将列表转换为3个列表

Question

get_plan= 
[('Depot', 'Addr_3'),
  ('Addr_3', 'Addr_4'),
  ('Addr_4', 'Depot'),
  ('Depot', 'Addr_1'),
  ('Addr_1', 'Addr_2'),
  ('Addr_2', 'Depot'),
  ('Depot', 'Addr_5'),
  ('Addr_5', 'Addr_6'),
  ('Addr_6', 'Depot')]

我怎样才能把这个计划转变成这样的路线列表(一种解决方案，每条路线都在仓库结束，另一条路线从仓库开始)

route_list= 
[[('Depot', 'Addr_3'),
  ('Addr_3', 'Addr_4'),
  ('Addr_4', 'Depot'),]
  [('Depot', 'Addr_1'),
  ('Addr_1', 'Addr_2'),
  ('Addr_2', 'Depot'),]
  [('Depot', 'Addr_5'),
  ('Addr_5', 'Addr_6'),
  ('Addr_6', 'Depot')]]

我一般需要它。

非常感谢！

Answer 1

假设输入数据是正确的，并且不需要进行检查，那么使用一个简单的循环。

依靠路线的尽头：

route_list = []
tmp = []
for t in get_plan:
    tmp.append(t)
    if t[1] == 'Depot':
        route_list.append(tmp)
        tmp = []

依靠路线的开始

route_list = []
tmp = []
for t in get_plan:
    if t[0] == 'Depot':
        if tmp:
            route_list.append(tmp)
        tmp = [t]
    else:
        tmp.append(t)
route_list.append(tmp)

产出：

[[('Depot', 'Addr_3'), ('Addr_3', 'Addr_4'), ('Addr_4', 'Depot')],
 [('Depot', 'Addr_1'), ('Addr_1', 'Addr_2'), ('Addr_2', 'Depot')],
 [('Depot', 'Addr_5'), ('Addr_5', 'Addr_6'), ('Addr_6', 'Depot')]]

Answer 2

您可以使用来自split_before的迭代工具

from more_itertools import split_before

list(split_before(get_plan, lambda x: x[0] == 'Depot'))

Answer 3

您可以使用这样的numpy来实现这一点。

import numpy as np

get_plan = [('Depot', 'Addr_3'),
            ('Addr_3', 'Addr_4'),
            ('Addr_4', 'Depot'),
            ('Depot', 'Addr_1'),
            ('Addr_1', 'Addr_2'),
            ('Addr_2', 'Depot'),
            ('Depot', 'Addr_5'),
            ('Addr_5', 'Addr_6'),
            ('Addr_6', 'Depot')]

# create a numpy array
np_arr = np.array(get_plan)
# Find the indices on which to split the array
indices, = np.nonzero(np_arr[:, 0] == "Depot")
# Split the array
route_list = [sub.tolist() for sub in np.split(np_arr, indices) if sub.size]
print(route_list)

输出：

[[['Depot', 'Addr_3'], ['Addr_3', 'Addr_4'], ['Addr_4', 'Depot']], 
[['Depot', 'Addr_1'], ['Addr_1', 'Addr_2'], ['Addr_2', 'Depot']],
[['Depot', 'Addr_5'], ['Addr_5', 'Addr_6'], ['Addr_6', 'Depot']]]

如果您想要元组而不是列表，可以将sub.tolist()替换为list(map(tuple, sub))

route_list = [list(map(tuple, sub)) for sub in np.splite(np_arr, indices) if sub.size]

产出：

[[('Depot', 'Addr_3'), ('Addr_3', 'Addr_4'), ('Addr_4', 'Depot')], 
[('Depot', 'Addr_1'), ('Addr_1', 'Addr_2'), ('Addr_2', 'Depot')], 
[('Depot', 'Addr_5'), ('Addr_5', 'Addr_6'), ('Addr_6', 'Depot')]]