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如何用元组中的一个项筛选元组列表?

python

Ejaz · 技术社区 · 6 年前

d = [('A', 'B', 1), ('C', 'D', 1), 
     ('B', 'D', 2), ('A', 'B', 3), 
     ('A', 'D', 3), ('B', 'C', 4), 
     ('A', 'C', 5), ('B', 'C', 8)]

在这里,元组中的前两项是节点,第三项是权重。我想删除具有相同第一和第二节点(两个元组之间的相同第一和第二节点)但权重更高的元组。

d = [('A', 'B', 1), ('C', 'D', 1), 
     ('B', 'D', 2), ('A', 'D', 3), 
     ('B', 'C', 4), ('A', 'C', 5)]

我试过类似的方法,但看起来不是很干净。

edge_dict = {}

for x in d:
    key = '%s%s' % (x[0], x[1])
    if not edge_dict.get(key):
        edge_dict[key] = x[2]
    else:
        if edge_dict[key] > x[2]:
            edge_dict[key] = x[2]       

final_list = []
for k, v in edge_dict.items():
    t = list(k)
    t.append(v)
    final_list.append(tuple(t))


final_list.sort(key=lambda x: x[2])
print final_list

4 回复 | 直到 6 年前

niraj 6 年前

sorted_res = sorted(d, key = lambda x:((x[0], x[1]), x[2]),reverse=True)
print(sorted_res)

[('C', 'D', 1),
 ('B', 'D', 2),
 ('B', 'C', 8),
 ('B', 'C', 4),
 ('A', 'D', 3),
 ('A', 'C', 5),
 ('A', 'B', 3),
 ('A', 'B', 1)]

dictionary 前两个元素的键和值将是最新的小元素:

my_dict = {(i[0], i[1]):i for i in sorted_res}
print(my_dict)

{('A', 'B'): ('A', 'B', 1),
 ('A', 'C'): ('A', 'C', 5),
 ('A', 'D'): ('A', 'D', 3),
 ('B', 'C'): ('B', 'C', 4),
 ('B', 'D'): ('B', 'D', 2),
 ('C', 'D'): ('C', 'D', 1)}

list(my_dict.values())

结果:

[('A', 'C', 5),
 ('A', 'B', 1),
 ('A', 'D', 3),
 ('B', 'D', 2),
 ('C', 'D', 1),
 ('B', 'C', 4)]

sorted 和 dictionary comprehension

result = list({(i[0], i[1]):i 
             for i in sorted(d, key = lambda x:((x[0], x[1]), x[2]),reverse=True)}.values())

wwii 6 年前

只是一点重构。

edge_dict = {}

for t in d:
    key = t[:2]
    value = t[-1]
    if key in edge_dict:
        edge_dict[key] = min(value, edge_dict[key]) 
    else:
        edge_dict[key] = value

final_list = [(q,r,t) for (q,r),t in edge_dict.items()]
final_list.sort(key=lambda x: x[2])

Ajax1234 6 年前

itertools.groupby ,并在每个分组中选择最小值:

import itertools
d = [('A', 'B', 1), ('C', 'D', 1), 
 ('B', 'D', 2), ('A', 'B', 3), 
 ('A', 'D', 3), ('B', 'C', 4), 
 ('A', 'C', 5), ('B', 'C', 8)]
new_d = [min(list(b), key=lambda x:x[-1]) for _, b in itertools.groupby(sorted(d, key=lambda x:x[:-1]), key=lambda x:x[:-1])]

[('A', 'B', 1), ('A', 'C', 5), ('A', 'D', 3), ('B', 'C', 4), ('B', 'D', 2), ('C', 'D', 1)]

Taohidul Islam 6 年前

使用节点对的排序形式作为字典 key

check_dict={} #This will store the minimum valued nodes

for i in d:
    if check_dict.get((i[0],i[1])) ==None: #if the node is absent then add it to the check_dict
        check_dict[tuple(sorted((i[0],i[1])))] = i[2]
    else: #if the node is present then compare with the previous value and store the minimum one
        check_dict[tuple(sorted((i[0],i[1])))] = min(check_dict[(i[0],i[1])],i[2]) #used sorted to treat ('A','B') as same as ('B',A')
expected_list = [tuple(key+(value,)) for key,value in check_dict.items()] #create your list of tuples
print(expected_list)

[('A', 'B', 1), ('C', 'D', 1), ('B', 'D', 2), ('A', 'D', 3), ('B', 'C', 4), ('A', 'C', 5)]