的元组的组元素

4 个答案: