我会先建立一本字典,以便按日查询费用:

>>> wd = [1,4,5,6]
>>> fees = [1.44, 1.17, 1.21, 1.26]
>>> fee_dict = dict(zip(wd, fees))

然后建造 correspDay 用一个简单的 range :

>>> correspDay = list(range(1, 7))

建造 result 通过迭代 记者日 ,使用 fee_dict 查找费用并在 费迪克特 结果是空的:

>>> result = []
>>> for i in correspDay:
...     result.append(fee_dict.get(i) or result[-1])
...
>>> result
[1.44, 1.44, 1.44, 1.17, 1.21, 1.26]

我会把这些值放到字典里,然后循环一周中的几天(1-7天),检查我们是否已经有了那一天的值。如果你那天确实有价值,那就把它储存起来,以防第二天没有价值。如果没有值,则在字典中创建一个条目,其中包含当天的密钥和最后一笔费用的值。
最后,我对列表进行了排序,但除了让输出更易于阅读之外,没有真正的理由这样做

fees_dict = {1: 1.44, 4: 1.17, 5: 1.21, 6: 1.26}
last_fee = 0
for i in range(1, 8):
    if i in fees_dict:
        last_fee = fees_dict[i]
    else:
        fees_dict[i] = last_fee

sorted_fees_dict = dict(sorted(fees_dict.items()))
print(sorted_fees_dict)

feeMap 地图 days 到 fee 这是必须支付的 prev 用于存储以前的值(费用)。

#!/usr/bin/env python3.10

wd = [1,4,5,6]
fees = [1.44, 1.17, 1.21, 1.26]

corresDay = list(range(1, 6))

feeMap = dict()
index = 0

for index, day in enumerate(wd):
    feeMap[day] = fees[index]

prev = None
for day in range(1, 6):
    if day not in feeMap.keys():
        feeMap[day] = prev
    else:
        prev = feeMap[day]
feeMap = sorted(feeMap.items())

print(feeMap)

输出:

$ ./working_days.py 
[(1, 1.44), (2, 1.44), (3, 1.44), (4, 1.17), (5, 1.21), (6, 1.26)]

我什么都没做,我看到了上面的好人,我试图让你们更清楚。

wd = [1,4,5,6]
fees = [1.44, 1.17, 1.21, 1.26]
fee_dict = dict(zip(wd, fees))
correspDay = list(range(1,8))
result = []
for i in correspDay:
    result.append(fee_dict.get(i) or result[-1])

我只是添加这一行,以便在字典中输出

result = dict(zip(correspDay, result))

输出:

{1: 1.44, 2: 1.44, 3: 1.44, 4: 1.17, 5: 1.21, 6: 1.26, 7: 1.26}