我正在尝试编写一个贪婪算法,并且有一个元组列表:
cow_list = [('Betsy',9),('Henrietta',9),('Herman',7),('Oreo',6),('Millie',5),('Maggie' ,3),('Moo Moo',3),('Milkshake',2),('Lola',2),('Florence',2)]
我正在尝试遍历列表:
def greedy_cow_transport(cow_list, maxWeight):
transport_name = []
taken_transport = []
not_taken = []
transport_weight = []
for i in range(len(cow_list)):
total_weight = sum(transport_weight)
if (total_weight+cow_list[i][1]) <= maxWeight:
transport_name.append(cow_list[i][0])
transport_weight.append(cow_list[i][1])
total_weight += cow_list[i][1]
elif (total_weight+cow_list[i][1]) > maxWeight:
taken_transport.append(cow_list[i][0])
我想做的是,如果> maxWeight,则删除列表中的第一个元组,以便for循环可以重新开始该过程,而无需使用该第一个元组。
所以我希望最终结果是:
transport_name = [['Betsy'],['Henrietta'],['Herman','Maggie'],['Herman','Moo Moo'],['Herman','奶昔']等,其中所有值的总和为<10。
如果我的问题没有道理,请告诉我。
答案 0 :(得分:0)
一种非常临时的方法,
def greedy_cow_transport(cow_list, maxWeight):
transport_name = []
taken_transport = []
not_taken = []
transport_weight = []
for i in range(len(cow_list)):
# Store an initial element
temp_name.append(cow_list[i][0])
temp_weight.append(cow_list[i][1])
total_weight = sum(temp_weight)
# Iterate on all the elements to the right
for j in range(i, len(cow_list)):
if (total_weight + cow_list[j][1])<= maxWeight:
temp_name.append(cow_list[j][0])
temp_weight.append(cow_list[j][1])
total_weight = sum(temp_weight)
else:
taken_temp_transport.append(cow_list[j][0])
transport_name.append(temp_name)
transport_weight.append(temp_weight)
taken_transport.append(taken_temp_transport)