我想要一個這樣的列表：['5', '0 1', '1 2', '1 8', '2 3'] 并返回一個由這樣的元組組成的新列表： [(0, [1]),(1,[0,2,8]),(2[1,3]),(3,[2]),(8,[1])]。每個元組的第一個元素是一個整數，第二個元素是一個整數列表，它出現在原始列表中。我不能使用字典、集合、雙端隊列、二等分模塊。def create_network(file_name): friends = open(file_name).read().splitlines() network=[] for strings in friends: relationship=strings.strip().split(' ') if len(relationship)==2: a,b=relationship a=int(a) b=int(b) if a>=len(network): network.append((a,[b])) else: wow=network[a] wow[1].append(b) network[a]=wow return network這是我到目前為止。我希望它返回： [(0, [1, 2, 3]), (1, [0, 4, 6, 7, 9]), (2, [0, 3, 6, 8, 9]) , (3, [0, 2, 8, 9]), (4, [1, 6, 7, 8]), (5, [9]), (6, [1, 2, 4, 8]) , (7, [1, 4, 8]), (8, [2, 3, 4, 6, 7]), (9, [1, 2, 3, 5])] 但它返回 [(0 , [1, 2, 3]), (1, [4, 6, 7, 9]), (2, [3, 6, 8, 9]), (3, [8, 9]), (4 , [6, 7, 8]), (5, [9]), (6, [8]), (7, [8])]。我不知道為什么它不起作用。
2 回答
白衣非少年
TA貢獻1155條經驗 獲得超0個贊
你可以這樣做：
from itertools import combinations
def create_network(lst):
seen = {}
for e in lst:
for s, t in combinations(map(int, e.split()), 2):
seen.setdefault(s, set()).add(t)
seen.setdefault(t, set()).add(s)
return [(k, sorted(values)) for k, values in seen.items()]
data = ['5', '0 1', '1 2', '1 8', '2 3']
result = create_network(data)
print(result)
輸出
[(0, [1]), (1, [0, 2, 8]), (2, [1, 3]), (3, [2]), (8, [1])]
一般的想法是創建一個字典，其中鍵是整數，值是它旁邊出現的一組整數。該語句map(int, e.split())創建一個整數迭代，然后使用組合從迭代中挑選每一個可能的對，將每一對添加到字典中，最后返回值排序的元組。
更新 （不使用任何內置模塊）
def combinations(m, lst):
if m == 0:
return [[]]
return [[x] + suffix for i, x in enumerate(lst) for suffix in combinations(m - 1, lst[i + 1:])]
def create_network(lst):
uniques = []
for s in lst:
for e in map(int, s.split()):
if e not in uniques:
uniques.append(e)
result = []
for number in uniques:
seen = []
for e in lst:
values = list(map(int, e.split()))
for s, t in combinations(2, values):
if s == number:
if t not in seen:
seen.append(t)
elif t == number:
if s not in seen:
seen.append(s)
if seen:
result.append((number, sorted(seen)))
return sorted(result)
data = ['5', '0 1', '1 2', '1 8', '2 3']
network = create_network(data)
print(network)
輸出
[(0, [1]), (1, [0, 2, 8]), (2, [1, 3]), (3, [2]), (8, [1])]
上面的代碼沒有使用 set、dictionary 和任何外部模塊。請注意，它可能會很慢。
智慧大石
TA貢獻1946條經驗 獲得超3個贊
您可以使用列表理解：
d = ['5', '0 1', '1 2', '1 8', '2 3']
def find_edges(_d, c):
return [(a if b == c else b) for a, b in _d if c in [a, b]]
new_d = [[int(c) for c in i.split()] for i in d if len(i) > 1]
_final = []
for i in [h for d in new_d for h in d]:
if not any(j == i for j, _ in _final):
_final.append((i, find_edges(new_d, i)))
輸出：
[(0, [1]), (1, [0, 2, 8]), (2, [1, 3]), (8, [1]), (3, [2])]
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