使用Python2.7,如何修复此代码,使程序在工作日将64乘以每个结果,在周末将80乘以每个结果?
这是我的代码:使用Python2.7,如何修复此代码,使程序在工作日将64乘以每个结果,在周末将80乘以每个结果?,python,python-2.7,Python,Python 2.7,这是我的代码: days=["Sunday: ", "Monday: ", "Tuesday: ", "Wednesday: ", "Thursday: ", "Friday: ", "Saturday: "] Wdays=["Monday: ", "Tuesday: ", "Wednesday: ", "Thursday: ", "Friday: "] Wends=["Sunday: ", "Saturday: "] demand=[] temp=[] demand=[15, 20, 25,
days=["Sunday: ", "Monday: ", "Tuesday: ", "Wednesday: ", "Thursday: ", "Friday: ", "Saturday: "]
Wdays=["Monday: ", "Tuesday: ", "Wednesday: ", "Thursday: ", "Friday: "]
Wends=["Sunday: ", "Saturday: "]
demand=[]
temp=[]
demand=[15, 20, 25, 18, 20, 22, 14]
while demand[0]>0 or demand[1]>0 or demand[2]>0 or demand[3]>0 or demand[4]>0 or demand[5]>0 or demand[6]>0:
min1=min(demand)
ind1=demand.index(min1)
temp=demand.copy()
temp[ind1]=100000
min2=min(temp)
ind2=temp.index(min2)
for i in range(7):
if i != ind1 and i != ind2:
demand[i]-=1
print(demand)
for e in range(7):
while demand[e]:
if Wdays:
demand[e]-=1*64
else:
demand[e]-=1*80
print(demand[e])`
for e in range(7):
while demand[e]:
if Wdays:
demand[e]-=1*64
else:
demand[e]-=1*80
print(demand[e])
[15, 19, 24, 17, 19, 21, 14]
Weekdays: $6400
Weekends: $2320
为什么要从值中减去64或80?
如果Wdays:
始终为True
因为Wdays
是非空列表。。您想在这里做什么?您从不打印“工作日:”
或“周末:”
。。。您只打印需求[e]。。不知道你认为你的输出应该如何完成-特别是因为你的最后一个而demand[e]:
是一个无止境的循环-demand[e]从来都不是0这将是一个错误的值我想修复代码的第二部分,因为它是错误的。我想对其进行组织/更改,使其首先输出数字(1-5)之和,然后再乘以64,然后输出数字(0和6)之和,然后再乘以80。如果需要,您可以完全删除代码的第二部分,因为它显然是错误的。那么打印(sum(demand[1:6])*64,sum(demand)*80)
?