Question

我有以下列表：

lines
['line_North_Mid', 'line_South_Mid',
 'line_North_South', 'line_Mid_South',
 'line_South_North','line_Mid_North' ]

关于它们的名称，我想将它们按照以下方式在元组列表中配对：

tuple_list
[('line_Mid_North', 'line_North_Mid'),
 ('line_North_South', 'line_South_North'),
 ('line_Mid_South', 'line_South_Mid')]

我想也许我可以在lines的元素中进行字符串搜索，但是效率不高。是否有更好的方式以lines

的形式对tuple_list元素进行排序

配对条件：

如果两个元素都具有相同的Area_name：('North', 'Mid', 'South')

例如：'line_North_Mid'应该与'line_Mid_North'结合

Answer 1

使用collections.defaultdict可以实现与订单无关的O（ n ）解决方案。想法是使用由from collections import defaultdict L = ['line_North_Mid', 'line_South_Mid', 'line_North_South', 'line_Mid_South', 'line_South_North', 'line_Mid_North'] dd = defaultdict(list) for item in L: dd[frozenset(item.rsplit('_', maxsplit=2)[1:])].append(item) res = list(map(tuple, dd.values())) # [('line_North_Mid', 'line_Mid_North'), # ('line_South_Mid', 'line_Mid_South'), # ('line_North_South', 'line_South_North')]分隔的字符串的最后2个部分作为我们的字典键，并在输入列表中附加值。然后提取值并转换为元组列表。

{{1}}

Answer 2

尝试一下：

from itertools import combinations

tuple_list = [i for i in combinations(lines,2) if i[0].split('_')[1] == i[1].split('_')[2] and i[0].split('_')[2] == i[1].split('_')[1]]

或者我认为这样更好：

[i for i in combinations(lines,2) if i[0].split('_')[1:] == i[1].split('_')[1:][::-1]]

Answer 3

您可以使用以下列表理解：

lines = ['line_Mid_North', 'line_North_Mid',
         'line_North_South', 'line_South_North',
         'line_Mid_South', 'line_South_Mid']

[(j,i) for i in lines for j in lines if j not in i 
       if set(j.split('_')[1:]) < set(i.split('_'))][::2]

[('line_Mid_North', 'line_North_Mid'),
 ('line_North_South', 'line_South_North'),
 ('line_Mid_South', 'line_South_Mid')]

Answer 4

我建议您有一个函数，该函数为应该在一起的字符串返回相同的键（分组键）。

def key(s):
    # ignore first part and sort other 2 parts, so they will always be in same order
    _, part_1, part_2 = s.split('_')
    return tuple(sorted([part_1, part_2]))

您必须使用某种分组方法；我以defaultdict为例：

import collections

lines = [
    'line_North_Mid', 'line_South_Mid',
    'line_North_South', 'line_Mid_South',
    'line_South_North','line_Mid_North',
]

dd = collections.defaultdict(list)
for s in lines:
    dd[key(s)].append(s)     # those with same key get grouped

print(list(tuple(v) for v in dd.values()))
# [
#     ('line_North_Mid', 'line_Mid_North'),
#     ('line_South_Mid', 'line_Mid_South'),
#     ('line_North_South', 'line_South_North'),
# ]

将str元素从列表耦合到元组列表

4 个答案: