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Birth_date 是 1 到 2359 之间的整数。每只动物也是列表的一部分。我想将出生日期转换为列表索引，但我不确定如何。

if (birth_time >= 2300) or (birth_time < 100):
    hour_animal = "RAT"
elif (birth_time >= 100) and (birth_time < 300):
    hour_animal = "OX"
elif (birth_time >= 300) and (birth_time < 500):
    hour_animal = "TIGER"
elif (birth_time >= 500) and (birth_time < 700):
    hour_animal = "RABBIT"
elif (birth_time >= 700) and (birth_time < 900):
    hour_animal = "DRAGON"
elif (birth_time >= 900) and (birth_time < 1100):
    hour_animal = "SNAKE"
elif (birth_time >= 1100) and (birth_time < 1300):
    hour_animal = "HORSE"
elif (birth_time >= 1300) and (birth_time < 1500):
    hour_animal = "SHEEP"
elif (birth_time >= 1500) and (birth_time < 1700):
    hour_animal = "MONKEY"
elif (birth_time >= 1700) and (birth_time < 1900):
    hour_animal = "ROOSTER"
elif (birth_time >= 1900) and (birth_time < 2100):
    hour_animal = "DOG"
elif (birth_time >= 2100) and (birth_time < 2300):
    hour_animal = "BOAR"

关于如何简化这方面的任何想法？

谢谢！

4

4 回答 4

2

如果您的列表以 开头RAT，则索引应该是(birth_time + 100) // 200 % 12。

但最好验证一下：

signs = ['rat', 'ox', 'tiger', 'rabbit', 'dragon', 'snake', 'horse', 'sheep', 'monkey', 'rooster', 'dog', 'boar']

while True:
    time = int(input('>> '))
    print(signs[(time + 100) // 200 % 12])
于 2013-12-10T14:56:48.843 回答
0

您可以设置一个具有最小和最大时间的字典，并默认返回“RAT”：

birth_times = {"OX": {"min":100, "max":300},
               "TIGER": {"min":300, "max":500},
               ...
               }

def animal(btime):
    return ([k for k in birth_times.keys()
             if birth_times[k]["min"] <= btime < birth_times[k]["max"]]
            + ["RAT"])[0]
于 2013-12-10T15:08:15.543 回答
0

步骤1：

if birth_time < 100:
    hour_animal = "RAT"
elif birth_time < 300:
    hour_animal = "OX"
elif birth_time < 500:
    hour_animal = "TIGER"
elif birth_time < 700:
    hour_animal = "RABBIT"
elif birth_time < 900:
    hour_animal = "DRAGON"
elif birth_time < 1100:
    hour_animal = "SNAKE"
elif birth_time < 1300:
    hour_animal = "HORSE"
elif birth_time < 1500:
    hour_animal = "SHEEP"
elif birth_time < 1700:
    hour_animal = "MONKEY"
elif birth_time < 1900:
    hour_animal = "ROOSTER"
elif birth_time < 2100:
    hour_animal = "DOG"
elif birth_time < 2300:
    hour_animal = "BOAR"
else:
    hour_animal = "RAT"

请注意，我在开始和结束时将 RAT 放入了两次，这让我删除了一半的条件。

第 2 步：请注意，这些数字是 100 的倍数，与 200 的倍数相差 100。因此，@Hyperboreus 的解决方案（他在我打字时提交）。

于 2013-12-10T15:01:20.940 回答
0

您可以在此处使用bisect模块。这里在O(log N)bisect.bisect_right时间内找到特定索引，因此您的- s链减少到只有一个：ifelifif

import bisect

def solve(lis1, lis2, n):
    ind = bisect.bisect_right(lis1, n) - 1
    if 0 <= ind < len(lis1)-1:
        return lis2[ind]
    return 'RAT'

birth_time = [100, 300, 500, 700, 900, 1100, 1300, 1500, 1700, 1900, 2100, 2300]

hour_animal = ['OX', 'TIGER', 'RABBIT', 'DRAGON', 'SNAKE', 'HORSE', 'SHEEP', 'MONKEY', 'ROOSTER', 'DOG', 'BOAR']

演示：

>>> solve(birth_time, hour_animal, 150)
'OX'
>>> solve(birth_time, hour_animal, 2400)
'RAT'
>>> solve(birth_time, hour_animal, 1300)
'SHEEP'
>>> solve(birth_time, hour_animal, 2300)
'RAT'
>>> solve(birth_time, hour_animal, 90)
'RAT'
>>> solve(birth_time, hour_animal, 1600)
'MONKEY'
于 2013-12-10T15:12:28.633 回答