我有以下包含元组的列表 A
,我想将 A
分割成一个列表列表,如 B
所示。逻辑是,如果元组的第一个和第四个元素重复,则将组打包为列表 A
内的列表。
A = [(1, 'C-30219', 'C-30060', 'C-6235d935d39c258876476e35a7acfd69-1-1', 2),
(1, 'C-30060', 'C-30022', 'C-6235d935d39c258876476e35a7acfd69-1-1', 3),
(1, 'C-30022', 'C-30205', 'C-6235d935d39c258876476e35a7acfd69-1-1', 4),
(3, 'C-30248', 'C-30260', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 4),
(3, 'C-30260', 'C-30108', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 5),
(3, 'C-30108', 'C-30240', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 6),
(5, 'C-30269', 'C-30285', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 9),
(5, 'C-30285', 'C-30109', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 10),
(5, 'C-30109', 'C-30211', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 11),
(5, 'C-30211', 'C-30289', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 12),
(5, 'C-30072', 'C-30375', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 15),
(5, 'C-30375', 'C-30095', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 16)]
出:
B = [[(1, 'C-30219', 'C-30060', 'C-6235d935d39c258876476e35a7acfd69-1-1', 2),
(1, 'C-30060', 'C-30022', 'C-6235d935d39c258876476e35a7acfd69-1-1', 3),
(1, 'C-30022', 'C-30205', 'C-6235d935d39c258876476e35a7acfd69-1-1', 4)],
[(3, 'C-30248', 'C-30260', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 4),
(3, 'C-30260', 'C-30108', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 5),
(3, 'C-30108', 'C-30240', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 6)],
[(5, 'C-30269', 'C-30285', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 9),
(5, 'C-30285', 'C-30109', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 10),
(5, 'C-30109', 'C-30211', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 11),
(5, 'C-30211', 'C-30289', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 12)],
[(5, 'C-30072', 'C-30375', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 15),
(5, 'C-30375', 'C-30095', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 16)]]
这是我的尝试,它在大量的笨拙的巨著之后给出了所需的输出。我正在寻找一种更有效和 Pythonic 的方式来实现这一目标。
inter = list(set([(i[0],i[3]) for i in A]))
B = {o_t: [] for o_t in inter}
for i in range(1, len(A)):
if (A[i][0] == A[i-1][0]
and A[i][3] == A[i-1][3]):
B[A[i][0],A[i][3]].append(A[i])
B[A[i][0],A[i][3]].append(A[i-1])
B = {key: sorted(list(set(B[key])), key = lambda x: x[-1]) for key in B.keys()}
list(B.values())
答案 0 :(得分:1)
来自groupby
的{{1}}的完美任务
itertools
输出
from itertools import groupby
A = [(1, 'C-30219', 'C-30060', 'C-6235d935d39c258876476e35a7acfd69-1-1', 2),
(1, 'C-30060', 'C-30022', 'C-6235d935d39c258876476e35a7acfd69-1-1', 3),
(1, 'C-30022', 'C-30205', 'C-6235d935d39c258876476e35a7acfd69-1-1', 4),
(3, 'C-30248', 'C-30260', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 4),
(3, 'C-30260', 'C-30108', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 5),
(3, 'C-30108', 'C-30240', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 6),
(5, 'C-30269', 'C-30285', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 9),
(5, 'C-30285', 'C-30109', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 10),
(5, 'C-30109', 'C-30211', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 11),
(5, 'C-30211', 'C-30289', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 12),
(5, 'C-30072', 'C-30375', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 15),
(5, 'C-30375', 'C-30095', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 16)]
B = [list(g) for _,g in groupby(A, key=lambda x: (x[0], x[3]))]
print(B)
注意:我假设 A 按第一个和第四个元素排序。 groupby 会将列表 [[(1, 'C-30219', 'C-30060', 'C-6235d935d39c258876476e35a7acfd69-1-1', 2),
(1, 'C-30060', 'C-30022', 'C-6235d935d39c258876476e35a7acfd69-1-1', 3),
(1, 'C-30022', 'C-30205', 'C-6235d935d39c258876476e35a7acfd69-1-1', 4)],
[(3, 'C-30248', 'C-30260', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 4),
(3, 'C-30260', 'C-30108', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 5),
(3, 'C-30108', 'C-30240', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 6)],
[(5, 'C-30269', 'C-30285', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 9),
(5, 'C-30285', 'C-30109', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 10),
(5, 'C-30109', 'C-30211', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 11),
(5, 'C-30211', 'C-30289', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 12)],
[(5, 'C-30072', 'C-30375', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 15),
(5, 'C-30375', 'C-30095', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 16)]]
分组到 [1,1,1,2,2,1,3,3]
。它不会将所有 [(1,1,1), (2,2), (1), (3,3)]
答案 1 :(得分:1)
下面
from collections import defaultdict
data = defaultdict(list)
A = [(1, 'C-30219', 'C-30060', 'C-6235d935d39c258876476e35a7acfd69-1-1', 2),
(1, 'C-30060', 'C-30022', 'C-6235d935d39c258876476e35a7acfd69-1-1', 3),
(1, 'C-30022', 'C-30205', 'C-6235d935d39c258876476e35a7acfd69-1-1', 4),
(3, 'C-30248', 'C-30260', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 4),
(3, 'C-30260', 'C-30108', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 5),
(3, 'C-30108', 'C-30240', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 6),
(5, 'C-30269', 'C-30285', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 9),
(5, 'C-30285', 'C-30109', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 10),
(5, 'C-30109', 'C-30211', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 11),
(5, 'C-30211', 'C-30289', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 12),
(5, 'C-30072', 'C-30375', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 15),
(5, 'C-30375', 'C-30095', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 16)]
for a in A:
data[(a[0], a[3])].append(a)
B = [v for v in data.values()]
for b in B:
print(b)
输出
[(1, 'C-30219', 'C-30060', 'C-6235d935d39c258876476e35a7acfd69-1-1', 2), (1, 'C-30060', 'C-30022', 'C-6235d935d39c258876476e35a7acfd69-1-1', 3), (1, 'C-30022', 'C-30205', 'C-6235d935d39c258876476e35a7acfd69-1-1', 4)]
[(3, 'C-30248', 'C-30260', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 4), (3, 'C-30260', 'C-30108', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 5), (3, 'C-30108', 'C-30240', 'C-ac19d0edcf4d4ebe071e8d43be1901e2-1-1', 6)]
[(5, 'C-30269', 'C-30285', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 9), (5, 'C-30285', 'C-30109', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 10), (5, 'C-30109', 'C-30211', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 11), (5, 'C-30211', 'C-30289', 'C-d0d36bb9f2a7e248638cff9a04065977-1-1', 12)]
[(5, 'C-30072', 'C-30375', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 15), (5, 'C-30375', 'C-30095', 'C-710c460e8dfc2b3a523e077b6c6bdb40-1-1', 16)]